A neutral atom of on element has two electrons with n=1, eight electrons with n=2, eight electrons with n=3 and one electron with n =4 and has mass number of 39 then the atomic number of element is 39 and number of neutron in the nucleus is 50 and total number of s electron is 2 and total number of p electron is 8 and the group is 3rd group
Yttrium is a metallic element with atomic number 39 usually included in the rare earth group that occur usually with other rare earth element in minerals and is used especially in phosphorous and YAG laser and alloy etc and element with mass number of 39 is yttrium and the atomic number of element is 39 and number of neutron in the nucleus is 50 and total number of s electron is 2 and total number of p electron is 8 and the group is 3rd group
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If 129.63grams of aluminum metal is reacted with767.44grams of iron (III) oxide, what mass ofaluminum oxide is likely to form?Reaction: 2Al(s) + Fe2O3(aq)--> Al2O3(aq) + 2Fe(s)
Step 1
The reaction must be written, completed, and balanced:
2Al(s) + Fe2O3(aq) => Al2O3(aq) + 2Fe(s)
----------------------
Step 2
Information provided:
The mass of Al = 129.63 g
The mass of Fe2O3 = 767.44 g
(because of this, the limiting reactant must be determined)
---
Information needed:
The molar masses:
Al) 26.981 g/mol
Fe2O3) 159.69 g/mol
---------------------
Step 3
The limiting reactant:
2Al(s) + Fe2O3(aq) => Al2O3(aq) + 2Fe(s)
By stoichiometry => 1 mole Al = 26.981 g and 1 mole Fe2O3 = 159.69 g
Procedure:
2 x 26.981 g Al --------- 159.69 g Fe2O3
129.63 g Al ---------- X
X = 129.63 g Al x 159.69 g Fe2O3/2 x 26.981 g Al = 383.61 g
(For 129.63 g of Al, 383.61 g of Fe2O3 is needed, but there is 767.44 g of Fe2O3. Therefore, the limiting reactant is the Al and the excess is the Fe2O3)
-----------------------
Step 4
The mass of Al2O3: (molar mass of Al2O3 = 101.96 g/mol)
2Al(s) + Fe2O3(aq) => Al2O3(aq) + 2Fe(s)
2 x 26.981 g Al ------- 101.96 g Al2O3
129.63 g Al ------- X = 244.93 g
Answer: mass of aluminum oxide = 244.93 g
Calculate the [OH−] of each aqueous solution with the following [H3O+].baking soda, 1.7×10−8M
Answer:
The [OH−] is 5.88x10^-7M.
Explanation:
1st) It is necessary to calculate the pH of the solution using the pH formula and replacing the concentration of H3O+:
[tex]\begin{gathered} pH=-log\left[H_3O+\right] \\ pH=-log(1.7×10−8) \\ pH=7.77 \end{gathered}[/tex]Now we know that the pH is 7.77.
2nd) Now we can calculate the pOH of the solution using the relation between pH and pOH:
[tex]\begin{gathered} pH+pOH=14 \\ 7.77+pOH=14 \\ pOH=14-7.77 \\ pOH=6.23 \end{gathered}[/tex]Now we know that the pOH of the solution is 6.23.
3rd) Finally, we can calculate the [OH−] using the pOH formula and replacing the value of pOH:
[tex]\begin{gathered} pOH=-log\left[OH−\right] \\ 6.23=-log\left[OH−\right] \\ 10^{(-6.23)}=\left[OH−\right] \\ 5.88*10^{-7}=\left[OH−\right] \end{gathered}[/tex]So, the [OH−] is 5.88x10^-7M.
Draw the electron dot diagram for H2CO and determine how many bonded pairs of electrons there are.
Pleasseeeeeee
The electron dot diagram is shown in the image attached to this answer.
What is the electron dot diagram?We know that an electron dot diagram is used to show the electron pairs that are present in a compound. the electron dot diagram would include the symbols of the elements as well as the number of electron pairs that can be found on the valence shells of each of the atoms.
In this case, we can see that the electrons that are involved in a bod are shown as a single dash where the dash shows the electrons that are involved in a covalent bonding situation.
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According to the graph above what is the substances melting temperature?
ANSWER
At 90 degrees Celcius the substance will melt completely
EXPLANATION
Melting point is the process in which solid is completely converted to the liquid phase.
In the given graph, the point A the solid particles are closely packed due to strong intermolecular forces binding the particles together. When temperature is applied to the system, the particles of the solid substance vibrate about a point and the particles are said to move independently. Further increase in the temperature will break the bond holding the particles together and turn to liquid at 90 degrees Celcius
Therefore, at 90 degrees Celcius the substance will melt completely and turn to liquid.
The universal indicator changes color throughout the pH range. How is this indicator produced?
Answer and Explanation:
The color change of a pH indicator is caused due to the dissociation of protons H+. The most common indicators are: phenolphthalein, methyl red, and bromothymol blue, which are used to indicate pH ranges of about 8 to 10, 4.5 to 6, and 6 to 7.5, respectively. For each indicator, phenolphthalein goes from colorless to pink, methyl red goes from red to yellow, and bromothymol blue goes from yellow to blue.
A tank contains 8.5 L of fluorine at 268K and 549mm Hg. The gas is then transferred to a 5.0L tank. What temperature is required to obtain a final pressure of 675 mm Hg?
The temperature required to obtain a final pressure of 675 mm Hg194 is .86 K.
An ideal gas is a theoretical gas composed of many randomly transferring factor particles that aren't difficult to interparticle interactions. the best gasoline idea is beneficial because it obeys the precise gas law, a simplified equation of country, and is amenable to evaluation under statistical mechanics.
Volume is a degree of occupied three-dimensional space. it's far more frequently quantified numerically the usage of SI-derived gadgets or by way of diverse imperial gadgets. The definition of length is interrelated with the extent.
An ideal gas is described as one for which both the extent of molecules and forces between the molecules are so small that they have got no effect at the behavior of the gas.
Using the ideal gas equation:-
Given;
P₁ = 549mm Hg = 0.722369 atm
V₁ = 8.5 L
T₁ = 268 K
P₂ = 675 mm Hg = 0.89 atm
V₂ = 5
T₂ = ?
P₁V₁/T₁ =P₂V₂/T₂
T₂ = P₂V₂T₁/P₁V₁
= 0.89 × 5 × 268 / 8.5 × 0.72
= 194 .86 K
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Describe the solution if 274 g of potassium iodide (KI) is dissolved in 200 g of H2O at 10oC.
A buret dispenses 16.4 mL of a 0.521 M NaOH solution. How many moles of NaOH were dispensed?
The number of moles will be 8.54 mole.
The concentration can be expressed as the ratio of number of mole and volume . It can be shown as:
C= n/V
where, C is concentration, n is number of moles and V is volume.
The given data:
C = 0.521 M
V = 16.4 mL
n = ?
The number of moles can be determined by using the formula:
C = n/v
n = C×v....(i)
Put the values of given data in above equation.
n = 0.521 × 16.4
n = 8.54
Therefore, the number of moles will be 8.54 mole.
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Which is the limiting reactant in the following chemical reaction: NH3 + 5 O2 —> 4 NO + 6 H2O, if 4 grams of O2 is used and 2 grams of NH3 is used ? Show work
The limitant reactant of a reaction is the reactant that we have the least number of mols considering its coefficient on the stoichiometry.
The reaction in this case is:
[tex]NH_3+5O_2\to4NO+6H_2O[/tex]The reactants are the one on the left side, so the answer can't be NO or H₂O.
The first thing to do is to calculate the number of moles of NH₃ and O₂, but for this we need their molar masses:
[tex]M_{NH_3}=1\cdot M_n+3\cdot M_H=(1\cdot14.0067+3\cdot1.00794)g/mol=17.03052g/mol[/tex][tex]M_{O_2}=2\cdot M_O=2\cdot15.9994g/mol=31.9988g/mol[/tex]Now, using them, let's calculate the number of moles of each:
[tex]\begin{gathered} M_{NH_{3}}=\frac{m_{NH_3}}{n_{NH_{3}}} \\ n_{NH_3}=\frac{m_{NH_3}}{M_{NH_{3}}}=\frac{2g}{17.03052g/mol}=0.11743\ldots.mol \end{gathered}[/tex][tex]\begin{gathered} M_{O_{2}}=\frac{m_{O_2}}{n_{O_{2}}} \\ n_{O_2}=\frac{m_{O_2}}{M_{O_{2}}}=\frac{4g}{31.9988g/mol}=0.12500\ldots mol \end{gathered}[/tex]Now, we can't compair directly, we need to consider their coefficients.
To do this, we divide the number of moles by the coefficient:
[tex]\begin{gathered} NH_3\colon\frac{n_{NH_3}}{1}=\frac{0.11743\ldots mol}{1}=0.11743\ldots mol \\ O_2\colon\frac{n_{O_{2}}}{5}=\frac{0.12500\ldots mol}{5}=0.025000\ldots mol \end{gathered}[/tex]Now, we can compair.
Since the value we have got to O₂ is less than NH₃, the limiting reactant is O₂
If the reaction in the given equation produced 0.364 moles of H2 gas at STP, how many liters would this amount of H2 occupy? [hint - the H2 gas is at STP! Is there a special conversion we can use?]
Equation: 6 HCl + 2 Al ---> 2 AlCl3 + 3 H2
In the given equation produced 0.364 moles of H₂ gas at STP, 22.4 liters volume would this amount of H₂ occupy.
What is moles?The mole, denoted by the sign "mol," is the volume of a system that has the same number of atoms in 0.012 kilograms of carbon-12 as there are elementary particles.
The equation becomes:
6 HCl + 2 Al → 2 AlCl₃ + 3 H₂
6 mol 2 mol 2 mol 3 mol
As we know,
one mole of a gas occupies 22.4 liters.
Here, in the above equation, 3 moles of H₂ gas is produced.
So, in STP condition 3 moles of H₂ gas occupy = 3 × 22.4 lit
= 67.2 lit
Thus, in the given equation produced 0.364 moles of H₂ gas at STP, 22.4 liters volume would this amount of H₂ occupy.
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Hello I’ve been working on this for about an hour and still don’t get it May Someone help? Thank you!
Intensive properties: These are those that do not depend on the mass of the sample.
Density is an intensive property. For example, the density of an ice cube and an iceberg is the same (0.92 g/cm3 at temperatures below 0°C at sea level), which is less than that of water (1.0 g/cm^3 , under a temperature of approximately 4°C and a pressure at sea level, which is equal to 1.0 atm). Therefore, both an ice cube and an iceberg float on water.
2. Extensive properties: These are those that depend on the mass (“extension”) of the sample.
So mass is an extensive property and length as well.
Answer:
Extensive physical properties: mass of 18.45 g and 3.50 cm in length
Intensive physical properties: Density of 2.70 g/cm^3
How much water must be added to 560 mL of 15M H2SO4 solution to prepare a 3.0M solution?
Answer
The volume of water that must be added = 2240 mL
Explanation
Given:
Initial concentration, C₁ = 15 M
Initial volume of solution, V₁ = 560 mL
Final concentration, C₂ = 3.0 M
What to find:
The volume of water added to 560 mL of 15M H2SO4 solution to prepare a 3.0M solution.
Step-by-step solution:
The solution involves two steps.
Step 1: Calculate the final volume, V₂ of the solution using the dilution formula.
[tex]\begin{gathered} C_1V_1=C_2V_2 \\ \\ 15\text{ }M\times560\text{ }mL=3.0\text{ }M\times V_2 \\ \\ V_2=\frac{15\text{ }M\times560\text{ }mL}{3.0\text{ }M} \\ \\ V_2=\frac{8400\text{ }mL}{3.0} \\ \\ V_2=2800\text{ }mL \end{gathered}[/tex]Step 2: Subtract the initial volume of the solution from the final volume in step 1 to determine the amount of water added.
The volume of water that must be added = V₂ - V₁
The volume of water that must be added = 2800 mL - 560 mL = 2240 mL
A person has a 1.43mol sample of carbon dioxide. If the pressure if the sample is 34.56kpa, and the volume is 440.99mL, what will the temperature of the sample be in Kelvin?
Answer:
[tex]1.28K\text{ }[/tex]
Explanation:
Here, we want to get the temperature of the gas
We can use the ideal gas equation here
Mathematically, we have this as:
[tex]PV\text{ = nRT}[/tex]P is the pressure which is given as 34.56 kPa
V is the volume which is given as 440.99 mL
R is the molar gas constant which is 0.08205 L atm/mol k
n is the number of moles which is given as 1.43 mol
We need to convert the pressure to the correct unit (considering the molar gas consnat value in atm)
1 kPa = 0.00986923 atm
34.56 kPa will be = 34.56 * 0.00986923 = 0.341 atm
We need to convert the volume to Liters by dividing by 1000
We have that as 440.99/1000 = 0.44099 L
Finally,we rewrite the equation in terms of temperature and substitute the values as follows:
[tex]\begin{gathered} T\text{ = }\frac{PV}{nR} \\ \\ T\text{ = }\frac{0.341\times0.44099}{1.43\times0.08205}\text{ =1.28 K } \end{gathered}[/tex]Answer the following question:Reminder R= .0822. If you have 5 moles of a gas at 400 K, what is the pressure if the volume is 15 L?
We are asked to find the pressure of a gas that is in certain conditions. To find it we must assume that there is no interaction between the gas molecules and the gas behaves like an ideal gas, in this way we can apply the ideal gas law that tells us:
[tex]\begin{gathered} PV=nRT \\ P=\frac{nRT}{V} \end{gathered}[/tex]Where,
P is the pressure of the gas
V is the volume of the gas = 15L
T is the temperature of the gas = 400K
n is the number of moles = 5mol
R is a constant = 0.08206 (atm L)//(mol K)
Now, we replace the known data in the equation:
[tex]P=\frac{5mol\times0.08206\frac{atm.L}{mol.K}\times400K}{15L}=11atm[/tex]Answer: The pressure of the gas will be 11 atm
Use a diagram to illustrate how a carbon-carbon double bond forms.
Carbon forms a double bond with another carbon by sharing two pairs of electrons.
The electron dot diagram of the carbon double bond is illustrated in the figure below.
calculate the mass solute and the mass solution when the mass solvent is 315.2g and the mass percent is 15.3 %.
Answer
mass of solute = 48.2 g
Mass of solution = 363.4 g
Explanation
Given:
Mass of solvent = 315.2 g
mass percentage = 15.3 %
Required
mass of the solute
mass of the solution
Solution:
mass % = mass of solute/mass of solvent
15.3/100 = mass of solute/315.2 g
0.153 = mass of solute/315.2 g
mass of solute = 0.153 x 315.2
mass of solute = 48.2 g
Mass of solution = mass of solute + mass of solvent
Mass of solution = 48.2 g + 315.2 g
Mass of solution = 363.4 g
Calculate the mass in grams of CO2 in 45.0 L of CO2 at STP
One of the options in the question is 22.4 L, which is the value for the volume of 1mol of gas in the old STP definition. Since there is no option for the updated STP conditions, we will need to assume it want in this old one.
In the old STP conditions, 1 mol of gas occupies 22.4 L, so we can use the rule of three to calculate how many moles are in 45.0L:
22.4L --- 1mol
45.0L --- n
[tex]\begin{gathered} \frac{22.4L}{45.0L}=\frac{1mol}{n} \\ n=45.0L\times\frac{1mol}{22.4L} \end{gathered}[/tex]Now, we want to convert to mass. Consulting the molar mass of CO₂, we can see that it is approximately 44.01g/mol, which is one of the optios. That is, in 1 mol of CO₂ there are 44.01 g. Using the rule of three:
1 mol --- 44.01 g
n --- m
[tex]\begin{gathered} \frac{1mol}{n}=\frac{44.01g}{m} \\ m=n\times\frac{44.01g}{1mol} \end{gathered}[/tex]Inputting the n we calculated earlier, we have:
[tex]m=45.0L\times\frac{1mol}{22.4L}\times\frac{44.01g}{1mol}\approx88.4g[/tex]So, there is approximately 88.4g of CO₂.
Complete the first row of the table.Express the volume in liters to three significant figures.
Answer:
[tex]V_2\text{ = 149 L}[/tex]Explanation:
Here, we want to get the missing volume
From Charles' law, we know that volume and temperature are directly proportional at constant pressure
Mathematically, we know that:
[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}_{}[/tex]We have to rewrite the formula in terms of the missing value as follows:
[tex]V_2\text{ = }\frac{V_1T_2}{T_1}[/tex]Before we proceed to input the values, we have to convert the temperatures in celsius to kelvin by adding 273.15 K to the celsius temperature
Thus, we have it that:
[tex]\begin{gathered} V_1\text{ = 112L} \\ V_2\text{ = ?} \\ T_1\text{ = 10.2 + 273.15 = 283.35 K} \\ T_2\text{ = 106.1 + 273.15 = 379.25 K} \end{gathered}[/tex]We can now proceed to substitute these values into the rewritten formula as follows:
[tex]V_2\text{ = }\frac{112\times379.25}{283.35}\text{ = 149 L}[/tex]A 5.0 L tank contains N2 at a pressure of 214 kPa. The gas is transferred to a second tank, where it’s pressure is measured at 148 kPa. What is the volume of the second tank
The volume of the second tank will be 7.2 L when the pressure is 148 kPa.
Boyle's Rule
Because the gas particles are compressed closer together as the pressure on a gas rises, the volume of the gas decreases. On the other hand, because the gas particles may now move further apart when the pressure on a gas lowers, the gas volume rises.
P 1 * V 1 = P 2 * V 2 can be used to calculate the pressure and volume of a given amount of gas at a fixed temperature.
P*V is a constant.
P stands for gas pressure.
V stands for gas volume.
P2 = 148 kPa
V2 = x
P1 = 214 kPa
V1 = 5.0 L
P1 * V1 = P2 * V2
214 * 5 = 148 * x
x = 7.2 L
V2= 7.2L
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Which statement best decribes the law of
conservation of mass?
d. The mass of the reactants and products is equal and is not dependent on the physical state of the substances explains the correct statement about the law of conservation of mass.
The law of conservation of mass is the law that states that matter or energy can neither be created nor destroyed. This means that the amount of matter in the universe is always the same. The law of conservation of mass is a fundamental law of physics and it is always obeyed by all physical processes.
The amount of matter in a closed system (one that is not subject to outside influences) will remain constant over time.
It is always obeyed even under the most extreme conditions, such as during nuclear reactions.
Hence, the correct option is d.
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If HCl (aq) is added to a saturated NaCl solution the Na+1 will a. Increase b. Decrease c remain the same
For this question, we have a solution with NaCl, and HCl is added to it, even though we have more Cl- in the solution, in terms of equilibrium this will shift the equilibrium to the left, but in terms of quantity of the sodium ion, it will remain the same, since no sodium was added or took away, letter C
The following reaction takes place in an acidic solution. MnO4–(aq) + Cl–(aq) --> Mn2+ + Cl2(g) (unbalanced).Do Each Step Individually**: Write the half reactions. Balance the equations for atoms (except O and H). Balance the equations for atoms O and H using H2O and H+. Balance the charge in the half reactions. Multiply each half reaction by the proper number to balance charges in the reaction. Add the equations and simplify to get a balanced equation.
Answer
2MnO₄⁻(aq) + 10Cl⁻(aq) + 16H⁺ → 2Mn²⁺(aq) + 5Cl₂(g) + 8H₂O
Explanation
What is given:
The following reaction takes place in an acidic solution:
MnO₄⁻(aq) + Cl⁻(aq) → Mn²⁺(aq) + Cl₂(g) (unbalanced).
What to find:
To identify the half-reactions and balance the redox reaction.
Step-by-step solution:
Step 1: Identify the half-reactions.
The oxidation half-reaction is: Cl⁻(aq) → Cl₂(g)
The reduction half-reaction is: MnO₄⁻(aq) → Mn²⁺(aq)
Step 2: Balance the equations for atoms (except O and H).
Oxidation: 2Cl⁻(aq) → Cl₂(g)
Reduction: MnO₄⁻(aq) → Mn²⁺(aq)
Step 3: Balance the equations for atoms O and H using H₂O and H⁺.
Oxidation: 2Cl⁻(aq) → Cl₂(g)
Reduction: MnO₄⁻(aq) + 8H⁺ → Mn²⁺(aq) + 4H₂O
Step 4: Balance the charge in the half-reactions.
Oxidation: 2Cl⁻(aq) → Cl₂(g) + 2e⁻
Reduction: MnO₄⁻(aq) + 8H⁺ + 5e⁻ → Mn²⁺(aq) + 4H₂O
Step 5: Multiply each half-reaction by the proper number to balance charges in the reaction.
Oxidation: 2Cl⁻(aq) → Cl₂(g) + 2e⁻ x 5
Reduction: MnO₄⁻(aq) + 8H⁺ + 5e⁻ → Mn²⁺(aq) + 4H₂O x 2
↓
Oxidation: 10Cl⁻(aq) → 5Cl₂(g) + 10e⁻
Reduction: 2MnO₄⁻(aq) + 16H⁺ + 10e⁻ → 2Mn²⁺(aq) + 8H₂O
Step 6: Add the equations and simplify to get a balanced equation.
2MnO₄⁻(aq) + 10Cl⁻(aq) + 16H⁺ + 10e⁻ → 2Mn²⁺(aq) + 5Cl₂(g) + 8H₂O + 10e⁻
Simplifying the equation, we have:
2MnO₄⁻(aq) + 10Cl⁻(aq) + 16H⁺ → 2Mn²⁺(aq) + 5Cl₂(g) + 8H₂O
What is the balanced equation for "magnesium reacts with phosphorus"
Magnesium reacts with Phosphorus and produce Magnesium phosphide. The equation is:
Mg + P ---> Mg₃P₂
The balanced equation is:
3 Mg + 2 P ---> Mg₃P₂
It need to have the same amount of atoms of each element on both sides of the equation.
Answer: 3 Mg + 2 P ---> Mg₃P₂
Given the balanced equation below:3Cu + 8HNO3 — 3Cu(NO3)2 + 2NO + 4H2OThe total number of grams of Cu neededto produce 188 grams of Cu(NO3)2 is?
Answer
Mass of Cu = 63.7 g
Explanation
Given:
A balanced chemical equation: 3Cu + 8HNO3 — 3Cu(NO3)2 + 2NO + 4H2O
Required: The total number of grams of Cu
Solution
[tex]\begin{gathered} 188\text{ g Cu\lparen NO}_3)_2\text{ x }\frac{1\text{ mole Cu\lparen NO}_3)_2}{187,56\text{ g Cu\lparen NO}_3)_2}\text{ x }\frac{3\text{ mole Cu}}{3\text{ mole Cu\lparen NO}_3)_2}\text{ x }\frac{63,546\text{ g Cu}}{1\text{ mole Cu}} \\ \\ =\text{ 63.7 g Cu} \end{gathered}[/tex]What would the products of a double-replacement reaction between NaF andMgS be? (Remember: In double-replacement reactions, the two cationsswitch places and the two anions switch places.)O A. NaF and MgsB. MgF and NashO C. F2S and MgNa2O D. MgF2 and Na2SSUBMIT
Step 1
The reaction between NaF and MgS:
2NaF + MgS => Na2S + MgF2
(Remember: In double-replacement reactions, the two cations switch places and the two anions switch places.)
Answer: D. MgF2 and Na2S
Which pair of formulae represents two alkanes?a) CH4 and C8H18b) C2H6 and C5H8c) C3H6 and C5H12d) C10H8 and C4H8
Answer: the best option to answer the question is letter A
Explanation:
The question requires us to choose the option that presents the formula of two alkanes.
Alkanes are hydrocarbons that are saturated and acyclic (i.e., the molecule of an alkane do not present double or triple bonds between carbons and it is a linear structure). The general formula of alkanes tells us that there are double plus 2 the amount of hydrogen atoms as there is carbon atoms, as it follows:
[tex]C_nH_{2n+2}[/tex]where n indicates the number of carbon atoms in the molecule.
For example: we can say that the compound C2H6 is an alkane because it contains 2 carbon atoms (n = 2) and 6 hydrogen atoms (2n +2= 6).
Considering the options given by the question, only option A presents the formula of two alkanes:
CH4: n = 1; 2n+2 = 2+2 = 4
C8H18: n = 8; 2n+2 = 16+2 = 18
Therefore, the best option to answer the question is letter A.
What is the Mass in grams of 1 formula unit or molecule of epsom salt MgSO4●7H2O?
MgSO4.7H2O
First of all, we need to calculate the molar mass of this molecule, using the periodic table.
Let's take each atomic mass of the elements that form this molecule from the periodic table.
Mg = 24.30 u
S = 32.06
O = 15.99 u
H = 1.007 u
Now, we have to do this:
Molar mass (g/mol) = 1 x 24.30 + 1 x 32.06 + 4 x 15.99 + 14 x 1.007 + 7 x 15.99
Molar mass = 246.3 g/mol
We have to be careful with this result because 246.3 g is the mass of 1 mole of MgSO4.7H2O and we want the mass of 1 formula unit or molecule.
Pay attention to this:
1 mole of MgSO4.7H2O = 246.3 g = 6.022 x 10^23 formula units or molecules of MgSO4.7H2O
(We want the mass of 1 molecule and we have just calculated the mass of 6.022x10^23 molecules)
----------------------------------------------------------------------------------------------------------------
For 1 formula unit or molecule:
246.3 g of MgSO4.7H2O ------------- 6.022 x 10^23 formula units or molec.
x ------------- 1 formula unit or molec.
[tex]x\text{ = }\frac{1formula\text{ unit x 246.3 g}}{6.022x10^{23}formula\text{ units}}=4.09x10^{-22}g[/tex]Answer: The mass of 1 formula unit or molecule of MgSO4.7H2O = 4.09x10^-22 g
Standard temperature andpressure (STP) is [?] K and[ ] átm.
Remember that standard temperature and pressure (STP) is when the pressure is 1 atm and the temperature is 0 °C (273 K).
The answer is 273 K and 1 atm.
Which of these elements are often made in a lab?
transuranium
noble gas
alkali metal
halogen
Answer:
I am sure it's Transuranium.
Explanation:
Hope it's right
5 F2 (g) + 2 NH3 (g) → N2F4 (g) + 6 HF (g)If you reacted 7.78 moles of F2 how many moles of HF would be produced?
5 F2 (g) + 2 NH3 (g) → N2F4 (g) + 6 HF (g)
2. Determine mole proportions , then solveFrom the above balanced reaction, we can see that :
5 moles of F2 reacts to form : 6 moles of HF ,
so, 7.78moles F2 will give :x moles of HF
Therefore,x mole HF =( 6 moles HF * 7.78 mole F2)/ 5 mole F2
= 6 * 7.78/5
=9.336
≈9.34moles HF
• This means that , if you reacted 7.78 moles of F2 , ,9.34 moles of HF would be produced,.