alkali metals has 6-p electrons
what are the properties of alkali metals ?
The alkali metals are elements of group 1 elements which are lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs), and francium (Fr).,
These elements are s-block elements because they have their outermost electron in an s-orbital.
The alkali metals are soft, shiny and highly reactive and lose their outermost electron to create cations with charge +1, it can tarnish rapidly in the air due to oxidation by atmospheric moisture and oxygen.
Alkali metals are good conductors of heat and electricity, these elements are specific, They are soft and can be cut by knife.
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4. __H2SO4 + __Cr(OH)3 --> __Cr2(SO4)3 + __H2OYou have 4 moles of H2SO4. How many moles of H2O are produced?
The number of moles of H2O produced when 4 moles of H2SO4 are there is 8 moles of H2O
The reaction is __H2SO4 + __Cr(OH)3 --> __Cr2(SO4)3 + __H2O. The reaction is as follows;
2Cr(OH)3 + 3H2SO4 → Cr2(SO4)3 + 6H2O
3 moles of H2SO4 produces 6 moles of H2O
1 mole of H2SO4 produces 2 moles of H2O
4 moles of H2SO4 produces 8 moles of H2O
The unitary method is the method used to find the value of a single unit using the value.The number of moles is the given mass by molecular mass.n=m/Mn,m,M is the number of moles, given mass, molecular mass respectivelyTo learn more about moles visit:
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What is the frequency of light having a wavelength of 843 nm?Speed of light 3.00x10^8
Explanation:
Frequency and wavelength are inversely proportional to each other. The equation that relates them is:
f = c/λ
Where f is the frequency, λ is the wavelength and c is the speed of light.
f = ? λ = 843 nm c = 3.00 * 10^8 m/s
Before we replace these values into the formula we have to convert the wavelength from nm to m.
1 m = 1 * 10^9 nm
λ = 843 nm = 843 nm * 1 m/(1 * 10^9 nm)
λ = 8.43 *10^(-7) m
Finally we can replace the values into the equation and get the answer to our problem.
f = c/λ
f = (3.00 * 10^8 m/s)/( 8.43 *10^(-7) m)
f = 3.56 * 10^14 1/s
f = 3.56 * 10^14 Hz
Answer: the frequency of the light is 3.56 * 10^14 Hz
Which solution has the highest boiling point?A. 0.1 m CH4N₂OB. 0.06 m HCIC. 0.05 m CaCl2D. 0.04 m (NH4)3PO4E. 0.07 m NaCl
Answer
D. 0.04 m (NH4)3PO4
Explanation
Boiling point elevation depends on the solvent, of course, but we're assuming water, and upon the concentration of the solute and how many particles into which the solute breaks. Colligative properties like boiling point elevation, depend on the total number of particles in solution. The solvent used here is water. So elevation in boiling point will be above a boiling point of water for all solutions.
ΔT= i x Kb x m,
Where i = van Hoff's factor, Kb = boiling point elevation constant of water = 0.512, m = molality
Higher ΔT will have a higher boiling point.
For 0.1 m CH4N₂O,
i = 1
ΔT = 1 x 0.512 x 0.1 = 0.0512
For 0.06 m HCI;
i = 2
ΔT = 2 x 0.512 x 0.06 = 0.06144
For 0.05 m CaCl2
i = 3
ΔT = 3 x 0.512 x 0.05 = 0.0768
For 0.04 m (NH4)3PO4;
i = 4
ΔT = 4 x 0.512 x 0.04 = 0.08192
For 0.07 m NaCl
i = 2
ΔT = 2 x 0.512 x 0.07 = 0.07168
0.04 m (NH4)3PO4 has the highest ΔT.
Therefore, the solution that has the highest boiling point would be D. 0.04 m (NH4)3PO4.
A sample of 63.6 grams of copper completely reacted with oxygen to form 71.6 grams of a copper oxide product. How many grams of oxygen must have reacted?
Let us first illustrate the reaction of Copper and Oxygen
• 2Cu +O2 → 2CuO
ratioon > 2 : 1
step 1 : determine moles of Cu and O2 .
(i) Moles Cu = mass Cu/Molar Mass Cu
= 63.6 g/ 63.546g/mol
=1.001 mol
(ii) moles O2 = mass Oxygen/Molar Mass Oxygen
= 71.6/32
= 2.238mol
Step 2 :
Experimental moles of Cu : O2
1.001moles : 2.238moles
take note that excess reactant is O2 and limiting reactant is Cu
Therefore moles of Oxygen that reacted = 1/2 moles of Cu
= 1/2 * 1.001 = 0.5moles
Mass of Oxygen reacted =0.5* 32 (oxygen Molar mass)
=16.01g
This means that 16.01 grams of Oxygen reacted , this value is not in the options provi
For a 4s orbital ,what are the possible values of n, l and ml.
In order to know the values of the quantum numbers n, l and ml for al s4 orbital we go to a quantum table:
And as we can see to de 4s orbital correspond only the values:
n=4
l=0
ml=0
) What is the most notable difference between particles in the solid phase and the liquid phase?
Answer: Particles in the solid phase are tightly compact and they vibrate, while particles in the liquid phase move freely, randomly, and with more room to move
Explanation:
Match the reaction to the type that best describes it.NaBr + KF → NaF+ KB?Combination2Cu + MnCl2 →Mn + 2Cuci?DecompositionC + 2H2 → CH4?Single replacementCO2 + O2 + c?Double replacement
Answer:
[tex]\begin{gathered} a:\text{ Double Displacement} \\ b:\text{ Single Replacement} \\ c:\text{ Combination} \\ d:\text{ Decomposition} \end{gathered}[/tex]Explanation:
Here, we want to get the type of reactions listed
We proceed to evaluate the reactions one after the other:
a) This is a double replacement reaction. It usually occurs when molecules exchange ions in the course of their reaction leading to replacement in a molecule by an ion or a group of ions in the other molecule
b) This is a single replacement reaction which in other words could be referred to as a displacement reaction. This occurs usually when a metallic solid reacts with a solution of the salt of less reactive metal
c) This is a combination reaction. It is usually useful in the laboratory and industrial synthesis of some molecules
d) This is a decomposition reaction. It occurs when a particular molecule splits into the constituent molecules that make it up
The Haber Process is performed: __N₂ + __H₂__NH,If I start with 35.0 grams of N₂ and 45.0 grams of H₂, how many grams of NH, are produced? HowMuch excess is left over?
We must first balance the equation, we have two atoms of nitrogen and hydrogen in the reactants, so we must put the coefficient 2 on the side of the products to balance the equation. The balanced equation of the reaction will be:
[tex]N_2+H_2\rightarrow2NH[/tex]Now, to find the grams of NH that will be produced, we will follow these steps:
1. We find the moles of N2 and H2 by dividing the grams by the molar mass of each element.
Molar Mass N2=28.0g/mol
Molar Mass H2=2.0g/mol
2. By stoichiometry we find out which is the limiting reactant, we see that for each mole of H2 one mole of N2 reacts, therefore, the limiting reactant will be the one with the least number of moles. The other reactant will be the excess reactant
3. We find the moles of NH using the limiting reactant.
4. We find the grams of NH by multiplying the moles by the molar mass of NH.
5. We find how much of the excess reactant is left by subtracting the initial moles minus the reacting moles.
Let's proceed with the calculations
1. Moles of N2 and H2
[tex]\begin{gathered} molH_2=givengH_2\times\frac{1molH_2}{MolarMassgH_2} \\ molH_2=45.0gH_2\times\frac{1molH_2}{2.0gH_2}=22.5molH_2 \end{gathered}[/tex][tex]\begin{gathered} molN_2=givengN_2\times\frac{1molN_2}{MolarMassgN_2} \\ molN_2=35.0gN_2\times\frac{1molN_2}{28.0gN_2}=1.25molN_2 \end{gathered}[/tex]2. Limiting reactant
We see that there are 22.5 moles of H2 and 1.25 moles of N2. The ratio N2 to H2 is 1/1, therefore, the limiting reactant will be the one with fewer moles, since it will be consumed faster. Therefore, the limiting reactant is nitrogen.
3. Moles of NH
The ratio NH to N2(Limiting reactant) is 2/1, therefore the moles of NH produced will be:
[tex]\begin{gathered} molNH=givenmolN_2\times\frac{2molNH}{1molN_2} \\ molNH=1.25molN_2\times\frac{2molNH}{1molN_{2}}=2.5molNH \end{gathered}[/tex]4. Mass of NH
[tex]\begin{gathered} MassNH=givenmolNH\times\frac{MolarMass,gNH}{1molNH} \\ MassNH=2.5molNH\times\frac{286gNH}{1molNH}=15.0gNH \end{gathered}[/tex]5. Excess of H2
[tex]\begin{gathered} molH_2(LeftOver)=molH_2initial-molH_2react \\ molH_2(LeftOver)=22.5H_2-1.25molhH_2=21.25molH_2 \end{gathered}[/tex]The mass of H2 left over will be:
[tex]\begin{gathered} MassH_2=21.25molH_2\times\frac{MolarMass,gH_2}{1mol} \\ MassH_2=21.25molH_2\times\frac{2.0gH_2}{1molH_2}=42.5gH_2 \end{gathered}[/tex]Answer: 15.0 grams of NH will be produced and the excess will be 42.5g of H2
A 4.00-L vessel contained 1.53×10-1 mol of gaseous PCl3, 1.43×10-1 mol of gaseous PCl5, and 1.20×10-1 mol of Cl2 gas at equilibrium at 245°C.
Calculate the value of Keq (expressed in terms of the molar concentrations) for the reaction:
PCl3(g) + Cl2(g) ⇌ PCl5(g)
The equilibrium constant Keq is the ratio of product of concentration of products in a reaction to that reactants. In the given reaction the equilibrium constant is 31.5.
What is equilibrium constant?The equilibrium constant in a reaction is the ratio of product of molar concentrations of products to the product of molar concentration of the reactants in the reaction.
In the given reaction, the product is PCl₅ and reactants are Cl₂ and PCl₃. Thus the expression for Keq is written as:
[tex]\rm Keq = \frac{[PCl_{5}]}{[PCl_{3}][Cl_{2}]}[/tex]
Apply the number of moles and volume of all the reactants and products in this expression.
Keq = (0.143/4 L ) [(0.12/4L) (0.15/ 4L)]
= 32.5
Hence, the equilibrium constant Keq is 32.5.
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if you have 3.2 moles of copper how many moles of sodium iodide are required Nal+Cu=Cul2+Na
Step 1 - "reading" the chemical equation
The given chemical equation is:
[tex]\text{NaI}_{(s)}+Cu_{(s)}\to\text{CuI}_{2(s)}+Na_{(s)}[/tex]The proportion in moles for any chemical equation is given by the "bigger" numbers, those that come before the formula of the substances. We can use them to "read" what the equation is telling us:
1 mole of NaI react with 1 mole of Cu thus producing 1 mole of CuI2 and 1 mole of Na
Since the exercise has specifically asked us about the proportion between sodium iodide (NaI) and copper (Cu), we can simplify the statement to:
1 mole of NaI react with 1 mole of Cu
Therefore, we always need the same quantitie, in moles, of NaI and Cu for this reaction to occur completely.
Step 2 - Discovering how many moles of sodium iodide are required
Since the proportion between NaI and Cu, as we saw, is 1:1 in moles, whenever we use, let's say, x moles of Cu, we will need x moles of sodium iodide as well.
Therefore, the amount needed of sodium iodide would be exactly 3.2 moles.
Arrange in order of increasing ionization energy. (Use the appropriate <, =, or > symbol to separate substances in the list.)
Explanation:
The 3 elements are As, Ga, and Se.
Please, look at the next picture:
According to the 2 pictures above, the right arrangement will be:
Ga < As < Se
Ga has the lowest ionization energy and Se has the highest ionization energy.
Answer: Ga < As < Se
How many grams of FeSO4 (molar mass = 151.9 g/mol) are present in a 200.0 mL sample of a 0.250 M solution? Enter your answer with 3 significant figures and do not include the unit
7.60grams
Explanations:The formula for calculating the molarity of a solution is expressed as;
[tex]molarity=\frac{mole}{volume}[/tex]Given the following parameters
molarity of solution = 0.250M
volume of sample = 200mL
Substitute to determine the mole of the solute
[tex]\begin{gathered} mole\text{ of }FeSO_4=molarity\times volume \\ mole\text{ of }FeSO_4=\frac{0.250mol}{L}\times0.2L \\ mole\text{ of }FeSO_4=0.05moles \end{gathered}[/tex]Determine the required mass of FesO4
[tex]\begin{gathered} Mass\text{ of FeSO}_4=mole\times molar\text{ mass} \\ Mass\text{ of FeSO}_4=0.05\times151.9 \\ Mass\text{ of FeSO}_4=7.595grams \\ Mass\text{ of FeSO}_4=7.60grams \end{gathered}[/tex]Hence the mass of FeSO4 is 7.60grams
Select the structure that correspondsto the name:pentanoic acidA. CH3CH₂CH₂CH₂COOHB.C. bothCOOH
Answer
C. both
Explanation
Structures in option A and option B both corresponds to the name: pentanoic acid.
The correct answer is C. both
Solutions are:Group of answer choicesPure substancesHomogeneous mixturesHeterogeneous mixturesAlways water-based
ANSWER
Solutions are homogenous mixtures
EXPLANATION
A solution is a mixture that contains two or more substances in the right proportions.
Solution = solute + solvent
The solution is always a homogenous mixture because the substances are mixed in the right proportion
Hence, solutions are homogenous mixtures
What is the oxidation state of the nitrogen inNaNO2?1+2+3+4+
ANSWER
The oxidation state of nitrogen in NaNO2 is +3
EXPLANATION
Given information
[tex]\text{ Chemical formula }\rightarrow\text{ NaNO}_2[/tex]To find the oxidation state of nitrogen in NaNO2, follow the steps below
Step 1: Write NaNO2 in the ionic form
[tex]\text{ NaNO}_2\text{ }\rightarrow\text{ Na}^+\text{ + N + O}_2^{2-}[/tex]From the above reaction, you will see that the oxidation number of sodium is +1, and the oxidation number of oxygen is -2.
Step 2: Let the oxidation state of nitrogen be x
Add the oxidation numbers together and equate all to zero
[tex]\begin{gathered} +1\text{ + x + 2\lparen-2\rparen = 0} \\ +1\text{ + x + \lparen-4\rparen = 0} \\ +1\text{ + x - 4 = 0} \\ +1\text{ - 4 + x = 0} \\ -3\text{ + x = 0} \\ Add\text{ 3 to the both sides} \\ -3\text{ + 3 + x = 0 + 3} \\ x\text{ = 3} \end{gathered}[/tex]Hence, the oxidation state of nitrogen in NaNO2 is +3
.calculate amount of heat required to rais the temperature of 50 g og copper by 70 K? (specific heat of copper is 0.385 J/g.K.
To determine the heat required (Q) to increase the temperature of a substance we can apply the following equation:
[tex]Q=mCp\Delta T[/tex]where,
m is the mass of the substance, 50g
Cp is the specific heat of the substance, 0.386J/g.K
deltaT is the temperature difference, 70K
Now, we have the values we need to calculate the heat required, we replace the known data into the equation:
[tex]Q=50g\times0.385\frac{J}{g.K}\times70K=1347.5J[/tex]The heat required to raise the temperature of 50 g of copper by 70 K will be 1347.5J
Which of the following is a measure of the amount of space that an object occupies?MassVolumeDensityWeight
Step 1 - Understanding the definition of each term
Mass: it measures the amount of matter of a body
Volume: it measures the space occupied by a body
Density: it's the mass divided by the volume
Weight: it is an indirect measure of the mass, since it also includes the action of gravity
Step 2 - Answering the exercise
As we can see, the definition that fits what is being asked is volume.
Answer: volume
What is the correct formula that wouldresult from the combination of the two ionic species?
In order to find which ionic compound will be made from these two ions, we need to match their charges:
NH4 has a +1 charge
N has a -3 charge
We will need 3 NH4 in order to match the -3 charge of N, therefore we will have a neutral compound:
(NH4)3N is the best option, this is Ammonium nitride
QUESTION 16What is the empirical formula that corresponds to a compound that contains 57.69% C, 11.54% H, and 30.77% O by mass?O C5H1202O C2.5H6OC3H1202O C58H12031С6HO3
Answer
[tex]C_5H_{12}O_2[/tex]Explanation
Given:
57.69% C, 11.54% H, and 30.77% O by mass,
Meaning:
Mass of carbon = 0.5769 g
Mass of Hydrogen = 0.1154 g
Mass of Oxygen = 0.3077 g
We know:
Molar masses of:-
C = 12.0107 g/mol
H = 1.00794 g/mol
O = 15.999 g/mol
Required: Empirical formula
Solution:
Step 1: Calculate the number of moles of the 3 atoms
C : 0.5769 g/12.0107 g/mol = 0.04826 mol
H : 0.1154 g/1.00794 g/mol = 0.1145 mol
O : 0.3077 g/15.999 g/mol = 0.0192
Step 2: Divide the moles by the lowest number of moles
In this case divide by 0.0192
C: 2.5
H: 6
O:1
Step 3: multiply by 2 to remove a decimal
C: 5
H: 12
O:2
Therefore the answer is: C5H12O2
Ascorbic acid (vitamin C, C6H8O6) is a diprotic acid (Ka1 = 8.0×10^–5 and Ka2 = 1.6×10^–12). What is the pH of a 0.240 M solution of ascorbic acid?
Explanation:
We are given: Ka1 = 8.0*10^-5
: Ka2 = 1.6*10^-12
: concentration = 0.240M
The value of Ka1>>Ka2, therefore Ka2 will not be considered in this calculations.
We then determine the concentration of H+ at equilibrium.
Let [C6H6O6-] = x, [H+] = x
[tex][/tex]
Consider an electron transition from n = 2 to n = 4 of a hydrogen atom.
What wavelength of light will the hydrogen atom absorb or emit in this electron transition?
The required formula for doing so is
Here λ= wavelength, Rh = Rhydberg constant n1 is the the no. of orbital the electron transitioned from and n2 is the no. of orbital the electron transitioned to (ignore the z^2 as the atom in itself isn't mentioned so we assume that it is the electron of a hydrogen atom, btw z means atomic number)
So input n1 = 2, n2 = 4, Rh = 10 973 731.6 m^(-1)
you'll find the value of wavelength by inversing the answer you get from this equation directly. I don't have my calculator with me rn so can't show exact values :(
What is the molarity of a solution that contains 98 g of H2SO4 in 0.05 L of solution?
Answer
20 mol/L
Explanation
Given:
Mass of H₂SO₄ = 98 g
Volume = 0.05 L
From the Periodic Table; Atomic mass (H = 1.0 g, O = 16.0 g, S = 32.0 g)
What to find:
Molarity of the solution.
Step-by-step solution
Step 1: Convert the given mass of 98 g to mole.
The formula to determine the mole is given by:
[tex]\text{mole }=\frac{Mass}{Molar\text{ mass}}[/tex]Molar mass of H₂SO₄ = (2 x 1.0 g) + 32.0 g + (4 x 16.0 g) = 98 g/mol
So mole is
[tex]\text{Mole }=\frac{98\text{ g}}{98\text{ g/mol}}=1.0\text{ mol}[/tex]Step 2: Determine the molarity
The molarity formula is given by:
[tex]\text{Molarity }=\frac{Mole}{Volume\text{ in L}}=\frac{1.0\text{ mol}}{0.05\text{ L}}=20\text{ mol/L}[/tex]Need help with chemistry 52.3 g of metal at a temperature of 126.2°C are placed in 71.2 g of water which abs an into Al temperature of 24.5°C. The final temperature of the metal and water is 35.5°C. What is the specific heat of the metal in units of J/g°C? Use 4.184 J/g°C for this problem?
Answer:
The specific heat of the metal is 0.69 J/g°C.
Explanation:
The given information from the exercise is:
• Metal,:
- Mass (massmetal): 52.3g
- Initial temperature (Tinitialmetal): 126.2°C
- Final temperature (Tfinalmetal): 35.5°C
• Water,:
- Mass (masswater):71.2g
- Initial temperature (Tinitialwater): 24.5°C
- Final temperature (Tfinalwater): 35.5°C
1st) It is necesary to write the Heat formula for both materials, the metal and water. In the case of water, we can solve the formula, because we have all the nedeed information:
• Metal,:
[tex]\begin{gathered} Q_m=m_{metal}*c_{metal}*(T_{fmetal}-T_{imetal}) \\ Q_m=52.3g*c_{metal}*(35.5_\degree C-126.2\degree C) \\ Q_m=-4,743.61g\degree C*c_{metal} \\ \end{gathered}[/tex]Water:
[tex]\begin{gathered} Q_w=m_{water}*c_{water}*(T_{fwater}-T_{\imaginaryI water}) \\ Q_w=71.2g*4.184\frac{J}{g\degree C}*(35.5\degree C-24.5\degree C) \\ Q_w=3,276.9J \end{gathered}[/tex]So, the heat absorbed by the water is 3,276.9 J.
2nd) In the end both materials reach the equilibrium temperature, because the heat released by the piece of metal is absorbed by the water, this is represented as:
[tex]-Q_m=Q_w[/tex]Finally, we have to replace the heat of the metal equation (Qm) and the heat of water (Qw) to calculate the specific heat of the metal:
[tex]\begin{gathered} -Q_{m}=Q_{w} \\ -(-4,743.61g)\degree C*c_{metal}=3,276.9J \\ 4,743.61g\degree C*c_{metal}=3,276.9J \\ c_{metal}=\frac{3,276.9J}{4,743.61g\degree C} \\ c_{metal}=0.69\frac{J}{g\degree C} \end{gathered}[/tex]So, the specific heat of the metal is 0.69 J/g°C.
The number of electrons in the L orbital of an element with atomic number 10 is?
The electronic distribution of the element is going to tell us how the electrons are distributed all across the atom.
There are some concepts to check:
Energy levels: They are numbered from 1 to 7 and correspond to the element period.
In each energy level, there are 4 sub energy levels: s, p, d, and f.
The quantic number l refers to the angular position of the angle that forms the element. The number of electrons present on this orbital is equal to the addition of the last energy level, in this case, 2.
The electronic configuration for this element is:
[tex]1s^2\text{ 2s}^2\text{ 2p}^6[/tex]Then, the number of electrons in the last energy level is 8.
H2N-(CH2)6 -NH2The compound above is one of the monomers of (A)Perupex(B)terylase(C)nylon(D)Urea
Answer
(C) nylon
Explanation
H2N-(CH2)6 -NH2 is hexamethylene diamine. The condensation reaction of hexamethylenediamine and adipic acid will produce a polymer, nylon.
Therefore, H2N-(CH2)6 -NH2 is one of the monomers of nylon
The correctanswer is (C)nylon
With all the steps please:A solution is prepared by dissolving 166.0 g of potassium iodide, Kl, in water to a total volume of 1,000 mL. Calculate the molarity of this solution.
Explanation:
Molarity, or concentration in amount of matter (mol/L), is the ratio between the amount of matter in the solute (n) and the volume of the solution in liters (V): M = n/V
Here we have:
m = 166 g of KI (solute)
V = 1000 mL = 1L (solution)
So we need to transform 166 g of KI into moles. We use the following formula:
n = m/MM
MM of KI is 166 g/mol
n = 166/166
n = 1 mol
So:
M = 1/1
M = 1 mol/L
Answer: M = 1 mol/L
estion 21 of 30hat types of bonds and atoms are in the molecule shown below?HH1H-C-C=C-C-H||||HHHHOA. Single bonds, double bonds, carbon atoms, and hydrogen atomSingle bonds double bonds and only carbon atoms
According to the given picture, in the molecules there are single bonds, double bonds, carbon atoms and hydrogen atoms. It means that the correct answer is A.
Calculate the molarity of a solution made by adding 150.0 mL of water to 85.00 mL of a 0.157 M solution
Answer:
0.089 M
Explanation:
M1V1=M2V2
Total volume = 150 ml + 85 ml = 235ml
0.157(.085 L) = M2(.235L)
M2 = 0.05678723404 or 0.057 M
What is the net ionic equation for the reaction between HCL(aq) and NaHCO3(aq)
We have to write the net ionic equation for the reaction between HCl and NaHCO₃.
The molecular equation will show us the products of this equation.
Molecular equation:
HCl (aq) + NaHCO₃ (aq) ------> NaCl (aq) + CO₂ (g) + H₂O (l)
Now we can split this compounds into their ions to get the total ionic equation:
Total ionic equation:
H+ (aq) + Cl- (aq) + Na+ (aq) + HCO3- (aq) ----> Na+ (aq) + Cl- (aq) + CO₂ (g) + H₂O (l)
And finally we can identify the spectator ions. The ones that we have on both sides, they are Na+ and Cl-. We can cancel them out.
Net ionic equation:
H+ (aq) + HCO3- (aq) ----> CO₂ (g) + H₂O (l)
Answer: H+ (aq) + HCO3- (aq) ----> CO₂ (g) + H₂O (l)
You are given a bulb, one metre copper wire, a switch and a battery. How could you use them to distinguish the samples of metals and non-metals? Write the usefulness of this test in distinguishing metals and non-metals.
With the given materials following experiment can be done
Electrical conductivity test:
Metals are good conductors of electricity, therefore when you join metals with a battery, wire, and bulb, you get a bulb.
Similarly, if nonmetals are poor electrical conductors, they may not ignite the bulb when attached to a wire and a battery.
Generally, the above method can be used to identify metals and non-metals. But there are some exceptions also,. Graphite, an allotrope of non-metal carbon, is a good conductor of electricity.
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The periodic chart has many different types of elements. Some elements are metals, some are nonmetals, and yet others are metalloids. Therefore, by electrical conductivity test, we can detect metal and non metal.
What is metal ?Metals are hard, conduct electricity, are ductile, lustrous, and malleable materials.
Since metals have free electrons that's why they can conduct electricity. Since atoms in metals are very closely packed in a definite crystal solid. It is not brittle that is it can not be broken down easily.
Metals carry electricity well, therefore when you combine metals with a battery, wire, and bulb, you get a light. Similarly, if nonmetals are poor electrical conductors, when connected to a wire and a battery, they may not light the bulb.
Therefore, by electrical conductivity test, we can detect metal and non metal.
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