Answer:
ΔH = -44.029 KJ/mol
Explanation:
We are told that 60.0 g of NaOH is dissolved in one litre of water.
From conversion, 1 litre = 1000 g
From tables, specific heat capacity of water; c = 4.18 J/g/°C
We are given ΔT = 15.8 °C
Formula for the heat absorbed by the solution is given as;
Q = mcΔT
Where;
Q = Heat gained by the water
m = Mass of the water
c = Specific heat of water
ΔT = change in temperature
Thus;
Q = 1000 × 4.18 × 15.8
Q = 66044 J
The solution absorbed 66044 J and therefore it means that the dissolution of the salt gave off 66044 J.
Thus;
Enthalpy is; ΔH = -66044 J
Now, Mass of NaOH is 60 g.
We know that molar Mass of NaOH is 39.997 g/mol. Thus;
Converting to moles, we have;
n = 60/39.997
n = 1.5 mol
Now, enthalpy when 1.5 moles of NaOH are dissolved in water will be;
ΔH = -66044 J/1.5 moles
ΔH = -44029.33 J/mol
ΔH = -44.029 KJ/mol
Cool air can hold less water vapor than warm air. Apply this fact to explain why clouds and precipitation form on the windward side of the mountain.
The pressure is directly proportional to temperature (when the pressure decrease the temperature decrease too). Because the air parcel expands so the molecules will not interact with each other as much.
The energy of the particles does not change but the fact that the particles are more spaced out means the parcel is cooler.
so now, the warmer a parcel of air the more water vapor it can hold. so, if a parcel of air cools it's ability to hold water vapor drops and if it drops to a low enough point that is when the water vapor will condense and turn back into liquid water. This is how clouds and precipitation form on the the windward side of the mountain.
Clouds and precipitation form on the windward side of the mountain due to change in temperature and pressure.
What is the relation between pressure and temperature?
Temperature of any substance is directly proportional to the pressure of that substance.
If the temperature of the wind increases so that winds get warmer and at the same time pressure of gas also increases, due to which particles will go far from each other. After this warm air rises up and cools down will condenses to form clouds, after this precipitate will falls on the windward side of the mountain.
Hence due to temperature and pressure precipitate will form on the windward side of the mountain.
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What is the mass of 0.063x10^-4 moles of aluminum sulphate ?
Answer:
The mass of 0.063*10⁻⁴ moles of aluminum sulphate is 2.15*10⁻³ grams.
Explanation:
Aluminum sulfate Al₂(SO₄)₃ has a molar mass of 342.15 g/mol.
Molar mass is the amount of mass that a substance contains in one mole of a substance, which can be an element or a compound.
So in this case you can apply the following rule of three: if 342.15 grams are present in 1 mole of aluminum sulfate, how much mass is present in 0.063*10⁻⁴ moles of the compound?
[tex]mass of aluminum sulphate=\frac{0.063*10^{-4}moles*342.15 grams }{1 mole}[/tex]
mass of aluminum sulphate= 2.15*10⁻³ grams
The mass of 0.063*10⁻⁴ moles of aluminum sulphate is 2.15*10⁻³ grams.
what energy is walking?
A. light energy
B. sound energy
C. Thermal energy
D. Mechanical energy
E. electrical energy
Answer:
Mechanical
Explanation:
In a single replacement reaction, the chloride ion in NaCl can be replaced
by
a. potassium
b. carbon
c.fluorine
d.neon
Answer:
c. fluorine
Explanation:
A single replacement reaction is defined as the chemical reaction in which a strong molecule replaces the weak molecule from a compound.
In a single replacement reaction, the chloride ion in NaCl can be replaced by fluorine and gives Sodium fluoride as fluorine (F) is stronger than chlorine (Cl) and from other given molecules also. So the single replacement reaction between NaCl and F will be:
NaCl + F2 => NaF + Cl2
Hence, the correct answer is "c. fluorine".
A chemist needs to know the mass of a sample of to significant digits. He puts the sample on a digital scale. This is what the scale shows:
0
0
0
7
6
.
3
g
If this measurement is precise enough for the chemist, round it to significant digits.
Otherwise, press the "No solution" button.
Suppose you want to make an acetic acid/acetate buffer to a pH of 5.00 using 10.0 mL of 1.00 M acetic acid solution. How many milliliters of 1.00 M sodium acetate solution would you need to add? The pKa for acetate buffer is 4.75.
Answer:
Explanation:
Molarity of NaOAc needed
Using the Henderson-Hasselbalch Equation calculate base molarity needed given [HOAc] = 1.00M and pKa(NaOAc) = 4.75 and [HOAc] = 1.00m.
pH = pKa + log [NaOAc]/[HOAc]
5.00 = 4.75 + log[NaOAc]/[1.00M]
5.00 - 4.75 = log [NaOAc] - log[1.00M]
log [NaOAc] = 0.25 => [NaOAc] = 10⁰·²⁵ M = 1.78
Given 10ml of HOAc, how much (ml) 1.78M NaOAc to obtain a buffer pH of 5.00.
Determine Volume of Base Needed
(M·V)acid = (M·V)base => V(base) = (M·V)acid / (M)base
Vol (NaOAc) needed = (1.00M)(0.010L)/(1.78M) = 0.0056 liter = 5.6 ml.
Checking Results:
5.00 = 4.75 + log [1.78M]/[1.00M] = 4.75 + 0.25 = 5.00 QED.
The volume of 1.00 M sodium acetate solution needed to prepare an acetic/acetate buffer of pH 5.00 using 10.0 mL of 1.00M acetic acid solution is 17.8 mL.
We can find the volume of the acetate solution with the Henderson-Hasselbalch equation:
[tex]pH = pka + log(\frac{[CH_{3}COONa]}{[CH_{3}COOH]})[/tex] (1)
Where:
[CH₃COOH] = 1.00 M
[CH₃COONa] =?
pH = 5.00
pKa = 4.75
From equation (1), we have:
[tex] log(\frac{[CH_{3}COONa]}{[CH_{3}COOH]}) = pH - pKa [/tex]
[tex] \frac{[CH_{3}COONa]}{[CH_{3}COOH]} = 10^{pH - pKa} [/tex]
[tex] \frac{[CH_{3}COONa]}{[CH_{3}COOH]} = 10^{5.00 - 4.75} = 1.78 [/tex]
Now, the volume of the acetate solution is:
[tex]\frac{n_{CH_{3}COONa}/Vt}{n_{CH_{3}COOH}/Vt} = 1.78[/tex]
Since the total volume is the same, we have:
[tex]\frac{n_{CH_{3}COONa}}{n_{CH_{3}COOH}} = 1.78[/tex]
[tex] \frac{[CH_{3}COONa]_{i}*V_{b}}{[CH_{3}COOH]_{i}*Va} = 1.78 [/tex]
Solving for Vb
[tex] Vb = \frac{1.78*[CH_{3}COOH]_{i}*Va}{[CH_{3}COONa]_{i}} = \frac{1.78*1.00M*10.0mL}{1.00 M} = 17.8 mL [/tex]
Therefore, we need to add 17.8 mL of sodium acetate solution.
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What is the molar mass of V(CO3)2?
Given the reaction
3 H3PO4 + 2 Ba(OH)2 ⇒ BaPO4 + 6 H20
How many grams of precipitate could you make if you reacted 286.0 mL of 3.000 M H3PO4 with 855.0 mL of 1.400 M Ba(OH)2
Answer:
240.17 g Ba3(PO4)2
Explanation:
1. Determine the limiting reactant.
2H3PO4 + 3Ba(OH)2 --> Ba3(PO4)2 + 6H2O
moles H3PO4 = M x V = 3 x 0.286 = .858 moles H3PO4
moles Ba(OH)2 = M x V = 1.4 x 0.855 = 1.197 moles Ba(OH)2
ratio Ba(OH)2 : H3PO4 = 1.197: .858 = 1.39: 1
stoichiometric ratio Ba(OH)2 : H3PO4 = 3:2
Ba(OH)2is the limiting reactant
MM Ba3(PO4)2 = 601.92 g/mol
g Ba3(PO4)2 = moles Ba(OH)2 x(1 mol Ba3(PO4)2/3 moles Ba(OH)2) x (MM Ba3(PO4)2/ 1mol Ba3(PO4)2) = 1.197 x 1/3 x 601.92 = 240.17 g Ba3(PO4)2
Convert 5.7 kcal to its value in calories.
Answer:
5700
Explanation:
Which type of rock does Florian predict these pieces will change into when subjected to heat and pressure?
Answer:
The Florian predict the metamorphic rocks because it changes when subjected to heat and pressure.
Explanation:
i hope this helps
Laura can use a toy car to demonstrate the force of gravity by doing which of these?
Letting the toy car roll down a ramp
Placing the toy car on the floor
Lifting the toy car onto a shelf
Pushing the toy car across a table
Answer:
letting the car fall down the ramp i got that right.
Explanation:
I guess this is free:)))): ((((
Answer:
tysm!!!! :D <<<33333
Explanation:
How many molecules are in 450.0 grams of aluminum fluoride(AIF3)?
Answer:
It would be exactly 5.3586262014272155. But if you were to round it up it would be 5.35.
why blood is separated into different parts
Answer:
Blood fractionation is the process of fractionating whole blood, or separating it into its component parts. This is typically done by centrifuging the blood. The resulting components are: a clear solution of blood plasma in the upper phase (which can be separated into its own fractions, see Blood plasma fractionation),
Answer: Centrifugal force is used to separate the components of blood – red blood cells, platelets and plasma – from each other. ... The red blood cells precipitate to the bottom of the bag, with the platelets above them, then the white blood cells and the plasma at the very top. Also because Each part of the blood has a different function. Separating the blood into parts lets patients get only the specific part or parts of the blood that they need. So a whole blood donation can be used for several patients.
Hope this helps have a awesome day/nigh❤️✨t
Explanation:
molar mass of Mg(CO3)
Answer:
[tex]\boxed {\boxed {\sf 84.313 \ g/mol}}[/tex]
Explanation:
The molar mass is the mass per mole of an element or substance. The values for atomic mass on the Periodic Table are used as the molar mass. Look up the molar masses for the elements in the formula.
Mg: 24.305 g/molC: 12.011 g/molO: 15.999 g/molAnalyze the formula. Magnesium and carbon have no subscript, so there is only one atom of each. Oxygen has a subscript of 3, so there are 3 atoms and the mass has to be multiplied by 3.
MgCO₃= 24.305 + 12.011+ 3(15.999) MgCO₃= 24.305+12.011+47.997=84.313 g/molThe molar mass of magnesium carbonate is 84.313 grams per mole.
The percent composition by mass of nitrogen in NH OH (gram-formula mass = 35 grams/mole) is equal to
Answer:
the percentage composition of mass of nitrogen in NH OH is 42.86 %
I have a balloon that can hold 100. liters
of air. If I blow up this balloon with 3.0
moles of oxygen gas at a pressure of 1.0
atmosphere, what is the temperature of
the balloon?
Answer:
T = 4.062V
Explanation:
from PV = nRT => T = PV/RT
P = 1 atm
V = Final Volume
n = 3 moles
R = 0.08206 L·atm/mol·K
T = ?
T = 1 atm · V(Liters)/(3 moles)(0.08206L·atm/mol·K) = 4.062·V(final) Kelvin
The temperature of the balloon is 406 K
We'll begin by listing out what was given from the question. This is shown below:
Volume (V) = 100 L
Mole of oxygen (n) = 3 moles
Pressure (P) = 1 atm
Temperature (T) =?We can obtain the temperature of the balloon by using the ideal gas equation as shown below:
PV = nRTNOTE:P is the pressure.
V is the volume.
n is the number of mole
R is the gas constant (0.0821 atm.L/Kmol)
T is the temperature.
Applying the ideal gas equation, we have:
PV = nRT
1 × 100 = 3 × 0.0821 × T
100 = 0.2463 × T
Divide both side by 0.2463
T = 100 / 0.2463
T = 406 KTherefore, the temperature of the balloon is 406 K
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How do the causes of surface and deep water currents differ?
A. Surface currents are caused by wind deep water currents are caused by difference in water density
B. Surface currents are caused by the Coriolis effect deep water currents are caused by differences in water density
C. Surface currents are caused by differences in water salinity deep water currents are caused by differences in water temperature
D. Surface currents are caused by differences in water density deep water currents are caused by wind
Answer:
The answer is A. Surface currents are caused by wind deep water currents are caused by difference in water density.
The smallest arteries in your body are called _____. arterioles capillaries veins bronchioles
Answer: arterioles
Explanation:
We obviously cannot wait for million years to determine the half-life of technetium, or even years to determine the half-life of plutonium. The half-life of a substance is determined using a Geiger counter, which is essentially a tube filled with an inert gas. When a particle decays and produces radiation, it briefly ionizes the gas and produces a detectable electric pulse, which is amplified and counted by a detector (before digital displays were available, you would hear clicks and see a needle register on a dial). Determine the formula for the half-life of a substance with an initial count of pulses and a count of pulses minutes later.
The question is incomplete, the complete question is;
We obviously cannot wait for 4 million years to determine the half-life of technetium, or even 87.7 years to determine the half-life of plutonium. The half-life of a substance is determined using a Geiger counter, which is essentially a tube filled with an inert gas. When a particle decays and produces radiation, it briefly ionizes the gas and produces a detectable electric pulse, which is amplified and counted by a detector (before digital displays were available, you would hear clicks and see a needle register on a dial). Determine the formula for the half-life of a substance with an initial count of No pulses and a count of N1 pulses M minutes later.
Answer:
M1/2 = 0.693/k
Explanation:
Given that the initial count rate is No pulses
The count rate after M minutes is N1
Then from;
A = Aoe^-kt
A= N1
Ao=No
I can now write;
N1 = Noe^-kM
The half life is the time taken for the activity of the radioactive nuclide to reach half its initial value. Hence;
N1/No =e^-kM = 1/2
Taking natural logarithm of both sides as shown below;
-kM1/2 = ln(1/2)
M1/2 = - (ln(1/2)/k)
M1/2 = 0.693/k
If you began a reaction with the following ions in solution, what would be the net ionic equation?
Na+, PO43–, Fe2+, and NO3–
Answer:
2 PO₄³⁻(aq) + 3 Fe²⁺(aq) ⇒ Fe₃(PO₄)₂
Explanation:
Let's consider the complete ionic equation between the ions present. It includes all the ions and the insoluble compounds (Fe₃(PO₄)₂ is insoluble).
Na⁺(aq) + 2 PO₄³⁻(aq) + 3 Fe²⁺(aq) + NO₃⁻(aq) ⇒ Fe₃(PO₄)₂ + Na⁺(aq) + NO₃⁻(aq)
The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and the insoluble compounds.
2 PO₄³⁻(aq) + 3 Fe²⁺(aq) ⇒ Fe₃(PO₄)₂
Pls someone help rn!! Pls pls
Answer:
A. Mass = 61kg, weight on moon = 97.8N
Explanation:
To calculate the weight (force) from Mass of a body, the following formula is used:
W = mg
Where;
W = weight (Newtons)
m = mass (kg)
g = acceleration due to gravity (9.8m/s²)
According to this question, a scale shows that I weigh 600N on Earth, the mass on Earth can be calculated thus:
Using W = mg
m = W/g
m = 600/9.8
m = 61.2
m = 61kg
However, this mass on Earth will weigh differently on the Moon because gravity in the moon is 1.6 N/kg. The weight of the 61kg on the Moon is:
W(moon) = mass × gravity
W = 61 × 1.6
W = 97.6 N
Therefore, Mass = 61kg, weight on moon = 97.8N
1. At an environmental testing lab, a lab technician calculated the known concentration of the calibrating solution three times too high. How will this affect the results of the samples run using this calibration solution
Answer:
The deviation in the results will be too large when compared with the actual result
Explanation:
The effect this will have on the results of the samples run using this Calibration solution is that the deviation in the results will be too large when compared with the actual result that was supposed to be obtained. and this will also lead to wrong analysis by the technician .
This will cause a result that is very different from the actual result, which will lead to errors in the analysis.
We can arrive at this answer because:
A laboratory test must be done accurately and all calibrations must follow the test protocol.Changing the calibration will cause inaccurate and incorrect results, which differ from the actual result.This will promote an analysis that is unreal, false, and far from true, which will leave the entire test inaccurate.As we can see from the question above, the lab technician used a much higher calibration than what was recommended in the lab. This will lead to inaccurate and incorrect results.
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Hydrofluoric acid and Water react to form fluoride anion and hydronium cation, like this HF(aq) + H_2O(l) rightarrow F(aq) + H_3O^+ (aq) At a certain temperature, a chemist finds that a 5.6 L reaction vessel containing an aqueous solution of hydrofluoric acid, water, fluoride anion, and hydronium cation at equilibrium has the following composition: Calculate the value of the equilibrium constant K_C for this reaction. Round your answer to 2 significant digits. K_C =
Answer:
Kc = 1.09x10⁻⁴
Explanation:
HF = 1.62g
H₂O = 516g
F⁻ = 0.163g
H₃O⁺ = 0.110g
To solve this question we need to find the moles of each reactant in order to solve the molar concentration of each reactan and replacing in the Kc expression. For the reaction, the Kc is:
Kc = [H₃O⁺] [F⁻] / [HF]
Because Kc is defined as the ratio between concentrations of products over reactants powered to its reaction coefficient. Pure liquids as water are not taken into account in Kc expression:
[H₃O⁺] = 0.110g * (1mol /19.01g) = 0.00579moles / 5.6L = 1.03x10⁻³M
[F⁻] = 0.163g * (1mol /19.0g) = 0.00858moles / 5.6L = 1.53x10⁻³M
[HF] = 1.62g * (1mol /20g) = 0.081moles / 5.6L = 0.0145M
Kc = [1.03x10⁻³M] [1.53x10⁻³M] / [0.0145M]
Kc = 1.09x10⁻⁴what does the word interact
Lift
Fly
Discuss
Work with
Answer:
The answer would be work with.
Explanation:
Because, if we know what the word interact means then we can look for a word similar to interact. hope this makes sense <3.
Cu + 2AgNO3 → Cu(NO3)2 + 2Ag
If a 40. g sample of Copper is used with an excess of silver nitrate, calculate the theoretical yield of silver. What is the percent yield of the silver, if 120 g was collected?
Answer:
[tex]m_{Ag}=135.8gAg[/tex]
[tex]Y=88.4\%[/tex]
Explanation:
Hello there!
In this case, according to the described chemical reaction, it is possible to compute the theoretical mass of silver as mass via the 1:2 mole ratio of copper to silver and their atomic mass in the periodic table, in order to perform the following stoichiometric setup:
[tex]m_{Ag}=40.gCu*\frac{1molCu}{63.55gCu}*\frac{2molAg}{1molCu}*\frac{107.87gAg}{1molAg}\\\\ m_{Ag}=135.8gAg[/tex]
Next, given the actual yield of 120 g, we compute the percent yield via:
[tex]Y=\frac{120g}{135.8g}*100\%\\\\Y=88.4\%[/tex]
Regards!
A granulated cylinder was filled to 30.0 mL with liquid. An unknown metal cylinder having a mass of 60.27 g was submerged in the liquid. The final volume of the liquid level was 45.2 mL.
Answer:
Explanation:
A graduated cylinder was filled to 30.0 mL with a liquid. An unknown metal cylinder having a mass of 60.27 g was submerged in the in the liquid. The final volume of the liquid level was 45.2 mL.
a. Calculate the density of the unknown metal cylinder.
b. The density of the liquid is 0.899 g/mL. What is the mass of the liquid in the graduated cylinder?
Balance the chemical equation
C2H6+O2 = CO2 +H2O
Help help help !!!!!!
Answer:
3. 116.5 V
4. 119.6 V
Explanation:
3. Determination of the voltage.
Resistance (R) = 25 Ω
Current (I) = 4.66 A
Voltage (V) =?
V = IR
V = 4.66 × 25
V = 116.5 V
Thus, the voltage is 116.5 V
4. Determination of the voltage.
Current (I) = 9.80 A
Resistance (R) = 12.2 Ω
Voltage (V) =?
V = IR
V = 9.80 × 12.2
V = 119.6 V
Thus, the voltage is 119.6 V
The particle on a ring is a useful model for the motion of electrons around the porphine ring, the conjugated macrocycle that forms the structural basis of the heme group and the chlorophylls. We may treat the group as a circular ring of radius 440 pm, with 22 electrons in the conjugation system moving along the perimeter of the ring. In the ground state of the molecules each state is occupied by two electrons.
A) Calculate the energy and angular momentum of an electron in the highest occupied level.
B) Calculate the frequency of radiation that can induce a transition between the highest occupied and lowest unoccupied levels.
Answer:
Explanation:
The formula for determining the energy of state [tex]m_l[/tex] can be computed by using the formula:
[tex]Em_i = \dfrac{m_1^2h^2}{2I}[/tex]
Also, the momentum is:
[tex]l_2 = m_i h[/tex]
There are 22 electrons with two electrons in each of the lowest II energy levels so that the highest occupied states are [tex]m_1 = \pm 5[/tex]
The moment of inertia of an electron on a ring of radius 440 ppm is:
[tex]I = mR^2 \\ \\ = 9.109 \times 10^{31} \ kg (440 \times 10^{-12}) ^2 \\ \\ = 1.76 \times 10^{-49} \ kgm^2[/tex]
[tex]E_{\pm 5 } = \dfrac{25h^2}{2I} \\ \\ = \mathbf{7.89 \times 10^{-19} \ J}[/tex]
The angular momentum is:
[tex]l_2 = \pm 5h \\ \\ = \pm 5 \tiems ( 1.05457 \times 10^{-34} \ Js)[/tex]
[tex]= \mathbf{5.275 \times 10^{-34} \ Js}[/tex]
B) Let's recall that:
The lowest occupied energy level is [tex]m_1 = \pm 6[/tex] which implies that the energy [tex]E_{\pm 6}= 1.14 \times 10^{-18} \ J[/tex]
Thus;
[tex]\Delta E = E_{\pm 6} - E_{\pm 5} \\ \\ = 1.14 \times 10^{-18 \ J} - 0.79 \times 10^{-18} \ J \\ \\ = 0.35 \times 10^{-18} \ J \\ \\ 0.35 \times 10^{-18} \ J = hv \\ \\ 0.35 \times 10^{-18} \ J = h \dfrac{c}{\lambda}[/tex]
Hence, the radiation which would induce a transition that relates to the wavelength of about 570nm, a wavelength of visible light.