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wolverine [178]
3 years ago
15

QUESTION 3

Chemistry
1 answer:
Blababa [14]3 years ago
4 0
These questions are all about indirect and direct variation with PV=nRT, the ideal gas equation

Q3.
false, because of PV=nRT, the ideal gas equation if V goes up, P has to go down to equal nRT

Q4. false, if V remains constant, and P and T are constant as moles of gas are added, then something is wrong becse something has to change when stuff is added (V has to go down)

Q5.
PV=nRT
when T and n are constant, (R is the gas constant)
PV=k, aka V=k/P which means inversly proportional

TRUE


Q6.
ggeasy
refer to past question
PV=k
if P is doubled then V has to halve in order to equal k
1/2 times 2=1
volume is halved


Q7. use charles law
V/T=k
so
given
V=4
T= kelvins, so 299
4/299=k
so when temp goes to 22 does V go to 3.95
4/299=3.95/295?
true
because they're equal



Q8
FALSE, must be used in kelvins
T=absolute tempurature in kelvins


Q9
PV=nRT
solve for T
(PV)/(nR)=T

use final volumes and pressures
P=5atm
V=24L
n=1
R=0.082057 atm L/(mol K)
(5atm*24L)/(1mol*0.082057 atm L/mol K)=T
see, if you didn't mess up, the units cancel nicely
T=1462.4
1200 K is closest



Q10
PV/T=constant because moles are constant (supposedly)

V=4L
P=2.08atm
T=275K

so find initial to final is constant
(2.08atm*4L)/(275K)=(Pfinal*2.5L)/(323K)
solve for Pfinal
Pfinal=3.92315 atm
answer is 3.9atm

Merry Christmas
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How many moles of water are in 1/4 cup of water
il63 [147K]

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3.4752 moles of water

Explanation:

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In which reaction is it possible for nitrogen to change into carbon?
sertanlavr [38]
C. ionization reaction...
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5 0
3 years ago
Ammonia gas will react with oxygen gas to yield nitrogen monoxide gas and water vapor.
ruslelena [56]

Answer:

a) <u>0.168 moles O2</u>

<u>b) </u> <u>9.50 grams O2</u>

<u>c) 0.01662 kg NO</u>

<u>d)</u>88.9 %

Explanation:

Step 1: Data given

Molar mass of NH3 = 17.03 g/mol

Molar mass of O2 = 32 g/mol

Molar mass of NO = 30.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: The balanced equation:

4NH3 + 5O2 → 4NO + 6H2O

a. How many moles of ammonia will react with 6.73g of oxygen?

Calculate moles of oxygen = mass O2/ molar mass O2

moles oxygen =  6.73 grams / 32.00 g/mol = 0.210 moles

Calculate moles of NH3

For 4 moles of NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

For 0.210 moles O2 we need 4/5 *0.210 = <u>0.168 moles O2</u>

<u />

b. If 6.42g of water is produced, how many grams of oxygen gas reacted?

Calculate moles of H2O = 6.42 grams / 18.02 g/mol = 0.356 moles

Calculate moles of O2:

For 4 moles of NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

For 0.356 moles H2O we'll need 5/6 * 0.356 = 0.297 moles O2

Calculate mass of O2 = moles O2 * molar mass O2

Mass O2 = 0.297 moles O2 * 32.00 g/mol =  <u>9.50 grams O2</u>

c. If the reaction uses up 9.43105 g of ammonia, how many kilograms of nitrogen monoxide will be formed?

Calculate moles of ammonia = 9.43105 grams / 17.03 g/mol =0.5538 moles

Calculate moles of NO:

For 4 moles of NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

For 0.5538 moles of NH3 we'll have 0.5538 moles NO

Calculate mass of NO

Mass NO = 0.5538 moles * 30.01 g/mol = 16.62 grams = <u>0.01662 kg NO</u>

<u />

<u />

d. When 2.51 g of ammonia react with 3.76 g of oxygen, 2.27 g of water vapor are produced. What is the percentage yield of water?

<em>Calculate moles of NH3</em> = 2.51 grams / 17.03 g/mol = 0.147 moles

<em>Calculate moles of O2 </em>= 3.76 grams / 32 g/mol = 0.118 moles

<em>Determine the limiting reactant</em>

O2 is the limiting reactant, it will completely be consumed (0.118 moles)

NH3 is in excess. There will react 4/5 * 0.118 = 0.0944 moles

There will remain 0.147 - 0.0944 = 0.0526 moles

<em>Calculate moles H2O</em>: For 0.118 moles O2 we'll have 6/5 * 0.118 = 0.1416 moles H2O

<em>Calculate mass H2O</em> = 0.1416 moles * 18.02 g/mol = 2.552 grams H2O

<em>Calculate % yield</em> = (2.27/2.552)*100 % = <u>88.9 %</u>

4 0
3 years ago
n a coffee cup calorimeter, 100.0 mL of 1.0M NaOH and 100.0 mL of 1.0M HCl are mixed. Both solutions were originally at 24.6 deg
RoseWind [281]

Answer:

Explanation:

mass of the solution = volume x density = 200 x 1 = 200 gm

heat absorbed = m x s x Δ t , s is specific heat , Δt is rise in temperature

= 200 x 4.18 x ( 31.3 - 24.6 )

= 5601 J .

This is the enthalpy change required.

3 0
3 years ago
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