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DaniilM [7]
3 years ago
10

A gas occupies 14.3 liters at a pressure of 45.0 mm Hg. What is the volume when the pressure is increased to

Chemistry
1 answer:
Simora [160]3 years ago
8 0

Answer:

P1V1= P2V2

Explanation:

Inverse relationship

V2 = V1 X P1/P2

V2= 14.3 L x 45.0 mm Hg/63.0 mmHg= 8.99

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Determine the grams of Iron(III) chloride produced if 22.5 g Iron reacts with Chlorine gas. First, balance the chemical equation
TEA [102]
22.5 because it still stays the same
4 0
3 years ago
Need chem help asap
gavmur [86]

Oh I’m so4+ds2=563

Which is the answer

5 0
3 years ago
Propane (c3h8) is burned in oxygen to produce carbon dioxide and water. the heat of combustion of propane is -2012 kj/mole. how
Olegator [25]
C_{3} H_{8} + 5 O_{2} ---\ \textgreater \  3CO_{2}  +4H_{2}O    (-2012 \frac{kJ}{mol} )


3 mol                10 mol


C_{3}H_{8} is /excess /reactant

because 3 mol propane require 15 mol oxygen (by reaction)

5 mol oxygen ---1 mol propane, so

10 mol oxygen ---2 mol propane

Only 2 mole propane will be burned,

so 

2012 ( kJ/mol)*2 mol =4024 KJ heat will be given off

Correct answer is number 4.



5 0
3 years ago
What volume of CH4(g), measured at 25oC and 745 Torr, must be burned in excess oxygen to release 1.00 x 106 kJ of heat to the su
anastassius [24]

Answer:

V=27992L=28.00m^3

Explanation:

Hello,

In this case, the combustion of methane is shown below:

CH_4+2O_2\rightarrow CO_2+2H_2O

And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:

n_{CH_4}=\frac{-1.00x10^6kJ}{-890.8kJ/mol}= 1122.6molCH_4

Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:

PV=nRT\\\\V=\frac{nRT}{P}=\frac{1122.6mol*0.082\frac{atm*L}{mol*K}*298K}{0.98atm}\\\\V=27992L=28.00m^3

Best regards.

8 0
3 years ago
Assume the molality of isoborneol in your product is 0.275 mol/kg. What is the melting point of your impure sample given that th
aliina [53]

Answer:

168°C is the melting point of your impure sample.

Explanation:

Melting point of pure camphor= T =179°C

Melting point of sample = T_f = ?

Depression in freezing point = \Delta T_f

Depression in freezing point  is also given by formula:

\Delta T_f=i\times K_f\times m

K_f = The freezing point depression constant

m = molality of the sample  = 0.275 mol/kg

i = van't Hoff factor

We have: K_f = 40°C kg/mol

i = 1 (  non electrolyte)

\Delta T_f=1\times 40^oC kg/mol\times 0.275 mol/kg

\Delta T_f=11^oC

\Delta T_f=T- T_f

T_f=T- \Delta T_f=179^oC-11^oC=168^oC

168°C is the melting point of your impure sample.

4 0
3 years ago
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