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2 years ago

GAS LAW:

Chemistry

1 answer:

ivann1987 [24]2 years ago

7 0

Answer:

bh yftd5xsdcfvgbhnjm

Explanation:

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Convert 4.500000000 to scientific notation

aniked [119]

I think it’s 4.5 x 10^9

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3 years ago

Exercise A mixture of 250 mL of methane, CH4, at 35 °C and 0.55 atm and 750 mL of propane, C3Hg, at 35° C and 1.5 atm, were intr

dalvyx [7]

Explanation:

The given data is as follows.

$V_{1}$ = 250 mL, $V_{2}$ = 750 mL

$T_{1}$ = $35^{o}C$ = 35 + 273 K = 308 K

$T_{2}$ = 35 + 273 K = 308 K

$P_{1}$ = 0.55 atm, $P_{2}$ = 1.5 atm

P = ? , V = 10.0 L

Since, temperature is constant.

So, $P_{1}V_{1} + P_{2}V_{2}$ = PV

Now, putting the given values into the above formula as follows.

$P_{1}V_{1} + P_{2}V_{2}$ = PV

$0.55 atm \times 250 mL + 1.5 atm \times 1.5 atm$ = $P \times 10.0 L$

P = 0.126 atm

As, 1 atm = 760 torr. So, $0.126 atm \times \frac{760 torr}{1 atm}$ = 95.76 torr.

Thus, we can conclude that the final pressure, in torr, of the mixture is 95.76 torr.

8 0

2 years ago

Explore how the identity of reactants and products can be used to classify types of reactions

aliya0001 [1]

Reactants are substances that start a chemical reaction, and products are substances that are produced in the reaction hope this helps

8 0

2 years ago

Which set of coefficients will balance this chemical equation?

Montano1993 [528]

Answer:

Option B is correct = 1,3

Explanation:

Chemical equation:

C₂H₄ + O₂ → CO₂ + H₂O

Balanced chemical equation:

C₂H₄ + 3O₂ → 2CO₂ + 2H₂O

Step 1:

Left side Right side

C = 2 C = 1

H = 4 O = 3

O = 2 H = 2

Step 2:

C₂H₄ + O₂ → 2CO₂ + H₂O

Left side Right side

C = 2 C = 2

H = 4 O = 5

O = 2 H = 2

Step 3:

C₂H₄ + O₂ → 2CO₂ + 2H₂O

Left side Right side

C = 2 C = 2

H = 4 O = 6

O = 2 H = 4

Step 4:

C₂H₄ + 3O₂ → 2CO₂ + 2H₂O

Left side Right side

C = 2 C = 2

H = 4 O = 6

O = 6 H = 4

5 0

2 years ago

The internal energy of 10 moles of helium (a monatomic gas) is 15 kJ. What is the rms speed of the molecules? (The molar mass of

Radda [10]

Answer:

$v_{rms}=866.32m/s$

Explanation:

Hello,

In this case, since the rms speed of the molecules is computed by:

$v_{rms}=\sqrt{\frac{3RT}{M} }$

Whereas the absolute temperature is computed from the internal energy (by using the Cp of helium (3.1156 J/g*K) as shown below:

$U=nCvT\\\\T=\frac{U}{nCv}=\frac{15kJ*\frac{1000J}{1kJ} }{10mol*\frac{4.00g}{1mol} *3.1156\frac{J}{g*K} } \\\\T=120.36K$

Thereby, the rms speed results:

$v_{rms}=\sqrt{\frac{3*8.314\frac{kg*m^2}{s^2*mol*K}*120.36K}{4.00\frac{g}{mol}*\frac{1kg}{1000} } } \\\\v_{rms}=866.32m/s$

Regards.

6 0

3 years ago