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Sunny_sXe [5.5K]
3 years ago
8

How many liters of methane gas (CH4) need to be combusted to produce 8.5 liters of water vapor, if all measurements are taken at

the same temperature and pressure? Show all of the work used to solve this problem. CH4 (g) + 2O2 (g) yields CO2 (g) + 2H2O (g)
Chemistry
1 answer:
zimovet [89]3 years ago
6 0
8.5L H2O (1 mol H2O/22.4L)=.37946 mol H2O
.3796 mol H2O(1 mol CH4/2 mol H2O)=.18973 mol CH4
.18973 mol CH4(22.4L/1mol CH4)=4.25L CH4 gas
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Which is an unreliable source?
vodomira [7]

Answer:

I believe CDC I s the most reliable

6 0
2 years ago
How much thermal energy (q) is required to heat 15.2g of a metal (specific heat=0.397J/C) from 21.0C to 40.3C? Show steps please
Elena L [17]
Q = ?

Cp = 0.397 J/ºC

Δt =  40.3 - 21.0<span> => 19.3</span><span> ºC</span>

m = 15.2 g

Q = m x Cp x Δt

Q = 15.2 x 0.397 x 19.3

Q ≈ 116.46 J

<span>hope this helps! </span>
7 0
3 years ago
How many hydrogen atoms are in 709 grams of water? Answer in units of atoms.
zimovet [89]

Answer:

26 Hydrogen atoms

Explanation:

H2O

Each hydrogen atom: 2+16 = 18g

Hence,

1 atom -> 18g

x atoms -> 709g

709/18 = 39 atoms

Therefore, 39 atoms give 709g

Hence, 26 Hydrogen atoms are used

<em>Feel free to mark it as brainliest :D</em>

4 0
3 years ago
33.6 grams of CaO should have been produced in the decomposition of
spayn [35]

Answer:

answer answer answer answer answer

6 0
2 years ago
A certain reaction with an activation energy of 185 kJ/mol was run at 505 K and again at 525 K . What is the ratio of f at the h
frosja888 [35]

Answer:

The ratio of f at the higher temperature to f at the lower temperature is 5.356

Explanation:

Given;

activation energy, Ea = 185 kJ/mol = 185,000 J/mol

final temperature, T₂ = 525 K

initial temperature, T₁ = 505 k

Apply Arrhenius equation;

Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]

Where;

\frac{f_2}{f_1}  is the ratio of f at the higher temperature to f at the lower temperature

R is gas constant = 8.314 J/mole.K

Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]\\\\Log(\frac{f_2}{f_1} ) = \frac{185,000}{2.303 \times 8.314} [\frac{1}{505} -\frac{1}{525} ]\\\\Log(\frac{f_2}{f_1} ) = 0.7289\\\\\frac{f_2}{f_1}  = 10^{0.7289}\\\\\frac{f_2}{f_1}  = 5.356

Therefore, the ratio of f at the higher temperature to f at the lower temperature is 5.356

5 0
3 years ago
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