Interesting problem. Thanks for posting.
C2H2 + (3/2)02 ====> H2O + 2CO2
CH4 + 2O2 =====> 2H2O + CO2
The molar mass of C2H2 = 2*12 + 2*1 = 26
The molar mass of CH4 = 1*12 + 4*1 = 16
The number of moles of C2H2 = x
The number of moles of CH4 = y
26x + 16y = 230.9 grams
For water we get (from the C2H2). Water has a molar mass of 2*1 + 16 = 18
x*18 See the balanced equation to see what it is the same number of moles as C2H2
From the methane we get
y*18
2*y* 18. Again see the balanced equation to see where that 2 came from.
18x + 36y is the total amount of water.
Now for the CO2. CO2 has a molar mass of 12 + 2*16 = 44
From C2H2 we get 2*44*x = 88x grams of CO2
From CH4 we get 1*y*44 grams of CO2
88x + 44y for CO2
Now we total to get the grand total of water and CO2
18x + 44y + 88x + 44y = 972.7 grams total.
106x + 88y = 972.7
Two equations, two unknowns, we should be able to solve this problem
26x + 16y = 230.9
106x + 88y = 972.7
I'm not going to go through the math unless you request me to do so.
x = 8.03 moles
y = 1.38 moles
The initial amount of C2H2 was 8.03 * 26 = 208.78
The initial amount of CH4 was 16*1.38 = 22.08
The total (as a check is 230.86 which is pretty close to the given amount.
So Methane's mass in the initial givens was 22.08 grams.
Answer:
The number of hydrogen atoms is 4.96x10²⁴.
Explanation:
The number of atoms can be found with the following equation:

Where:
N: is the Avogadro's number = 6.022x10²³ atoms/mol
η: is the number of moles of hydrogen
n: is the number of hydrogen atoms
First, we need to find the number of hydrogen moles. The number of moles of CH₄ is:

Where:
m: is the mass of methane = 33 g
M: is the molar mass of methane = 16.04 g/mol

Now, since we have 4 hydrogen atoms in 1 mol of methane, the number of moles of hydrogen is:

Hence, the number of hydrogen atoms is:

Therefore, the number of hydrogen atoms is 4.96x10²⁴.
I hope it helps you!
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Answer: 70.0°C
Explanation:
Quantity of heat = Mass * Specific heat * Change in temperature
Quantity of heat = 104.6 KJ
Mass = 500.0 g
Specific heat of water is 4.18 J/g°C
Change in temperature assuming final temperature is x = x - 20
Units should be in grams and joules:
104,600 = 500 * 4.18 * (x - 20)
104,600 = 2,090 * (x - 20)
x - 20 = 104,600/2,090
x = 104,600/2,090 + 20
x = 69.8
= 70.0°C
Zn = 28.15%
Cl = 30.53%
O = 41.32%
<h3>Further explanation</h3>
Given
Zn(CIO3)2 compound
Required
The % composition
Solution
Ar Zn = 65.38
Ar Cl = 35,453
Ar O = 15,999
MW Zn(CIO3)2 = 232.3
Zn = 65,38/232.3 x 100% = 28.15%
Cl = (2 x 35.453) / 232.3 x 100% = 30.53%
O = (6 x 15.999) / 232.3 x 100% = 41.32%