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Dima020 [189]
3 years ago
12

श्रीरयाफल्सन यसका लागि तियरलप्रतिस्थापन स्तर को के हो​

Chemistry
1 answer:
jeka943 years ago
3 0
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What mass of water is formed when 16g of hydrogen react with excess oxygen​
artcher [175]

Hello!

To start off, we must look at atomic masses. Atoms all have different weights, so we must first find hydrogen and oxygen's atomic masses.

Oxygen: 16.00 amu

Hydrogen: 1.01 amu

Now, moving on to the weight of water itself. Water has the formula of H20, with two hydrogen atoms and one oxygen. Therefore, <u>add up the amus to get the weight of one molecule of water.</u>

1.01 + 1.01 + 16.00 = 18.02 amu

Now, to see the ratio of each component. Since hydrogen weighs a total of 2.02 amu (1.01 + 1.01) in the entire atom, we can state that hydrogen makes up about 0.112 of the weight of water. Now apply that ratio to 16 g, and solve.

0.112x = 16

142.857143 = x

So therefore, about 143 grams of water are made when 16g of hydrogen reacts with excess oxygen.

Hope this helps!

8 0
3 years ago
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madreJ [45]

Answer:

B) The group the atom is located

C) There needs to be 2 Na atoms and 1O atom to form Na2O

6 0
3 years ago
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Determine the concentration of an unknown acid or base in a neutralization reaction
ryzh [129]
Titration is the process by which the concentration of an unidentified analyte is found
8 0
4 years ago
Hello, <br><br> So I was wondering if this is correct... Is it?
Sonja [21]
Looks correct but the second to last I would of put abiotic and biotic factors but I don’t know what’s right for you
4 0
3 years ago
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A gaseous fuel mixture stored at 747 mmHg and 298 K contains only methane (CH4) and propane (C3H8). When 11.1 L of this fuel mix
Alisiya [41]

Answer:

M_f=38.8\%

Explanation:

From the question we are told that:

Pressure P=747mmHg

Temperature T=298K

Volume V=11.1

Heat Produced Q=780kJ

Generally the equation for ideal gas is mathematically given by

 PV=nRT

 n= (747/760) *11.1/ (0.0821*298)

 n=0.446mol

Therefore

 x+y=0.446

 x=0.446-y .....1

Since

Heat of combustion of Methane=889 kJ/mol

Heat of combustion of Propane=2220 kJ/mol

Therefore

 x(889) + y(2220) = 760 ...... 2

Comparing Equation 1 and 2 and solving simultaneously

 x=0.446-y .....1

 x(889) + y(2220) = 760 ...... 2

 x=0.173

 y=0.273

Therefore

Mole fraction 0f Methane is mathematically given as

 M_f=\frac{x}{n}*100\%

 M_f=\frac{1.173}{0.446}*100\%

 M_f=38.8\%

7 0
3 years ago
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