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Leni [432]
2 years ago
10

Formation of Water Reaction

Chemistry
2 answers:
Naddika [18.5K]2 years ago
5 0

Answer:

Oxygen had 2 before and 2 after.

Hydrogen had 4 before and 4 after.

Explanation:

The atoms are transferred, not transformed.

lilavasa [31]2 years ago
3 0

Answer:

hydrogen and oxygen combine to water by the formula 2h2+02 the formation reaction of this process is h2+1\2 02 how to make water from hydrogen and oxygen limiting reactant definition limiting reagent

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A 6.175 gram sample of an organic compound containing only C, H, and O is analyzed by combustion analysis and 13.30 g CO2 and 5.
Bingel [31]

Answer:

Empirical and molecular formulas are the same, C₅H₁₀O₂.

Explanation:

Hello!

In this case, when determining the empirical and molecular formulas of organic compounds via combustion analysis, we first need to compute the moles of carbon and hydrogen via the yielded mass of carbon dioxide and water:

n_C=13.30gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.30molC\\\\n_H=5.447gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}=0.60molH

Next, we need to compute the mass of oxygen by subtracting the mass of carbon and hydrogen to the mass of the sample of the compound:

m_O =6.175g-0.3molC*12.01gC/molC-0.6molH*1.01gH/molH =1.966gO

And consequently the moles:

n_O=0.12molO

Now, we need to divide the moles of each atom by the fewest moles, it in this case, those of oxygen to obtain the subscripts in the empirical formula:

C=\frac{0.30}{0.12} =2.5\\\\H=\frac{0.60}{0.12} =5\\\\O=\frac{0.12}{0.12} =1

Thus, the empirical formula, taken the nearest whole number is:

C_5H_{10}O_2

Now, if we divide the molar mass of the molecular formula (102.1 g/mol) by that of the empirical formula (102.1 g/mol) we infer they are both the same.

Best regards!

6 0
2 years ago
There are two different compounds of sulfur and fluorine.
weqwewe [10]
Atomic mass of F: 19.0 g/mol

Atomic mass of S: 32.1 g/mol

1.18 g F = [1.18 g / 19.0 g / mol] = 0.062  mol F

1 g S =  1 g/ 32.1 g/mol = 0.031  mol S

Divide by 0.031

0.062 mol F / 0.031 = 2  mol F

0.031 mol S / 0.031 = 1 mol S

SF2 Then X = 2

Verification:
F2 = 2*19.0 g = 38 g F
S = 32.1 g

36 gF / 32.1 g S = 1.18 g F / g S



7 0
3 years ago
I need help with unit conversion​
saw5 [17]

Answer:

what is it?

Explanation:

you haven't said?

3 0
3 years ago
What mass of ammonia can be produced if 13.4 grams of nitrogen gas reacted ?
aivan3 [116]

Answer:

If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia

Explanation:

Step 1: Data given

Mass of nitrogen gas (N2) = 13.4 grams

Molar mass of N2 = 28 g/mol

Molar mass of NH3 = 17.03 g/mol

Step 2: The balanced equation

N2 + 3H2 → 2NH3

Step 3: Calculate moles of N2

Moles N2 = Mass N2 / molar mass N2

Moles N2 = 13.4 grams / 28.00 g/mol

Moles N2 = 0.479 moles

Step 4: Calculate moles of NH3

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

For 0.479 moles N2 we'll produce 2*0.479 = 0.958 moles

Step 5: Calculate mass of NH3

Mass of NH3 = moles NH3 * molar mass NH3

Mass NH3 = 0.958 moles * 17.03 g/mol

Mass NH3 = 16.3 grams

If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia

3 0
3 years ago
A . liquid and solid<br>b. liquid only <br>c. liquid and gas <br>d. gas only <br><br>​
MArishka [77]

c. liquid and gas

Explanation:

its obvious, lol.

4 0
2 years ago
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