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sladkih [1.3K]
3 years ago
8

How many water molecules are created when 32 grams of oxygen react completely with 4.0 grams of hydrogen?

Chemistry
1 answer:
ch4aika [34]3 years ago
3 0

Answer:

d. 1.2 × 1024

Explanation:

From the equation of reaction

2H2 + O2= 2H2O

i.e 2mole(4g) of hydrogen requires 1 mole(32g) of oxygen to produce 2mole (2×6.02×10^23 molecules) of H2O= 1.2×20^24 molcules of water.

NB: 1 mole of H2O contains 6.02×10^23 molecules of H2O

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Convert 2.345 centimeters cubed to inches cubed. 1inch = 2.54cm
GalinKa [24]

Answer:

0.92322835

Explanation:

0.92322835 rounded= 0.9 or 0.92

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Name one other organ system that would be affected if it had no marrow.Explain?
muminat
It can possible be you're arteries or also you're intestines with is large and small.

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Iron is extracted from iron oxide in the Blast Furnace: Fe 2 O 3 + 3 CO → 2 Fe + 3 CO 2
arsen [322]

a. mass of iron = 69.92 g

b. percent yield = 93%

<h3>Further eplanation </h3>

Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated

General formula:

Percent yield = (Actual yield / theoretical yield )x 100%

An actual yield is the amount of product actually produced by the reaction. A theoretical yield is the amount of product that you calculate from the reaction equation according to the product and reactant coefficients

a.

Reaction

Fe₂O₃+3CO⇒2Fe+3CO₂

MW Fe₂O₃ :  159.69 g/mol

mol Fe₂O₃

\tt \dfrac{100}{159,69}=0.626

mol Fe₂O₃ : mol Fe = 1 : 2

mol Fe :

\tt \dfrac{2}{1}\times 0.626=1.252

mass of Fe(Ar=55.845 g/mol) :

\tt 1.252\times 55.845=69.92~g

b.

actual yield = 65 g

theoretical yield = 69.92 g

percent yield :

\tt =\dfrac{65}{69.92}=0.93=93\%

8 0
3 years ago
4. A student started with a 0.032 g sample of copper which he took through the series of reactions described in this experiment.
Elodia [21]

Answer:

Y=48.6\%

Explanation:

Hello,

In this case, we can consider the following chemical reaction for the oxidation of copper which only occurs at high temperatures:

2Cu+O_2\rightarrow 2CuO

In such a way, for 0.032 grams of copper, the following grams of copper (II) oxide (black product) are yielded:

m_{CuO}=0.032gCu*\frac{1molCu}{63.546gCu} *\frac{2molCuO}{2molCu}*\frac{79.546gCuO}{1molCuO}  =0.078gCuO

Therefore, the percent yield is:

Y=\frac{0.038g}{0.078g}*100\%\\ \\Y=48.6\%

Best regards.

6 0
3 years ago
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