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Len [333]
3 years ago
15

What is the mass of H2O produced when 14.0 grams of H2 reacts completely with 2.0 grams of O2?

Chemistry
1 answer:
Vladimir79 [104]3 years ago
4 0
Balanced chemical equation:

2 H2 + 1 O2 = 2 H2O

4 g H2  -------> 32 g O2 -----------> 36 g H2O
   ↓                       ↓                             ↓
14.0 g ---------> 2.0 g O2 ----------> mass H2O ?

32 * mass H2O = 2.0 * 36

32 * mass H2O = 72

mass of H2O = 72 / 32

mass of H2O = 2.25 g

hope this helps!.


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(3.45 x 10^20 molecules of sulfur dioxide ) x \frac{1 mole sulfur dioxide}{6.022x10^23 molecules sulfur dioxide} = 5.73x10^-4 mole of sulfur dioxide
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Which one is it answer the question
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Explanation:

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The________ cation will precipitate after the_________ step. Then the _________cation will precipitate after the________ step.
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Answer:

f. Sn^4+

c. second

e. Al^3+

d. third

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This question comes from a quantitative analysis showing the flowchart of a common scheme for identifying cations.

Now, from the separation scheme, Let's assume that Sn⁴⁺ & Al³⁺ were given; Then, Yes, the separation will work.

However, there will be occurrence of precipitation after the 1st step1.

So,  the <u>Sn⁴⁺</u> cation will precipitate after the <u>second </u>step. Then the <u>Al³⁺</u> cation will precipitate after the <u>third</u> step.

8 0
2 years ago
What type of reaction is Pb(NO3)2 (aq) + 2KI (aq) →PbI2 + 2KNO3 (aq)
kifflom [539]

Answer:

Precipitation Reactions

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4 0
3 years ago
Current passes through a solution of sodium chloride. In 1.00 second, 2.68×1016Na+ ions arrive at the negative electrode and 3.9
EleoNora [17]

Answer : The current passing between the electrodes is, 1.056\times 10^{-2}A

Explanation :

First we have to calculate the charge of sodium ion.

q=ne

where,

q = charge of sodium ion

n = number of sodium ion = 2.68\times 10^{16}

e = charge on electron = 1.6\times 10^{-19}C

Now put all the given values in the above formula, we get:

q=(2.68\times 10^{16})\times (1.6\times 10^{-19}C)=4.288\times 10^{-3}C

Now we have to calculate the charge of chlorine ion.

q'=ne

where,

q' = charge of chlorine ion

n = number of chlorine ion = 3.92\times 10^{16}

e = charge on electron = 1.6\times 10^{-19}C

Now put all the given values in the above formula, we get:

q'=(3.92\times 10^{16})\times (1.6\times 10^{-19}C)=6.272\times 10^{-3}C

Now we have to calculate the current passing between the electrodes.

I=\frac{q}{t}+\frac{q'}{t}

I=\frac{4.288\times 10^{-3}}{1.00}+\frac{6.272\times 10^{-3}}{1.00}

I=1.056\times 10^{-2}A

Thus, the current passing between the electrodes is, 1.056\times 10^{-2}A

4 0
3 years ago
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