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vitfil [10]
3 years ago
10

If 1,000 mL = 1 L, which of the following are possible conversion factors for liters and milliliters? Check all that apply.

Chemistry
2 answers:
eimsori [14]3 years ago
8 0

Answer:

the second questions answers are A=1,000 B=1 C=5,400

Fudgin [204]3 years ago
7 0

Answer:

Two conversion factors:

         1=\dfrac{1L}{1,000ml}\\ \\ \\ 1=\dfrac{1,000mL}{1L}

Explanation:

You can create two possible <em>conversion factors</em>, one to convert from mL to L, and one to convert from L to mL

<u />

<u>a) From mL to L</u>

To convert mL to L you need to multiply by a conversion factor that has mL on the denominator and L in the numerator.

Your starting point is: 1,000mL=1L

Then, divide both sides by 1,000mL (this will be on the denominator of the fraction);

       1,000mL=1L\\\\ \\\dfrac{1,000ml}{1,000mL}=\dfrac{1L}{1,000mL}\\ \\ \\ 1=\dfrac{1L}{1,000mL}

<u>b) From L to mL</u>

Divide both sides by 1 L:

            1,000mL=1L\\\\ \\\dfrac{1,000ml}{1L}=\dfrac{1L}{1L}\\ \\ \\ 1=\dfrac{1,000mL}{1L}

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6 0
3 years ago
How many liters of CH, gas are in 5.40 moles of CH, gas at STP?<br> (3 sig figs)
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Answer:

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"Heat of combustion"

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Cu(s) + 4 HNO3 (aq) --&gt; Cu(NO3)2 (aq) + 2NO2 (g) + 2H2O(l)
GREYUIT [131]

Answer:

The percent by mass of copper in the mixture was 32%

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The ammount of HNO₃ used is:

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mol HNO₃ = 0.015 l * 15.8 mol/l

mol HNO₃ = 0.237 mol

According to the reaction, 4 mol HNO₃ will react with 1 mol Cu and produce 1 mol Cu²⁺. Since we have 0.237 mol HNO₃, the amount of Cu that could react would be (0.237 mol HNO₃ * 1 mol Cu / 4 mol HNO₃) 0.06 mol. This reaction would produce 0.060 mol Cu²⁺, however, only 0.010 mol Cu²⁺ were obtained, indicating that only 0.010 mol Cu were present in the mixture. This means that the acid was in excess, so we can assume that all copper present in the mixture has reacted.

Since 0.010 mol of Cu²⁺ were produced, the amount of Cu was 0.01 mol.

1 mol of Cu has a mass of 63.5 g, then 0.01 mol has a mass of:

0.01 mol Cu * 63.5 g / 1 mol = 0.635 g.

Since this amount was present in 2.00 g mixture, the amount of copper in 100 g of the mixture will be:

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Explanation:

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