What is the percent yield of NaCl if 31g of CuCl2 reacts with excess NaNo3 to produce 21.2g of NaCl

A fish tank holds 29 gal of water. Using the density of 1.00 g/mL for water, determine the number of pounds of water in the fish

vova2212 [387]

1. Convert gallons to mL. 1 gal = 3785.4117840007 mL, multiply that by 29 and get 109776.94173602 mL.

2. Since there is one gram per every mL, there are 109776.94173602 g of water in the fish tank.

3. Convert g to pounds. 1 g = 0.0022 pounds. Multiply 109776.94173602 by 0.0022 and end up with about 241.5 pounds of water.

3 0

3 years ago

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Private land is generally used for _______. a. homes b. businesses c. factories d. all of the above Please select the best answe

alekssr [168]

HM, I think the answer would be D. This is just a guess, so please use it if ou want to answer D it's ok :D

5 0

3 years ago

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Use the equation to answer the question

vaieri [72.5K]

CH4 + 2O2→CO2 + 2H2O

Explanation:

CH4 + O2→CO2 + H2O

First look at the C atoms. At first glance they are balanced with 1 C on each side.

Now look at the H atoms. They are not balanced. There are 4 H atoms on the left side and 2 H atoms on the right. Place a coefficient of 2 in front of the H2O. We now have 4 H atoms on both sides.

CH4 + O2→CO2 + 2H2O

Now look at the O atoms. They are not balanced. There are 2 O atoms on the left side and 4 on the right. Place a coefficient of 2 in front of the O2. We now have 4 O atoms on both sides.

CH4 + 2O2→CO2 + 2H2O

The equation is now balanced. Each side has 1 C atom, 4 H atoms, and 4 O atoms.

6 0

3 years ago

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A nitrogen oxide, containing 53.85% N, acts as a vasodilator, lowering blood pressure in the human body. What is its empirical f

jenyasd209 [6]

The empirical formula of the nitrogen oxide that acts as a vasodilator and lowers human blood pressure is $N_4O_3$ . Empirical formula is the smallest whole number ratios of elements in a compound.

FURTHER EXPLANATION

To get the empirical formula from mass percent data, the following steps are followed:

1. Get the mass percent data for all the elements that make up the compound.

2. Assume that there is 100 grams of sample compound. Determine the equivalent masses of each element using the mass percent given.

3. Convert the mass of each element to number of moles.

4. Divide each calculated mole by the smallest mole value.

5. The quotient will be the subscript of the element in the compound. If a decimal is obtained, round off the answers to the nearest whole number. Some exceptions to this are the following decimals: x.25, x.33, x.50, and x.75. When these decimals are obtained, all the subscripts are multiplied by a number that will result in a whole number.

Applying the steps to the problem,

STEP 1: Get the mass percent.

Mass % of N = 53.85

Mass % of O = (100 - 53.85) = 46.15

STEP 2: Assume that 100 g sample is used and convert the mass % to mass in grams.

mass of N = 53.85% x 100 g = 53.85 g

mass of O= 46.15% x 100 g = 46.15 g

STEP 3: Convert mass to moles by dividing the given mass by the formula mass.

$moles \ of \ N \ = 53.85 \ g (\frac{1 \ mol}{14 \ g}) = 3.85 \\\\moles \ of \ O \ = 46.26 \ g \ (\frac{1 \ mol}{16 \ g}) \ = 2.89$

The moles may be written as the temporary subscripts of the elements in the compound as follows:

$N_{3.85}O_{2.89}$

STEP 4: Divide the moles by the smallest mole value.

$N_{3.85}O_{2.89}\\N_{ \frac{3.85}{2.89}} \ O _{ \frac{2.89}{2.89}} \\ N_{1.33}O_{1}\\$

<u>STEP 5:</u> Multiply the subscripts by a number that would give the smallest whole number ratio.

Since the decimal is 1.33 it cannot be rounded off to 1. It should be multiplied by 3 to get the smallest whole number.

NOTE: All subscripts must be multiplied by the same factor, 3.

$N_{1.33}O_{1}\\3(N_{1.33}O_{1})\\\boxed {N_{4}O_{3}}$

To check if the empirical formula is correct, calculate the mass % of each element based on the formula and compare with the given in the problem.

$mass \ \%\ = (\frac{mass \ of \ element}{mass \ of \ compound}) x 100\\\\mass \ \% \ of \ N = \ \frac{(4)(14)}{(4)(14) \ + \ (3)(16)}$

$mass \ \% \ of \ N \ = 53.85%\\mass \ \% \ of \ O \ = (\frac{(3)(16)}{(4)(14)+(3)(16)}) \ 100\\mass \ \% \ of \ O \ = 46.15%$

Since the values are the same from the given, the answer is correct.

LEARN MORE

Writing Chemical Formula brainly.com/question/4697698
Naming Compounds brainly.com/question/8968140
Mole Conversion brainly.com/question/12979299

Keywords: empirical formula, compounds

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3 years ago

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How many moles of water 9.31 x 10^22

san4es73 [151]

No calculator will show the answer

5 0

3 years ago