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Marysya12 [62]
4 years ago
6

36.8 g of CuSO4(s) was added to water to prepare a 2.00 L solution. What is the

Chemistry
1 answer:
Yuki888 [10]4 years ago
5 0

Answer:

Option D. 0.115 M

Explanation:

The following data were obtained from the question:

Mass of CuSO4 = 36.8 g

Volume of solution = 2 L

Molar mass of CuSO4 = 159.62 g/mol

Molarity of CuSO4 =..?

Next, we shall determine the number of mole in 36.8 g of CuSO4.

This can be obtained as shown below:

Mass of CuSO4 = 36.8 g

Molar mass of CuSO4 = 159.62 g/mol

Mole of CuSO4 =.?

Mole = mass /Molar mass

Mole of CuSO4 = 36.8 / 159.62

Mole of CuSO4 = 0.23 mole

Finally, we shall determine the molarity of the CuSO4 solution as illustrated below:

Mole of CuSO4 = 0.23 mole

Volume of solution = 2 L

Molarity of CuSO4 =..?

Molarity = mole /Volume

Molarity of CuSO4 = 0.23 / 2

Molarity of CuSO4 = 0.115 M

Therefore, the molarity of the CuSO4 solution is 0.115 M.

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How many ml of 2.50 M NaOH solution are required to make a 525 mL of 0.150 M?
larisa [96]

Answer:

31.5 mL of a 2.50M NaOH solution

Explanation:

Molarity (M) is an unit of concentration defined as moles of solute (In this case, NaOH), per liter of solvent. That is:

Molarity = moles solute / Liter solvent

If you want to make 525mL (0.525L) of a 0.150M of NaOH, you need:

0.525L × (0.150mol / L) = <em>0.07875 moles of NaOH</em>

<em />

If you want to obtain these moles from a 2.50M NaOH solution:

0.07875mol NaOH × (1L / 2.50M) = 0.0315L = <em>31.5 mL of a 2.50M NaOH solution</em>

7 0
3 years ago
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4 0
3 years ago
Calculate the volume in ml of 0.200 M na2co3 needed to produce 2.00 g of caco3 there is an excess of cacl2
UkoKoshka [18]

Answer:

V=100mL

Explanation:

Hello.

In this case, since the chemical reaction is:

Na_2CO_3+CaCl_2\rightarrow  CaCO_3+2NaCl

We next compute the moles of sodium carbonate from the 2.00 grams of calcium carbonate via their 1:1 mole ratio in the chemical reaction:

n_{Na_2CO_3}=2.00gCaCO_3*\frac{1molCaCO_3}{100.09gCaCO_3}*\frac{1molNa_2CO_3}{1molCaCO_3}  \\\\n_{Na_2CO_3}=0.0200molNa_2CO_3

Thus, by knowing the molarity, we compute the volume:

M=\frac{n}{V}\\ \\V=\frac{n}{M}=\frac{0.0200mol}{0.200mol/L}\\  \\V=0.100L*\frac{1000mL}{1L}\\ \\V=100mL

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3 years ago
3. Two different compounds are formed by the elements carbon and oxygen. The first compound contains 42.9% by mass carbon and 57
mars1129 [50]

Answer:

The law of multiple proportions is the third postulate of Dalton's atomic theory. It states that the masses of one element which combine with a fixed mass of the second element are in a ratio of whole numbers.

Therefore, the masses of oxygen in the two compounds that combine with a fixed mass of carbon should be in a whole number ratio. In 100 grams of the first compound (100 is chosen to make calculations easier), there are 57.1 grams oxygen and 42.9 grams carbon. The mass of oxygen (O) per gram of carbon (C) is:

57.1 g O / 42.9 g C = 1.33 g O per g C

In the 100 grams of the second compound, there are 72.7 grams of oxygen (O) and 27.3 grams of carbon (C). The mass of oxygen per gram of carbon is:

72.7 g O / 27.3 g C = 2.66 g O per g C

Dividing the mass O per g C of the second (larger value) compound:

2.66 / 1.33 = 2

This means that the masses of oxygen that combine with carbon are in a 2:1 ratio. The whole-number ratio is consistent with the law of multiple proportions.

Explanation:

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3 years ago
Convert the following names to balanced formulas.
yulyashka [42]
You can reduce the charges and subscripts so it is PbS2
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