All categories

3 years ago

Calculate the ratio of moles of H2O to moles of anhydrous CuSO4 in CuSO4.

Chemistry

1 answer:

rjkz [21]3 years ago

7 0

Answer:

CuSO₄.H₂O = 1

CuSO₄.3H₂O = 3

CuSO₄.5H₂O = 5

CuSO₄7.H₂O = 7

CuSO₄.9H₂O = 9

Explanation:

Some salts as CuSO₄ are presented in the hydratated form to give some stability in their uses.

Ratio of moles represents moles of H₂O / moles of CuSO₄.

In CuSO₄.H₂O you have 1 mole of water per mole of CuSO₄, Ratio is 1/1 = 1.

For CuSO₄.3H₂O are 3 moles of water per mole of CuSO₄. Ratio is 3/1 = 3

For CuSO₄.5H₂O are 5 moles of water per mole of CuSO₄. Ratio is 5/1 = 5

For CuSO₄.7H₂O are 7 moles of water per mole of CuSO₄. Ratio is 7/1 = 7

For CuSO₄.9H₂O are 9 moles of water per mole of CuSO₄. Ratio is 9/1 = 9

You might be interested in

A chemist adds of a M zinc nitrate solution to a reaction flask. Calculate the mass in grams of zinc nitrate the chemist has add

weqwewe [10]

corrected question: A chemist adds 135mL of a 0.21M zinc nitrate solution to a reaction flask. Calculate the mass in grams of zinc nitrate the chemist has added to the flask. Round your answer to significant digits.

Answer:

5.37g

Explanation:

0.21M means ; 0.21mol/dm³

1dm³=1L , so we can say 0.21mol/L

if 0.21mol of Zinc nitrate is contained in 1L of water

x will be contained in 135mL of water

x= 0.21*135*10³/1

=0.02835moles

number of moles= mass/ molar mass

mass= number of moles *molar mas

molar mass of Zn(NO₃)₂=189.36 g/mol

mass= 0.02835 *189.36

mass=5.37g

3 0

3 years ago

A 12.0-liter sample of air at 25°C and 1.00 atm pressure was collected and dried. After drying, the volume of the sample was exa

zmey [24]

Answer: The mass percent of water in the original air sample is 4.16 %

Explanation:

We are given:

Volume of moist air = 12.0 L

Volume of dry air = 11.50 L

Volume of water lost = (12.0 - 11.50) L = 0.50 L

Here, the percent by mass ratio will be equal to the percent by volume ratio as the number of moles is directly related to the volume.

To calculate the volume percentage of water in the air, we use the equation:

$\text{Volume percent of water}=\frac{\text{Volume of water}}{\text{Volume of air}}\times 100$

Putting values in above equation, we get:

$\text{Volume percent of water}=\frac{0.50L}{12.00L}\times 100\\\\\text{Volume percent of water}=4.16\%$

Hence, the mass percent of water in the original air sample is 4.16 %

8 0

3 years ago

When converting from grams to the must be used as a conversion factor

Anna35 [415]

Moles=grams/molar mass

4 0

3 years ago

Identify whether longhand notation or noble-gas notation was used in each case below. Potassium (K): 1s22s22p63s23p64s1

Vaselesa [24]

Answer: The given electronic configuration is long hand notation.

Explanation:

Long-hand notation of representing electronic configuration is defined as the arrangement of total number of electrons that are present in an element.

Noble-gas notation of representing electronic configuration is defined as the arrangement of valence electrons in the element. The core electrons are represented as the previous noble gas of the element that is considered.

The given electronic configuration of potassium (K): $1s^22s^22p^63s^23p^64s^1$

The above configuration has all the electrons that are contained in the nucleus of an element. Thus, this configuration is a long-hand notation.

3 0

3 years ago

Read 2 more answers

Calculate the volume of a sample of mercury with a density of 14.6 g/mL and a mass of 1.00 g.

Zinaida [17]

Answer:

hiiiiiiiijsiosoaplq

4 0

3 years ago