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3 years ago

10

Your window thermometer shows that the temperature outdoors is 87.3°f. how should you report this temperature to your friend in

israel (i.e., on the celsius scale) and to your si-enthusiastic physicist neighbor (i.e., on the kelvin scale)?

1 answer:

Lorico [155]3 years ago

5 0

30.72 Cel
303.872 Kelvin

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The concentrated sulfuric acid we use in the laboratory is 98.0% sulfuric acid by weight. Calculate the molality and molarity of

timama [110]

Answer : The molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

Solution : Given,

Density of solution = $1.83g/cm^3=1.83g/ml$

Molar mass of sulfuric acid (solute) = 98.079 g/mole

98.0 % sulfuric acid by mass means that 98.0 gram of sulfuric acid is present in 100 g of solution.

Mass of sulfuric acid (solute) = 98.0 g

Mass of solution = 100 g

Mass of solvent = Mass of solution - Mass of solute = 100 - 98.0 = 2 g

First we have to calculate the volume of solution.

$\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{100g}{1.83g/ml}=54.64ml$

Now we have to calculate the molarity of solution.

$Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution}}=\frac{98.0g\times 1000}{98.079g/mole\times 54.64ml}=18.29mole/L$

Now we have to calculate the molality of the solution.

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{98.0g\times 1000}{98.079g/mole\times 2g}=499.59mole/Kg$

Therefore, the molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

7 0

3 years ago

A chemist has to prepare 250.0 mL of a 0.300 M Na2SO4(aq) solution. What mass, in grams, of sodium sulfate (formula mass 142.05

Sedbober [7]

The mass of sodium sulphate, Na₂SO₄, required to prepare the solution is 10.65 g

<h3>How to determine the mole of sodium sulphate Na₂SO₄</h3>

Volume = 250 mL = 250 / 1000 = 0.25 L
Molarity = 0.3 M

Mole of Na₂SO₄ =?

Mole = Molarity x Volume

Mole of Na₂SO₄ = 0.3 × 0.25

Mole of Na₂SO₄ = 0.075 mole

<h3>How to determine the mass of sodium sulphate Na₂SO₄</h3>

Molar mass of Na₂SO₄ = 142.05 g/mol
Mole of Na₂SO₄ = 0.075 mole

Mass of Na₂SO₄ =?

Mass = mole × molar mass

Mass of Na₂SO₄ = 0.075 × 142.05

Mass of Na₂SO₄ = 10.65 g

Thus, 10.65 g of Na₂SO₄ is needed to prepare the solution.

Learn more about molarity:

brainly.com/question/15370276

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1 year ago

zloy xaker [14]

Answer:

1. They lack a nucleus

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5 0

2 years ago

Compare the densities of hydrogen, oxygen, and water.

riadik2000 [5.3K]

Answer:

Water is a compound made up of two elements—hydrogen and oxygen. ... They are both colorless, odorless gases , and they both readily react with other elements—making them "reactive" elements. But in many ways they are very different from each other. Hydrogen has the lowest density of all the elements.

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3 years ago

Manganese sulfate forms a pale pink hydrate with the formula MnSO 4 ⋅ n H 2 O ( s ) . If this hydrate is heated to a high enough

Mama L [17]

Answer:

Value of n in MnSO₄.nH₂O is one.

Explanation:

The n represents the number of moles of water attached to the formula unit manganese sulfate. These moles (n) can be determined by taking the ratio of the moles of anhydrous salt and the moles of water. The moles of water can be determined by taking the difference of final and initial mass of the salt. This difference is equal to the mass of the water, mathematically it can be represented as,

Mass of H₂O = initial mass of the salt (g) - final mass of the salt (g)

Mass of H₂O = 16.260 g - 14.527 g

Mass of H₂O = 1.733 g

moles of H₂O = (1.733 g) ÷ (18.015 g/mole)

moles of H₂O = 0.0962

For the moles of anhydrous salt:

moles of MnSO₄ = mass of MnSO₄ ÷ molar mass of MnSO₄

moles of MnSO₄ = 14.5277 ÷ 151.001

moles of MnSO₄= 0.0962

Now for n:

n = moles of water ÷ moles of MnSO₄

n = 0.0962 ÷ 0.0962

n = 1

The above calculations show that one mole of H₂O is attached to the one formula unit of MnSO₄

8 0

3 years ago