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4 years ago

The jet stream is one of the key winds that moves air masses in which direction does the jet stream in the United

Chemistry

2 answers:

Blababa [14]4 years ago

4 0

Answer:

west to east

Explanation:

Yuri [45]4 years ago

3 0

Answer: West to East

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How many cm3 are there in 0.25 dm3?

aniked [119]

250cm3 i think im not sure about it though

8 0

3 years ago

Which of these is a carbohydrate?

Paladinen [302]

Answer:

C6H12O6

Glucose

Explanation:

Glucose is a monocarbohydrate

carbohydrates always have only C, H, and O

brainliest plz

6 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

6 0

2 years ago

A solution of 100.0 mL of 0.200 M KOH is mixed with a solution of 200.0 mL of 0.150 M NiSO4. (a) Write the balanced chemical equ

NISA [10]

Answer:

a) 2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂

b) Ni(OH)₂

c) KOH

d) 0.927 g

e) K⁺=0.067 M, SO₄²⁻=0.1 M, Ni²⁺=0.067 M

Explanation:

a) The equation is:

2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂ (1)

b) The precipitate formed is Ni(OH)₂

c) The limiting reactant is:

$n_{KOH} = V*M = 100.0 \cdot 10^{-3} L*0.200 mol/L = 0.020 moles$

$n_{NiSO_{4}} = V*M = 200.0 \cdot 10^{-3} L*0.150 mol/L = 0.030 moles$

From equation (1) we have that 2 moles of KOH react with 1 mol of NiSO₄, so the number of moles of KOH is:

$n = \frac{2}{1}*0.030 moles = 0.060 moles$

Hence, the limiting reactant is KOH.

d) The mass of the precipitate formed is:

$n_{Ni(OH)_{2}} = \frac{1}{2}*n_{KOH} = \frac{1}{2}*0.020 moles = 0.010 moles$

$m = n*M = 0.010 moles*92.72 g/mol = 0.927 g$

e) The concentration of the SO₄²⁻, K⁺, and Ni²⁺ ions are:

$C_{K^{+}} = \frac{2*\frac{1}{2}*n_{KOH}}{V} = \frac{0.020 moles}{0.300 L} = 0.067 M$

$C_{SO_{4}^{2-}} = \frac{\frac{1}{2}*n_{KOH + (0.03 - 0.01)}}{V} = \frac{0.030 moles}{0.300 L} = 0.1 M$

$C_{Ni^{2+}} = \frac{0.020 moles}{0.300 L} = 0.067 M$

I hope it helps you!

5 0

4 years ago

Of the concentration units below, only ________ uses kg of solvent in its calculation.

oksano4ka [1.4K]

Answer:

E) molality

Explanation:

Molality -

Molarity of a substance , is the number of moles present in a Kg of solvent .

Hence , the formula for molality is given as follow -

m = n / s

m = molality

s = mass of solvent in Kg ,

n = moles of solute ,

Hence , from the given information of the question,

The concentration unit which have Kg of solvent , is molality.

6 0

3 years ago