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2 years ago

To change state from gas to a liquid

Chemistry

1 answer:

Nostrana [21]2 years ago

4 0

I believe the answer is condensation, correct me if I'm wrong

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What is the density of a block of gold that occupies 1000 ml and has a mass of 3.5 kg? Show your work

tiny-mole [99]

Answer:

Density of block of gold is <em>3.5 g/cm³.</em>

Explanation:

Given data:

Volume of block = 1000 mL

Mass of block = 3.5 kg (3.5×1000 = 3500 g)

Density of block = ?

Solution:

Density of substance is calculated by dividing the mass of substance over its volume.

Formula:

d = m/v

d = 3500 g/ 1000 mL

d = 3.5 g/mL

d = 3.5 g/cm³ (1ml = 1cm³)

5 0

3 years ago

Calculate how many moles of carbon dioxide are contained in 5.57 x 10²¹ molecules of CO₂

Alex777 [14]

<h2>Answer: 0.00925 mol </h2>

Explanation:

Moles = # of molecules ÷ Avogadro's Number

= (5.57 × 10²¹ molecules) ÷ (6.022 × 10²³ molecules/mol)

= 0.00925 mol

<h3>∴ the 5.57 x 10²¹ molecules of carbon dioxide ≡ 0.00925 mol</h3>

3 0

2 years ago

What does the p mean in 1s22s22p63s??

larisa86 [58]

Answer:

subshel

or orbital

7 0

2 years ago

Question 1<br> 1 pts<br> How many mols of bromine are present in 35.7g of<br> Tin(IV) bromate?

sleet_krkn [62]

Answer:

n = 0.0814 mol

Explanation:

Given mass, m = 35.7g

The molar mass of Tin(IV) bromate, M = 438.33 g/mol

We need to find the number of moles of bromine. We know that,

No. of moles = given mass/molar mass

So,

$n=\dfrac{35.7}{438.33}\\\\n=0.0814\ mol$

So, there are 0.0814 moles of bromine in 35.7g of Tin(IV) bromate.

3 0

2 years ago

If you combine 27.1 g of a solute that has a molar mass of 27.1 g/mol with 100.0 g of a solvent, what is the molality of the res

baherus [9]

<u>Answer:</u> The molality of the solution is 0.1 m.

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute = 27.1 g

$M_{solute}$ = Molar mass of solute = 27.1 g/mol

$W_{solvent}$ = Mass of solvent = 100 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{27.1\times 1000}{27.1\times 100}\\\\\text{Molality of solution}=0.1m$

Hence, the molality of the solution is 0.1 m.

8 0

3 years ago

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