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3 years ago

How many moles of chlorine gas are in 1.2m^3 at room temperature and pressure?

Chemistry

1 answer:

marshall27 [118]3 years ago

5 0

Answer:

look it up lol

Explanation:

quizlet

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Is there a relationship between CO2 levels in average global temperatures

Vilka [71]

Answer:

Yes

Explanation:

A small part of the correspondence is due to the relationship between temperature and the solubility of carbon dioxide in the surface ocean, but the majority of the correspondence is consistent with a feedback between carbon dioxide and climate.

8 0

2 years ago

True or False: Particles that are moving faster have a higher temperature

allochka39001 [22]

Answer:

true

Explanation:

I'm not sure why cause I dont know how to explain but it's TRUE

4 0

3 years ago

Read 2 more answers

PORFA AYUDAA !!!

matrenka [14]

Explanation:

The problem here is to find the atomic number of each of the element given.

Sum the powers of the configuration.

a- 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹

Atomic number is = 2 + 2 + 6 + 2 + 6 + 1 = 19

b- 1s² 2s² 2p⁶ 3s² 3p⁴

Atomic number = 2 + 2 + 6 + 2 + 4 = 16

c- 1s¹

Atomic number = 1

6 0

2 years ago

What are the elements of PCl5 and the name

prisoha [69]

Phosphorus, Chlorine. The IUPAC name is Phosphorus pentachloride

8 0

3 years ago

One solution has a formula C (n) H (2n) O (n) If this material weighs 288 grams, dissolves in weight 90 grams, the solution will

saw5 [17]

Explanation:

The given data is as follows.

Boiling point of water ( $T^{o}_{b}) = 100^{o}C$ = (100 + 273) K = 323 K,

Boiling point of solution ( $T_{b}) = 101.24^{o}C$ = (101.24 + 273) K = 374.24 K

Hence, change in temperature will be calculated as follows.

$\Delta T_{b} = (T_{b} - T^{o}_{b})$

= 374.24 K - 323 K

= 1.24 K

As molality is defined as the moles of solute present in kg of solvent.

Molality = $\frac{\text{weight of solute \times 1000}}{\text{molar mass of solute \times mass f solvent(g)}}$

Let molar mass of the solute is x grams.

Therefore, Molality = $\frac{\text{weight of solute \times 1000}}{\text{molar mass of solute \times mass f solvent(g)}}$

m = $\frac{288 g \times 1000}{x g \times 90}$

= $\frac{3200}{x}$

As, $\Delta T_{b} = k_{b} \times molality$

$1.24 = 0.512 ^{o}C/m \times \frac{3200}{x}$

x = $\frac{0.512 ^{o}C/m \times 3200}{1.24}$

= 1321.29 g

This means that the molar mass of the given compound is 1321.29 g.

It is given that molecular formula is $C_{n}H_{2n}O_{n}$ .

As, its empirical formula is $CH_{2}O$ and mass is 30 g/mol. Hence, calculate the value of n as follows.

n = $\frac{\text{Molecular mass}}{\text{Empirical mass}}$

= $\frac{1321.29 g}{30 g/mol}$

= 44 mol

Thus, we can conclude that the formula of given material is $C_{44}H_{88}O_{44}$ .

4 0

2 years ago