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3 years ago

Anyone wannna be friends

Chemistry

2 answers:

grin007 [14]3 years ago

6 0

Answer: no

Explanation: I don’t know u

Nataliya [291]3 years ago

6 0

Answer:

Uh... I guess o_o

Explanation:

Mark me as a Brianalist first then sure

How old r u?

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6.33 x 1024 atoms of C are how many moles?

Readme [11.4K]

Answer:

Explanation:

just use avogardo's number which show that 1 mole has 6.022×10^23 atoms. so if you have 6.33 x 10^24 atoms of C, just divide that by 6.022x 10^23 and plug it into a calculator.

(6.33x10^24) / (6.022 x 10^23)

7 0

4 years ago

What volume would be occupied by 100. g of oxygen gas at a pressure of 1.50 atm and a temperature of

PSYCHO15rus [73]

Answer:

Use PV = nRT. Don't forget T must be in kelvin.

Explanation:

8 0

3 years ago

What is the mass percent when 144 grams of NaCl is dissolved in 255 grams of water?

Daniel [21]

Answer:

36.09% approximately

Explanation:

I'm not good at English so I don't know if I get it right

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3 years ago

Q13. Which of the following is a balanced chemical equation for the reaction between potassium carbonate soletion and hyelrochlo

klasskru [66]

Answer:

Below

Explanation:

2HCl + K2CO3 -> 2KCl + CO2 + H2O

4 0

3 years ago

The equilibrium constant, Kp, has a value of 6.5 × 104 at 308 K for the reaction of nitrogen monoxide with chlorine. 2NO(g) + C

Juliette [100K]

<u>Answer:</u> The $K_c$ for the given reaction is $2.67\times 10^{-2}$

<u>Explanation:</u>

For the given chemical equation:

$2NO(g)+Cl_2\rightleftharpoons 2NOCl(g)$

Relation of $K_p\text{ with }K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta n_g}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = $6.5\times 10^{-4}$

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 308 K

$\Delta n_g$ = change in number of moles of gas particles $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$6.5\times 10^{-4}=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{6.5\times 10^{-4}}{(0.0821\times 500)^{-1})}=2.67\times 10^{-2}$

Hence, the $K_c$ for the given reaction is $2.67\times 10^{-2}$

6 0

3 years ago