I need help with this question ASAP

pentagon [3]

<span>
It contains 6.02 mc001-1.jpg 1023 grams of sodium chloride.

It is the mass of 12 carbon atoms.

It contains 6.02 mc001-2.jpg 1023 particles of a given substance.</span>

5 0

3 years ago

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you just got home from a run on a hot Atlanta afternoon. you grab a 1.00-liter bottle of water and drink three-quarters of it in

nikitadnepr [17]

Answer:

41.67 mol

Explanation:

1 Litre of water = 1000g

Mole = mass / molar mass

Mass of 1 L of water = 1000 g

Molar mass of water (H2O) :

(H = 1, O = 16)

H2O = (1 * 2) + 16 = (2 + 16) = 18g/mol

Amount of water consumed = (3/4) of 1 litre

= (3/4) * 1000g

= 750g

Therefore mass of water consumed = 750g

Mole = 750g / 18g/mol

Mole of water consumed = 41.6666

= 41.67 mol

3 0

2 years ago

If 125 ml of o2 gas exerts a pressure of 1.0 atm inside a cylinder, what will the pressure be if the cylinder compressed the vol

Lelu [443]

<h2>Hello!</h2>

The answer is: The new pressure is 1.67 atm.

From the statement, we know that the temperature remains constant and the gas volume is changing, meaning that the new pressure will be different than the first pressure.

Since the temperature remains constant, we can calculate the new pressure using the Boyle's Law.

The Boyle's Law states that:

$P_{1}V_{1}=P_{2}V_{2}$

Where,

P is the pressure of the gas.

V is the volume of the gas.

Then, the given information is:

$V_{1}=125ml=0.125L\\P_{1}=1atm\\V_{2}=75ml=0.075L$

Remember, 1 L is equal to 1000 mL.

So,

$125mL*\frac{1L}{1000mL}=\frac{125mL*1L}{1000mL}=0.125L\\\\75mL*\frac{1L}{1000mL}=\frac{75mL*1L}{1000mL}=0.075L$

So, calculating the new volume, we have:

$P_{1}V_{1}=P_{2}V_{2}\\\\1atm*0.125L=P_{2}*0.075L\\\\P_{2}=\frac{1atm*0.125L}{0.075l}=1.67atm$

Hence, the new pressure is 1.67 atm.

Have a nice day!

7 0

2 years ago

I need help solving this chemistry question

Arturiano [62]

Answer:

I think the answer is 22.2

Explanation: What i DID was adding 12.0 + 10.0 and than that gave me 22 the I had added the to 0.200 and that how i got 22.2. Sorry if i got is wrong. :(

3 0

3 years ago

For the reaction: 3 H2(g) + N2(g) <--> 2 NH3(g), the concentrations at equilbrium were [H2] = 0.10 M, [N2] = 0.10 M, and [

masya89 [10]

The equilibrium constant, k of the reaction in which case, the concentrations of the given reactants and products are as indicated is; Choice A; K = 3.1 x 10⁵

<h3>What is the equilibrium constant , k of the reaction as described in the task content?</h3>

It follows from above that the concentrations of the reactants and products are as follows; [H2] = 0.10 M, [N2] = 0.10 M, and [NH3] = 5.6 M at equilibrium.

Hence, the equilibrium constant of the reaction in discuss is;

K = [5.6]²/[0.10]³[0.10]

k = 5.6² × 10⁴

k = 3.136 × 10⁵

K = 3.1 × 10⁵.