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3 years ago

Which of the following is not the same as 0.032 liters?

1 answer:

3 0

This question needs the answer choices.

I found these choices for you:
0.00032hL
320cL
32mL

Then you need to make the conversion of 0.032 liters to hectoliters, centiliters and milimilters to check which is not equivalent.

1) 0.032 liters to hectoliters:

 0.032 liter * 1 hectoliter / 100 liter = 0.00032 hecoliter

2) 0.032 liter to centiliters:

 0.032 liter * 100 centiliters / 1 liter = 3.2 centiliter

3) 0.032 liter to mililiter:

 0.032 liter * 1000 mililiter / liter = 32 mililiter

Then, the answer is 320 cL: 320 cL is not the same as 0.032 liters

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How to solve for enthalpy for changes

Natasha_Volkova [10]

Always use this method !!! always

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3 years ago

Consider the reaction: N2(g) 2 O2(g)N2O4(g) Write the equilibrium constant for this reaction in terms of the equilibrium constan

Valentin [98]

Answer : The equilibrium constant for this reaction is, $K=\frac{(K_b)^2}{K_a}$

Explanation :

The given main chemical reaction is:

$N_2(g)+2O_2(g)\rightarrow N_2O_4(g)$ ; $K$

The intermediate reactions are:

(1) $N_2O_4(g)\rightarrow 2NO_2(g)$ ; $K_a$

(2) $\frac{1}{2}N_2(g)+O_2(g)\rightarrow NO_2(g)$ ; $K_b$

We are reversing reaction 1 and multiplying reaction 2 by 2 and then adding both reaction, we get:

(1) $2NO_2(g)\rightarrow N_2O_4(g)$ ; $\frac{1}{K_a}$

(2) $N_2(g)+2O_2(g)\rightarrow 2NO_2(g)$ ; $(K_b)^2$

Thus, the equilibrium constant for this reaction will be:

$K=\frac{1}{K_a}\times (K_b)^2$

$K=\frac{(K_b)^2}{K_a}$

Thus, the equilibrium constant for this reaction is, $K=\frac{(K_b)^2}{K_a}$

5 0

3 years ago

Please anyone answer this question

Ainat [17]

Answer:

A, 18.4 and 34!

Explanation:

The Burette seems to have ~18.4 of the liquid? substance in it, and the Measuring Cylinder seems to have ~34 of the liquid? substance in it. So, it is A! :]

Hope this helps!!

7 0

3 years ago

Number of grams of hydrogen than can be prepared from 6.80g of aluminum

Anastaziya [24]

Answer:

0.7561 g.

Explanation:

The hydrogen than can be prepared from Al according to the balanced equation:

2Al + 6HCl → 2AlCl₃ + 3H₂,

It is clear that 2.0 moles of Al react with 6.0 mole of HCl to produce 2.0 moles of AlCl₃ and 3.0 mole of H₂.

Firstly, we need to calculate the no. of moles of (6.8 g) of Al:

no. of moles of Al = mass/atomic mass = (6.8 g)/(26.98 g/mol) = 0.252 mol.

Using cross multiplication:

2.0 mol of Al produce → 3.0 mol of H₂, from stichiometry.

0.252 mol of Al need to react → ??? mol of H₂.

∴ the no. of moles of H₂ that can be prepared from 6.80 g of aluminum = (3.0 mol)(0.252 mol)/(2.0 mol) = 0.3781 mol.

Now, we can get the mass of H₂ that can be prepared from 6.80 g of aluminum:

mass of H₂ = (no. of moles)(molar mass) = (0.3781 mol)(2.0 g/mol) = 0.7561 g.

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3 years ago

Vinil7 [7]

Answer:

c. a question or problem statement

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3 years ago