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lina2011 [118]
2 years ago
7

a liquid that occupies a volume of 5.0L gas a mass of 6.0 kg. what is the density of the liquid in kg/L?

Chemistry
1 answer:
AysviL [449]2 years ago
7 0

Answer:

1.2Kg/l

Explanation:

Density=Mass/Volume

Density=6.0/5.0

Density=1.2Kg/l

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Boron<br> B<br> 10.811<br> Number of neutrons
andrew-mc [135]

Answer:

6

Explanation:

To determine the number of neutrons we round 10.8 to 11 and subtract the atomic number (5) and get 6; therefore, boron has 6 neutrons.

7 0
3 years ago
The substance argon has the following properties: normal melting point: 83.9 K normal boiling point: 87.4 K triple point: 0.68 a
Vsevolod [243]

Answer:

The final state of the substance is a gas.

The sample is initially a liquid. One or more phase changes will occur.

Explanation:

Let's consider the phase diagram for Argon (not to scale).

<em>A sample of argon is initially at a pressure of 49.6 atm and a temperature of 101.4 K. The pressure on the sample is reduced to 0.680 atm at a constant temperature of 101.4 K. Which of the following are true? Choose all that apply </em>

<em>The final state of the substance is a gas.</em> TRUE. At 0.680 atm and 101.4 K, the substance is a gas.

<em>The gas initially present will solidify.</em> FALSE. Initially, Ar is present as a liquid.

<em>The final state of the substance is a solid.</em> FALSE.

<em>The sample is initially a liquid. One or more phase changes will occur.</em> TRUE. The sample is initially liquid and only one phase change will occur.

5 0
3 years ago
The answer to question 2
Bad White [126]

Answer:


12


Explanation:


You will need a chemical equation with masses and molar masses, so let’s gather all the information in one place.


M_{r}:                           258.21       18.02


                 KAl(SO₄)₂·xH₂O ⟶ KAl(SO₄)₂ + xH₂O


Mass/g:             4.74                                       2.16


Step 1. Calculate the mass of the KAl(SO₄)₂.


Mass = 4.74 g – 2.16 g = 2.58 g.


Step 2. Calculate the moles of each product.


\text{Moles of KAl(SO}_{4})_{2} = \text{2.58 g} \times \frac{\text{1 mol} }{\text{258.21 g}} = 9.992 \times 10^{-3} \text{ mol}

\text{Moles of H}_{2}\text{O} = \text{2.16 g} \times \frac{\text{1 mol} }{\text{18.02 g}} = \text{ 0.1200 mol}

Step 3. Calculate the molar ratio of the two products.


\frac{\text{Moles of KAl(SO}_{4})_{2}}{\text{Moles of H}_{2}\text{O}} = \frac{ 9.992 \times 10^{-3} \text{ mol}}{\text{ 0.1200 mol} } = \frac{1 }{12.01} \approx \frac{ 1}{ 12}

1 mol of KAl(SO₄)₂ combines with 12 mol H₂O, so x = 12.



3 0
3 years ago
This is the substance(s) formed in a chemical reaction.
Evgesh-ka [11]

Answer:A chemical reaction is the process in which atoms present in the starting substances rearrange to give new chemical combinations present in the substances formed by the reaction. These starting substances of a chemical reaction are called the reactants, and the new substances that result are called the products.

Explanation:

7 0
2 years ago
Read 2 more answers
G you have 57.0 ml of a 0.400 m stock solution that must be diluted to 0.100 m. assuming the volumes are additive, how much wate
Amanda [17]
Answer:
added water = 171 ml

Explanation:
Assuming volumes are additive, the rule that we will use to solve this question is:
M1V1 = M2V2
where:
M1 is the initial concentration = 0.4 m
V1 is the initial volume = 57 ml
M2 is the final concentration = 0.1 m
V2 is the final volume that we want to calculate
Substitute with the given in the above equation to get V2 as follows:
M1V1 = M2V2
(0.4)(57) = (0.1)V2
22.8 = 0.1V2
V2 = 228 ml

Now, the final volume is equal to the initial volume plus the amount of added water. So, to get the amount of added water, we will subtract the initial volume from the final volume as follows:
V2 = V1 + added water
228 = 57 + added water
added water = 228 - 57 = 171 ml

Hope this helps :)
3 0
3 years ago
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