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3 years ago

Using the element above answer the following question:

Chemistry

1 answer:

olga_2 [115]3 years ago

7 0

Answer:

20.180

Explanation:

Atomic mass is the sum of the masses of the protons, neutrons, and electrons in an atom, or the averagemass, in a group of atoms

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Melting can be best described as a process in which molecules

statuscvo [17]

Melting can be best described as a process in which molecules gain enough kinetic energy to be able to pass to each other.

3 0

3 years ago

12.70 L of a gas has a pressure of 0.63 atm. What is the volume of the gas at 105kPa?

Zielflug [23.3K]

Answer: 7.693 L

Explanation:

To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.

The equation given by this law is:

$P_1V_1=P_2V_2$

where,

$P_1\text{ and }V_1$ are initial pressure and volume.

$P_2\text{ and }V_2$ are final pressure and volume.

We are given:

$P_1=0.63atm\\V_1=12.70L\\P_2=105kPa=1.04atm(1kPa=0.009atm)\\V_2=?$

Putting values in above equation, we get:

$0.63\times 12.70mL=1.04\times V_2\\\\V_2=7.693L$

Thus new volume of the gas is 7.693 L

6 0

3 years ago

Read 2 more answers

Need help !!!!! ASAP

gavmur [86]

<h2>Hello!</h2>

The answer is:

The new temperature will be equal to 4 K.

$T_{2}=4K$

We are given the volume, the first temperature and the new volume after the gas is compressed. To calculate the new temperature after the gas was compressed, we need to use Charles's Law.

Charles's Law establishes a relationship between the volume and the temperature at a gas while its pressure is constant.

Now, to calculate the new temperature we need to assume that the pressure is kept constant, otherwise, the problem would not have a solution.

From Charle's Law, we have:

$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$

So, we are given the following information:

$V_{1}=500mL\\T_{1}=20K\\V_{2}=100mL$

Then, isolating the new temperature and substituting the given information, we have:

$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$

$T_{2}=\frac{T_{1}}{V_{1}}*V_{2} \\$

$T_{2}=\frac{20.00K}{500mL}*100mL\\$

$T_{2}=4K$

Hence, the new temperature will be equal to 4 K.

$T_{2}=4K$

Have a nice day!

7 0

3 years ago

Gaseous ICl (0.20 mol) was added to a 2.0 L flask and allowed to decompose at a high temperature:

Ne4ueva [31]

Answer:

The Kc is 1.36 (but this is not an option, may be the options are wrong, or may be I was .. Thanks!)

Explanation:

Let's think all the situation.

2 ICl(g) ⇄ I₂(g) + Cl₂(g)

Initially 0.20 - -

Initially I have only 0.20 moles of reactant, and nothing of products. In the reaction, an x amount of compound has reacted.

React x x/2 x/2

Because the ratio is 2:1, in the reaction I have the half of moles.

So in equilibrium I will have

(0.20 - x) x/2 x/2

Notice that I have the concentration in equilibrium so:

0.20 - x = 0.060

x = 0.14

So in equilibrium I have formed 0.14/2 moles of I₂ and H₂ (0.07 moles)

Finally, we have to make, the expression for Kc and remember that must to be with concentration in M (mol/L).

As we have a volume of 2L, the values must be /2

Kc = ([I₂]/2 . [H₂]/2) / ([ICl]/2)²

Kc = (0.07/2 . 0.07/2) / (0.060/2)²

Kc = 1.225x10⁻³ / 9x10⁻⁴

Kc = 1.36

8 0

3 years ago

What mass of HCL, in grams, is required to react with 0.610 g of al(oh)3 ?

kompoz [17]

Answer: 0.8541 grams of HCl will be required.

Explanation: Moles can be calculated by using the formula:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of $Al(OH)_3$ = 0.610 g

Molar mass of $Al(OH)_3$ = 78 g/mol

$\text{Number of moles}=\frac{0.610g}{78g/mol}$

Number of moles of $Al(OH)_3$ = 0.0078 moles

The reaction between $Al(OH)_3$ and HCl is a type of neutralization reaction because here acid and base are reacting to form an salt and also releases water.

Chemical equation for the above reaction follows:

$Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O$