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3 years ago

How many calcium atoms in Ca(ClO3)2?

Chemistry

1 answer:

Serjik [45]3 years ago

8 0

Answer:

Explanation:

There is only one calcium atom because the subscript 3 applies only to the oxygen. Outside of the parentheses, the subscript 2 only applies to the chlorate ion. Therefore, there is only one calcium atom because there are no coefficients and subscripts. (Also drawing it out will help)

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You have a gas at a volume of 9.71 L at a pressure of 209 torr and at 10.1 °C. What

Usimov [2.4K]

Answer:

219.95 °C

Explanation:

Given data:

Volume of gas = 9.71 L

Initial pressure = 209 torr (209/760 = 0.275 atm)

Initial temperature = 10.1 °C (10.1 +273 = 283.1 K)

Final temperature = ?

Final pressure = 364 torr (364/760 =0.479 atm)

Solution:

According to Gay-Lussac Law,

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

Now we will put the values in formula:

0.275 atm / 283.1 K = 0.479 atm/T₂

T₂ = 0.479 atm × 283.1 K/ 0.275 atm

T₂ = 135.6 atm. K /0.275 atm

T₂ = 493.1 K

Kelvin to °C:

493.1 K - 273.15 = 219.95 °C

8 0

3 years ago

Which list ranks the gasses in the correct order from the

JulijaS [17]

Answer:

D. chlorine, oxygen, nitrogen, hydrogen.

Explanation:

Thomas Graham found that, at a constant temperature and pressure the rates of effusion of various gases are inversely proportional to the square root of their masses.

where ν is the rate of effusion and M is the atomic or molecular mass of the gas particles.

The molecular mass for the listed gases are:

O₂: 32.0 g/mol,

Cl₂: 70.906 g/mol,

N₂: 28.0 g/mol,

H₂: 2.0 g/mol.

Hence, the smallest molecular mass of the gas, the fastest rate of effusion.

So, the order from the slowest to the fastest rate of effusion is:

<em>Chlorine, oxygen, nitrogen, hydrogen.</em>

6 0

4 years ago

Read 2 more answers

An oxide ion, O2–, has:

sergij07 [2.7K]

Answer:

the answer A- 8 protons and 10 electrons

6 0

3 years ago

Read 2 more answers

A vessel of 120ml capacity contains a certain amount of gas at 35°C and 1.2 bar pressure. The gas is transferred to another vess

Rainbow [258]

Given parameters:

Initial volume = 120ml

Initial temperature = 35°C

Initial pressure = 1.2bar

Final volume = 180ml

Final temperature = 35°C

Unknown:

Final pressure = ?

To solve this problem, we apply the combined gas law. The expression is given below;

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

Where P₁ is the initial pressure

P₂ is the final pressure

V₁ is the initial volume

V₂ is the final volume

T₁ is the initial temperature

T₂ is the final temperature

We need to convert the parameters to standard units

take the volume to dm³;

1000ml = 1dm³

120ml = $\frac{120}{1000}$ dm³ = 0.12dm³ = initial volume

Final volume;

1000ml = 1dm³

180ml = $\frac{180}{1000}$ dm³ = 0.18dm³

Now, the temperature;

K = 273 + °C

Initial temperature = 273 + 35 = 308k

Final temperature = 308k

We then input the parameters into the equation;

$\frac{1.2bar x 0.12 }{308} = \frac{P_{2} x 0.18 }{308}$

Solving for P₂;

P₂ = 0.8bar

The new pressure or final pressure in the vessel is 0.8bar

5 0

3 years ago

Substances like krypton, which is a gas at room temperature and pressure, can often be liquified or solidified only at very low

inysia [295]

Answer:

The boiling and freezing temperatures of krypton at absolute scale are 119.95 K and 116.05 K, respectively.

Explanation:

The absolute temperature on SI units corresponds to Kelvin scale, whose conversion formula in terms of the Celsius scale is:

$T_{K} = T_{C} + 273.15$

Where:

$T_{K}$ - Absolute temperature, measured in Kelvins.

$T_{C}$ - Relative temperature, measured in Celsius.

Finally, freezing and boiling temperatures are converted into absolute scale:

Boiling temperature

$T_{K} = (-153.2 + 273.15)\,K$

$T_{K} = 119.95\,K$

Freezing temperature

$T_{K} = (-157.1 + 273.15)\,K$

$T_{K} = 116.05\,K$

The boiling and freezing temperatures of krypton at absolute scale are 119.95 K and 116.05 K, respectively.

8 0

3 years ago