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2 years ago

The mass of single iridium atom is 3.19*10^-22 grams how many iridium atoms would be in 56.4 milligrams of iridium

Chemistry

2 answers:

g100num [7]2 years ago

8 0

Answer:162626363637272772.8766

Explanation:

13322/726513/272727-277373

earnstyle [38]2 years ago

3 0

First convert milligrams to grams by dividing it by 1000. This gets us 0.0564 grams of iridium. Since we know the ratio between one iridium atom and grams is 1 atom/3.19*10^22 grams, use that ratio as the conversion factor in dimensional analysis. 0.0564 grams * (1 atom/3.19*10^22 grams). Grams cancel out and we are left with 1.77*10^20 atoms.

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A gas has a volume of 25.0 mL when under a pressure of 525 mmHg. What is the new pressure when the volume has been increased to

Fudgin [204]

Answer:

152.26 mmHg

Explanation:

pv=p'v'

525× 25=p'×86.2

p'=525×25÷ 86.2

p'=152.26

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2 years ago

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mote1985 [20]

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2 years ago

Read 2 more answers

Is radium flammable?

rewona [7]

Answer:

Explanation:

Radium is silvery, lustrous, soft, intensely radioactive. It readily oxidizes on exposure to air, turning from almost pure white to black. Radium is luminescent, corrodes in water to form radium hydroxide. Although is the heaviest member of the alkaline-earth group it is the most volatile.

4 0

3 years ago

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

How many grams Kl are needed to prepare 100.mL of 0.55M solution?

Makovka662 [10]

Answer:

Explanation:

100mL = 0.1L

0.55 M = mol/0.1 L

mol = 0.055 mol

molar mass of KI = 165.998 g

0.055 * 165.998 = 9.13 g of KI

7 0

1 year ago