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pav-90 [236]
3 years ago
15

“True or False”

Chemistry
1 answer:
Sliva [168]3 years ago
3 0
True, I’m sorry if I’m wrong
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A scientist has two samples of gas: The first sample contains one mole of argon atoms and has a mass of 39.948 g; the second sam
V125BC [204]

There are 6.022 × 10²³ atoms in 39.948 g of argon and 4.0026 g of helium.

Explanation:

39.945 g/mole is the molar mass of argon so 39.948 g of argon are equal to 1 mole of argon.

4.0026 g/mole is the molar mass of helium so 4.0026 g of helium are equal to 1 mole of helium.

We know that Avogadro's number tell us the number of particles in 1 mole of substance which is 6.022 × 10²³.

So in 39.948 g of argon and 4.0026 g of helium contains the same number of atoms, 6.022 × 10²³.

Learn more about:

Avogadro's number

brainly.com/question/14148121

brainly.com/question/1445383

brainly.com/question/1528951

#learnwithBrainly

4 0
3 years ago
Exactly 5000 mL of air at 223K is warmed and has a new volume of 8.36 liters. What is the new temperature?
34kurt

Answer:

The new temperature is 373 K

Explanation:

Step 1: Data given

Volume air = 5000 mL = 5.0 L

Temperature = 223K

New volume = 8.36 L

Step 2: Calculate the new temperature

V1/T1 = V2/T2

⇒V1 = the initial volume = 5.0 L

⇒T1 = the initial temperature = 223 K

⇒V2 = the new volume = 8.36 L

⇒T2 = the new temperature

5.0/223 = 8.36 /T2

T2 = 373 K

The new temperature is 373 K

7 0
3 years ago
Help me answer this please
Vlada [557]
<span>1=H, 2=B, 3=F, 4=A,5=C,6=E, 7=D, 8=G
</span>9: 69Ga=60.12% and 71Ga=39.88%; total=69.797amu
10: 27  27.977  92.23; 28  28.976  4.67; 29  29.974  3.10; abundance =28.07 Silicon

I hope this helps!
8 0
3 years ago
URGENT!! I will have brainliest!!!
Vlad [161]

Answer:

sorry idk

Explanation:

3 0
3 years ago
How many grams of silver chromate, Ag2CrO4, are produced from 57.7
Anastasy [175]
48.3 g AgNO3 / 169.9 g/mol = 0.284 moles AgNO3
0.284 mol AgNO3 X (1 mol Ag2CrO4/2 mol AgNO3) = 0.142 mol Ag2CrO4
0.142 mol Ag2CrO4 X 331.7 g/mol = 47.1 g Ag2CrO4
5 0
3 years ago
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