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Llana [10]
3 years ago
6

The number of electrons in 1.6 gram CH, is:​

Chemistry
2 answers:
Alla [95]3 years ago
8 0

Answer:

6.022 x 10²³ electrons

Explanation:

Wewaii [24]3 years ago
8 0
Oh gosh oh I thought had the wrong
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How much mass do 3.4 moles of silver contain?
tensa zangetsu [6.8K]

367.2 g of silver

Explanation:

To find the mass of a substance knowing the number of moles we use the following formula:

number of mole = mass / molecular weight

In the case of silver we use the atomic weight of 108 g/mole.

mass = number of moles × molecular weight

mass of silver = 3.4 moles × 108 g/mole

mass of silver = 367.2 g

Learn more about:

moles

brainly.com/question/2293005

#learnwithBrainly

5 0
3 years ago
Predict how many grams of KCI is produced from 40 grams of K?
Nutka1998 [239]

Explanation:

firstly find for the molar mass of kcl and molar mass of k

and then

molar mass of k = x

molar mass of kcl= 40

cross mutiply and then simplify you will get your answer

5 0
3 years ago
A sample of nitrogen occupies 10.0 liters at 25°C what would be the new volume at 20°C? 7.9 L Ob 9.8 L 10.2 L 10.6 L
PtichkaEL [24]

Answer:

  9.4 liter

Explanation:

1) Data:

V₁ = 10.0 L

T₁ = 25°C = 25 + 273.15 K = 298.15 K

 P₁ = 98.7 Kpa

 T₂ = 20°C = 20 + 273.15 K = 293.15 K

  P₂ = 102.7 KPa

  V₂ = ?

2) Formula:

Used combined law of gases:

  PV / T = constant

  P₁V₁ / T₁ = P₂V₂ / T₂

3) Solution:

Solve the equation for V₂:

  V₂ = P₁V₁ T₂ / (P₂ T₁)

Substitute and compuite:

V₂ = P₁V₁ T₂ / (P₂ T₁)

V₂ = 98.7 KPa × 10.0 L × 293.15 K / (102.7 KPa × 298.15 K)

V₂ =  9.4 liter ← answer

You can learn more about gas law problems reading this other answer on

Explanation:

7 0
2 years ago
Which of the following best explains why solids do not change
Lesechka [4]

Answer: The particles in a solid are packed very close to each other.

8 0
3 years ago
How many liters of space will 7.80 moles of methane gas (CH4) occupy at STP
Alchen [17]

Answer:

V CH4(g) = 190.6 L

Explanation:

assuming ideal gas:

  • PV = RTn

∴ STP: T =298 K and P = 1 atm

∴ R = 0.082 atm.L/K.mol

∴ moles (n) = 7.80 mol CH4(g)

∴ Volume CH4(g) = ?

⇒ V = RTn/P

⇒ V CH4(g) = ((0.082 atm.L/K.mol)×(298 K)×(7.80 mol)) / (1 atm)

⇒ V CH4(g) = 190.6 L

4 0
3 years ago
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