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Fynjy0 [20]
3 years ago
9

An

Chemistry
1 answer:
stepan [7]3 years ago
5 0

Answer: atom

Explanation:

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PLEASE HELP!!!
algol [13]
Question 1 answer: A

Question 2 answer: H

Question 3 answer: J

Question 4 answer: T
8 0
3 years ago
How many hydrogen molecules are there in 1 ton of hydrogen?​
zaharov [31]
Hydrogen gas(H2) has a molar mass of 2 g. Molar mass of a substance is defined as the mass of 1 mole of that substance. And by 1 mole it is meant a collection of 6.022*10^23 particles of that substance.

So number of moles of H2 are 0.5 in this case. And thus it means there are (6.022*10^23)*0.5 particles( here they are molecules) in 1g of H2.
6 0
3 years ago
How many moles of gas occupy 98L at a pressure of 2.8atm and a temperature of 292k? and what law would you use?
algol13
Let's assume that the gas has ideal gas behavior.

Then we can use ideal gas equation,
PV = nRT

Where, P is Pressure of the gas (Pa), V is volume of the gas (m³), n is the number of moles of gas (mol), R is the Universal gas constant (8.314 J mol⁻¹ K⁻¹) and T is the temperature in Kelvin (K)

The given data for the gas is,
P = 2.8 atm = 283710 Pa
V = 98 L = 98 x 10⁻³ m³
T = 292 K
R = 8.314 J mol⁻¹ K⁻¹
n = ?

By applying the formula,
283710 Pa x 98 x 10⁻³ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 292 K
                                       n = 11.45 mol

Hence,moles of gas is 11.45 mol.
8 0
3 years ago
What was the original pressure in a rigid container if the new pressure is 82.5atm and the temperature was raised from 44C to 67
zhenek [66]

Answer: 54 atm

Explanation:

I did 67/82.5 then got 0.8121212121. I them divided 44 by 0.81212121 and got 54.1791044776

4 0
3 years ago
Argon has a pressure of 34.6 atm. It is transferred to a new tank with a volume of 456 L and pressure of 2.94 atm. What was the
NemiM [27]

Answer:

38.75 L

Explanation:

From the question,

Applying Boyles Law,

PV = P'V'....................... Equation 1

Where P = Original pressure of the Argon gas, V = Original Volume of Argon gas, P' = Final pressure of Argon gas, V' =  Final Volume of Argon gas.

make V the subject of the equation

V = P'V'/P.................... Equation 2

Given: P = 34.6 atm, V' = 456 L, P' = 2.94 atm.

Substitute these values into equation 2

V = (456×2.94)/34.6

V = 38.75 L

3 0
3 years ago
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