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Butoxors [25]
2 years ago
9

What is the volume of 4.78g of O2 gas at STP?

Chemistry
2 answers:
My name is Ann [436]2 years ago
8 0

Answer:

Explanation:

The trick here is to realize that if you know the volume of a gas at STP, you can use the fact that

1

mole of any ideal gas occupies

22.7 L

under STP conditions to calculate how many moles of gas you have in your sample.

Under STP conditions:

1 mole of an ideal gas = 22.7 L

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

In your case, you know that your sample of gas occupies

2.28 L

under STP conditions, which are currently defined as a pressure of

100 kPa

and a temperature of

0

∘

C

.

This means that your sample will contain

2.28

L

⋅

molar volume of a gas at STP



1 mole gas

22.7

L

=

0.10044 moles gas

Now, the molar mass of the gas is the mass of exactly

1

mole of the gas. In your case, you know that you get

3.78 g

for every

0.10044

moles, which means that you have

1

mole

⋅

3.78 g

0.10044

moles

=

37.6 g

Since this is the mass of

1

mole of gas, you can say that the molar mass of the gas is

molar mass = 37.6 g mol

−

1

−−−−−−−−−−−−−−−−−−−−−−−

Nonamiya [84]2 years ago
6 0
I’m not surreeeeeeeeeee
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Lady_Fox [76]

Answer:

1,300,000,000,000

Explanation:

1.3 x 10^12

We want to convert this from scientific notation.

Tip: in scientific notation the exponent tells you how many place you move the decimal point over. If the exponent is negative you move the decimal point to the left. Ex. For, 4.1 x 10^-8, we would move the decimal point over 8 times to the left to get .00000041. When the exponent is positive we move over to the right. Ex. For, 7.6 x 10^7 we would move the decimal point over 7 times to the right to get 76,000,000

So to convert 1.3 x 10^12 we simply move over the decimal point over 12 times to the right.

1.3 x 10^12 ------> 1,300,000,000,000

Our answer is 1,300,000,000,000

6 0
2 years ago
A particular first-order reaction has a rate constant of 1.35 × 102 s-1 at 25.0°C. What is the magnitude of k at 95.0°C if Ea =
never [62]

Answer:

k ≈ 9,56x10³ s⁻¹

Explanation:

It is possible to solve this question using Arrhenius formula:

ln\frac{k2}{k1} = \frac{-Ea}{R} (\frac{1}{T2} -\frac{1}{T1} )

Where:

k1: 1,35x10² s⁻¹

T1: 25,0°C + 273,15 = 298,15K

Ea = 55,5 kJ/mol

R = 8,314472x10⁻³ kJ/molK

k2 : ???

T2: 95,0°C+ 273,15K = 368,15K

Solving:

ln\frac{k2}{k1} = 4,257

\frac{k2}{k1} = 70,593

{k2} = 9,53x10^3 s^{-1}

<em>k ≈ 9,56x10³ s⁻¹</em>

I hope it helps!

5 0
2 years ago
How are stars and comets different?
ehidna [41]
The answer is letter C.

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3 0
2 years ago
If 5,800 j of energy are applied to a 15.2 kg piece of lead, by how much does the temp change if the specific heat of lead is 0.
scoray [572]

Answer:

The answer will be 2.98K

Explanation:

Using the formula:

Q = mc∆T

Q= 5,800 (heat in joules)

m= convert 15.2kg to g which is 15200g (mass in grams)

c= 0.128 J/g °c (Specific heat capacity)

∆T=  what we need to find (temperature change)

5800J = 15200g x 0.128 x ∆T

= 2.98K  

7 0
3 years ago
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sergey [27]

Answer:

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Explanation:

8 0
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