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Nikitich [7]
3 years ago
9

GIVING BRAINLIEST!! PLEASE HELP

Chemistry
1 answer:
LenaWriter [7]3 years ago
8 0

Answer:

huh

Explanation:

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BRAINLIEST WILL BE GIVEN!! FAST PLEASE!
WARRIOR [948]

Answer:

1) THE AGE OF THE SAMPLE 2) URANIUM- LEAD DATING

Explanation:

3 0
3 years ago
What is the standard notation for 7.934 x 10-4 ?
maks197457 [2]

Answer:

It is

Explanation:

75.34 Im hopeing this correct.Very sorry if wrong.

4 0
3 years ago
If a mixture of gases contained 78% nitrogen at a pressure of 984 torr and 22% carbon dioxide at 345 torr, what is the total pre
AysviL [449]

Answer:

P(total) = 1329 torr

Explanation:

Given data:

Pressure of nitrogen = 984 torr

Pressure of carbon dioxide = 345 torr

Total pressure of system = ?

Solution:

The given problem will be solve through the Dalton law of partial pressure of gases.

According to the this law,

The total pressure exerted by the mixture of gases is equal to the sum of partial pressure of individual gas.

Mathematical expression,

P(total) = P₁ + P₂ +.......+ Pₙ

Here we will put the values in formula,

P₁ = partial pressure of nitrogen

P₂ = partial pressure of carbon dioxide

P(total) = 984 torr + 345 torr

P(total) = 1329 torr

8 0
3 years ago
What is the energy of a battery?
nadya68 [22]
A battery is a device that is able to store electrical<span> energy in the form of chemical energy, and convert that energy into </span>electricity

8 0
2 years ago
Read 2 more answers
The rate constant for a first order reaction
Lena [83]

Answer:

E_a = 103.626 × 10³ KJ/mol

Explanation:

Formula to solve this is given by;

Log(k2/k1) = (E_a/2.303R)((1/T1) - (1/T2))

Where;

k2 is rate constant at second temperature

k1 is rate constant at first temperature

R is universal gas constant

T1 is first temperature

T2 is second temperature

We are given;

k1 = 2.8 × 10^(-3) /s

k2 = 4.8 × 10^(-4) /s

R = 8.314 J/mol.k

T1 = 60°C = 333.15 K

T2 = 45°C = 318.15 K

Thus;

Log((4.8 × 10^(-4))/(2.8 × 10^(-3))) = (E_a/(2.303 × 8.314))((1/333.15) - (1/318.15))

We now have;

-0.76592 = -0.00000739121E_a

E_a = -0.76592/-0.00000739121

E_a = 103.626 × 10³ KJ/mol

8 0
3 years ago
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