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Kruka [31]
2 years ago
12

A liquid element whose molecules consists of one atom​

Chemistry
2 answers:
kifflom [539]2 years ago
8 0

i think argon is the answer

AlexFokin [52]2 years ago
8 0
......................
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If nitrogen-13 has a half life of 2.5 years, how much remains from a 100g sample after 7.5 years
Sonbull [250]

Answer:

12.50g

Explanation:

T½ = 2.5years

No = 100g

N = ?

Time (T) = 7.5 years

To solve this question, we'll have to find the disintegration constant λ first

T½ = In2 / λ

T½ = 0.693 / λ

λ = 0.693 / 2.5

λ = 0.2772

In(N/No) = -λt

N = No* e^-λt

N = 100 * e^-(0.2772*7.5)

N = 100*e^-2.079

N = 100 * 0.125

N = 12.50g

The sample remaining after 7.5 years is 12.50g

5 0
3 years ago
Why can some metals be extracted from compounds by heating with carbon and why can some not
pentagon [3]

Answer:

Some metals can be extracted from compounds by heating with carbon atom because they are less reactive than carbon and some metals cannot be extracted because they are more reactive than carbon atom.

Explanation: If the metal is less reactive than carbon atom so the carbon atoms make bond with oxide or other atom present with metal and the metal is free from that oxide or that element. But if the reactivity of metal is higher than carbon is unable to remove the oxide or element.

8 0
3 years ago
What is the independent variable?
olganol [36]
X is always the independent variable
6 0
3 years ago
What is the mass of 5.84 × 10^21 atoms of xenon?
podryga [215]

Answer:

5.84e+21

Explanation:

8 0
2 years ago
If the concentration of the stock (provided) Cu(NH3)42 was 0.041 M, what concentration will the Cu2 be in beaker?
kodGreya [7K]

Answer:

[Cu^{2+}]=0.041 M

Explanation:

Hello!

In this case, since the molarity of a solution is defined in terms of the moles of the solute and the volume of solution, given that the concentration of Cu(NH₃)₄²⁺ is 0.041 M, and there is only one copper atom per Cu(NH₃)₄²⁺ ion, we can compute the concentration of Cu²⁺ as shown below:

[Cu^{2+}]=0.041\frac{molCu(NH_3)_4^{2+}}{L}*\frac{1molCu^{2+}}{1molCu(NH_3)_4^{2+}} =0.041 \frac{molCu(NH_3)_4^{2+}}{L}

[Cu^{2+}]=0.041 M

Best regards!

6 0
2 years ago
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