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andriy [413]
3 years ago
9

Is nitrogen gas a pure substance or a mixture

Chemistry
1 answer:
Verizon [17]3 years ago
3 0
Its a mixture because nitrogen gas is air. Air is homogeneous mixture
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Alternative Titles: nuclear force, strong interaction, strong nuclear force. Strong force, a fundamental interaction of nature that acts between subatomic particles of matter. The strong force binds quarks together in clusters to make more-familiar subatomic particles, such as protons and neutrons.

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3 common uses for salt
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7 0
2 years ago
The half-life of sr-90 is 28 years. after 56 years of decay only 0. 40 g of a sample remains. what was the mass of the original
Svetradugi [14.3K]

Half-life is the time taken for the concentration of the substance to reduce by 50%. The original sample of strontium had a mass of 1.6 gms. Thus, option d is correct.

<h3>What is half-life?</h3>

The half-life of any radioactive substance is the time period at which the concentration will get reduced to half the initial amount. The initial mass of Sr-90 is calculated as,

N(t) = N_{0} (\dfrac{1}{2})^{ \frac{t }{t 1/2}}

Given,

Quantity of the remaining substance N (t) = 0.40 gm

Initial radioactive substance quantity N_{0} =?

Time duration (t) = 56 years

Half-life = 28 years

Substituting values above:

\begin{aligned} 0.40 &= N_{0} (\dfrac{1}{2}) ^{{\frac{56}{28}}\\\\0.40 &= N_{0} (\dfrac{1}{2})^{2}\end{aligned}

= 1.6 gm

Therefore, option d. the initial mass of Sr is 1.6 gm.

Learn more about half-life here:

brainly.com/question/16145921

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7 0
2 years ago
Students in Ms. Nye's science class were using the microscopes to make slides of various organisms: algae, moss, onion cells, hu
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8 0
3 years ago
Read 2 more answers
Consider the following unbalanced reaction: P4(s) + F2(g) → PF3(g) What mass of fluorine gas is needed to produce 120. g of PF3
Studentka2010 [4]

Answer:

44.28 grams.

Explanation:

Let us write the balanced reaction:

P_{4}+6F_{2}-->4PF_{3}

As per balanced equation, six moles of fluorine gas will give four moles of PF₃.

The mass of PF₃ required = 120 g

The molar mass of PF₃ = 88g/mol

Moles of PF₃ required =\frac{mass}{molarmass}=\frac{120}{88}=1.364mol

The moles of fluorine gas required = \frac{4X1.364}{6}=0.91

the mass of fluorine gas required = moles X molar mass = 0.91x38 = 34.58g

Now this much mass will be required if the reaction is of 100% yield

But as given that the yield of reaction is only 78.1%

The mass of fluorine required = \frac{massX100}{78.1} =\frac{34.58X100}{78.1} =44.28g

4 0
3 years ago
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