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Tatiana [17]
3 years ago
6

What is an intensive property? *

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

Answer:A property that changes if the amount of substance changes

Explanation:

This is the answer

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Calculate the molarity of 22.5 g of MgS in 829 mL of solution
krek1111 [17]

Answer:18652.5

Explanation

5 0
3 years ago
Blast furnaces extract pure iron from the iron(III) oxide in iron ore in a two step sequence. In the first step, carbon and oxyg
zavuch27 [327]

Answer:

6 C(s) +  3 O₂(g) + 2 Fe₂O₃(s) →  4 Fe(s) + 6 CO₂(g)

Explanation:

Iron can be formed in two steps.

Step 1: 2 C(s) + O₂(g) → 2 CO(g)

Step 2: Fe₂O₃(s) + 3 CO(g) → 2 Fe(s) + 3 CO₂(g)

In order to get the net chemical equation, we will multiply the first step by 3, the second step by 2, and then add them.

6 C(s) +  3 O₂(g) → 6 CO(g)

+

2 Fe₂O₃(s) + 6 CO(g) → 4 Fe(s) + 6 CO₂(g)

--------------------------------------------------------------------------------------------------

6 C(s) +  3 O₂(g) + 2 Fe₂O₃(s) + 6 CO(g) → 6 CO(g) + 4 Fe(s) + 6 CO₂(g)

6 C(s) +  3 O₂(g) + 2 Fe₂O₃(s) →  4 Fe(s) + 6 CO₂(g)

6 0
3 years ago
The half-life for beta decay of strontium-90 is 28.8 years. a milk sample is found to contain 10.3 ppm strontium-90. how many ye
ryzh [129]

Answer : The correct answer is 96.68 yrs

Radioactivity Decay :

it is a process in which a nucleus of unstable atom emit energy in form of radiations like alpha particle , beta particle etc .

Radioactive decay follows first order kinetics , so its rate , rate constant , amount o isotopes can be calculated using first order equations .

The first order equation for radioactive decay can be expressed as :

ln \frac{N}{N_0}  = - k*t ----------- equation (1)

Where : N = amount of radioisotope after time "t"

N₀ = Initial amount of radioisotope

k = decay constant and t = time

Following steps can be used to find time :

1) To find deacy constant :

Decay constant can be calculated using half life . Decay constant and half life can be related as :

T _\frac{1}{2} = \frac{ln2}{k} ---------equation (2)

Given : Half life of Strontium -90 = 28.8 years

Plugging value of T_\frac{1}{2} in above formula (equation 2) :

28.8 yrs = \frac{ln 2}{ k }

Multiply both side by k

28.8 yrs * k = \frac{ln 2 }{k} * k

Dividing both side by 28.8 yrs

\frac{28.8 yrs * k}{28.8 yrs} = \frac{ln 2}{28.8 yrs}

(ln 2 = 0.693 )

k = 0.0241 yrs⁻¹

Step 2 : To find time :

Given : N₀ = 10.3 ppm N = 1.0 ppm k = 0.0241 yrs⁻¹

Plugging these value in equation (1) as :

ln (\frac{1.0 ppm}{10.3 ppm} ) = - 0.0241 yrs^-^1 * t

ln (0.0971 ) = -0.0241 yrs ^-^1 * t

(ln 0.0971 = - 2.33 )

Dividing both side by - 0.0241 yrs⁻¹

\frac{-2.33}{-0.0241 yrs^-^1} = \frac{-0.0241 yrs^-^1 * t}{-0.0241 yrs^-^1}

t = 96.68 yrs

Hence the concentration of Strontium-90 will drop from 10.3 ppm to 1.0 ppm is 96.68 yrs

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3 years ago
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How many significant figures are in 120 miles?
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2 significant zeros.

1 and 2 are the significant zeros.

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