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miss Akunina [59]

3 years ago

15

Cesium-137 is part of the nuclear waste produced by uranium-235 fission. The half-life of cesium-137 is 30.2 years. How much tim

e is required for the activity of a sample of cesium-137 to fall to 8.32 percent of its original value

1 answer:

Aleonysh [2.5K]3 years ago

5 0

Answer:

la primera va con la última

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Somebody tell me- why do cows go Moo?

m_a_m_a [10]

Answer:

To communicate with each other

Explanation:

Cows use sound (mooing) to communicate with each other and their environment. Cows are herd animals and have complex social structures. Mooing is one way that they interact and how they express their emotions

5 0

3 years ago

Read 2 more answers

Calculate the moles of Cu in 7.4×1021 atoms of Cu.

IceJOKER [234]

1 moles -------- 6.02x10²³ atoms
?? moles ----- 7.4x10²¹ atoms

moles = 7.4x10²¹ / 6.02x10²³

= 0.01229 moles

hope this helps!

5 0

4 years ago

What volume of 15.6M NH4OH is needed to make 500. mL of 3.00M solution?

Mariana [72]

Answer:

V₁ = 96.2 mL

Explanation:

Given data:

Initial volume of NH₄OH required = ?

Initial molarity = 15.6 M

Final molarity = 3.00 M

Final volume = 500.0 mL

Solution:

Formula:

M₁V₁ = M₂V₂

M₁ = Initial molarity

V₁ = Initial volume of NH₄OH

M₂ =Final molarity

V₂ = Final volume

Now we will put the values.

15.6 M ×V₁ = 3.00 M×500.0 mL

15.6 M ×V₁ = 1500 M.mL

V₁ = 1500 M.mL /15.6 M

V₁ = 96.2 mL

6 0

3 years ago

If you know any answer to this pls help out I'll give brainiest plus a lot of points

sukhopar [10]

Answer:

I cant see it clearly

Explanation:

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5 0

3 years ago

a 1.25g sample of ore containing iron pyrite (FeS2) was pulverized and ignited in air, converting the FeS2 to Fe2O3 and SO2(g).

svp [43]

Answer:

28.9%

Explanation:

Let's consider the following balanced equation.

2 FeS₂ + 11/2 O₂ ⇒ Fe₂O₃ + 4 SO₂

We can establish the following relations:

The molar mass of Fe₂O₃ is 159.6 g/mol
1 mole of Fe₂O₃ is produced per 2 moles of FeS₂
1 mole of Fe is in 1 mole of FeS₂
The molar mass of Fe is 55.84 g/mol

The amount of Fe in the sample that produced 0.516 g of Fe₂O₃ is:

$0.516gFe_{2}O_{3}.\frac{1molFe_{2}O_{3}}{159.6gFe_{2}O_{3}} .\frac{2molFeS_{2}}{1molFe_{2}O_{3}} .\frac{1molFe}{1molFeS_{2}} .\frac{55.84gFe}{1molFe} =0.361gFe$

The percent of Fe in 1.25 g of the ore is:

$\frac{0.361g}{1.25g} .100\%=28.9\%$

4 0

3 years ago