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3 years ago

Help please!!

Chemistry

1 answer:

frez [133]3 years ago

7 0

Answer:

6.25%

Explanation:

Given data:

Half life of lutetium-117 = 6.75 days

Percentage remaining after 27 days = ?

Solution;

Number of half lives = Time elapsed / half life

Number of half lives = 27 days / 6.75 days

Number of half lives = 4

At time zero = 100%

At first half life = 100%/2 = 50%

At second half life = 50%/2 = 25%

At 3rd half life = 25%/2 = 12.5%

At 4th half life = 12.5%/2 = 6.25%

You might be interested in

Rounded to the nearest whole number, how many electrons are in an atom of zirconium?

Dafna1 [17]

Answer:

40 electrons

Explanation:

The element of zirconium (Zr) has an atomic number of 40, which means it has 40 protons and 40 electrons. It also has a molar mass of 91.224 g/mol.

6 0

2 years ago

A 1.25 g sample of aluminum is reacted with 3.28 g of copper (II) sulfate. What is the limiting reactant? 2Al(s) + 3CuSO4(aq) →

vova2212 [387]

Answer:

Copper (II) sulfate

Explanation:

Given reaction is

2Al(s) + 3CuSO4(aq) → Al2(SO4)3(aq) + 3Cu(s)

Amount of aluminum = 1·25 g

Amount of copper (II) sulfate = 3·28 g

Atomic weight of Al = 26 g

Molecular weight of CuSO4 ≈ 159·5

Number of moles of Al = 1·25 ÷ 26 = 0·048

Number of moles of CuSO4 = 3·28 ÷ 159·5 = 0·021

From the above balanced chemical equation for every 2 moles of aluminum, 3 moles of copper (ll) sulfate will be required

So for 1 mole of Al, 1·5 moles of copper (ll) sulfate will be required

For 0·048 moles of Al, 1.5 × 0·048 moles of copper (ll) sulfate will be required

∴ Number of moles of copper (ll) sulfate required = 0·072

But we have only 0·021 moles of copper (ll) sulfate

As copper (ll) sulfate is not there in required amount, the limiting reactant will be copper (ll) sulfate

∴ The limiting reactant is copper (ll) sulfate

7 0

3 years ago

What is the percent composition of silicon dioxide, SiO2 ? (Atomic masses : O = 16 . 0 Si =28 .1

weeeeeb [17]

Answer:

60.1 i think

Explanation:

4 0

3 years ago

Can I find a tutor to help me with this question?

Eddi Din [679]

INFORMATION:

We have the following statements

And we must complete them

STEP BY STEP EXPLANATION:

To complete the statements, we need to classify matter according to its state:

- Solid:

there is not enough thermal energy to overcome the intermolecular interactions between the particles. As a result, solids have a definite shape and volume.

- Liquid:

That describes the liquid state. In a liquid, the particles are still in close contact, so liquids have a definite volume. However, because the particles can move about each other rather freely, a liquid has no definite shape and takes a shape dictated by its container.

- Gas:

That describes the gas state, which we will consider in more detail elsewhere. Like liquids, gases have no definite shape, but unlike solids and liquids, gases have no definite volume either.

Finally, we know that:

- A solid has a definite volume and has a definite shape

- A liquid has a definite volume and has not a definite shape

- A gas has not a definite volume and has not a definite shape

ANSWER:

- A solid has a definite volume and has a definite shape

- A liquid has a definite volume and has not a definite shape

- A gas has not a definite volume and has not a definite shape

8 0

1 year ago

A student wants to make a 0.150 M aqueous solution of silver (I) nitrate but only has 11.27 g of AgNO3. What volume of the 0.150

Usimov [2.4K]

Answer:

442.3 mL

Explanation:

Remember that Molarity is a measure of concentration in Chemistry and it's defined as the number of moles of the substance divided by liters of the solution:

$M=\frac{Moles of substance X}{Volume of the solution}$

Then, you can express 11.27 g of AgNO3 as moles of AgNO3 using the molar mass of the compound:

$11.27 g AgNO_{3} *\frac{1 mole AgNO_{3}}{169.87 g AgNO_{3}} = 0.06634 moles AgNO_{3}$

Then you can solve for the volume of the solution:

$Volume of the solution=\frac{Moles of AgNO_{3}}{M} =\frac{0.06634 mol AgNO_{3}}{0.150 M} =0.4423 L = 442.3 mL$

Hope it helps!

3 0

4 years ago