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3 years ago

If you have 100 g of radioactive isotope with a half-life of 10 years:

Chemistry

1 answer:

labwork [276]3 years ago

6 0

<h3>Answer:</h3>

#1. 50 g

#2. 25 g

#3. 4 half lives

<h3>Explanation:</h3>

We are given;

Original mass of a radioisotope as 100 g
Half life of the radioisotope as 10 years

We need to answer the questions:

#a. Mass remaining after 10 years

Using the formula;

Remaining mass = Original mass × 0.5^n , where n is the number of lives.

In this case, since the half life is 10 years then n is 1

Therefore;

Remaining mass = 100 g × 0.5^1

= 50 g

Therefore, 50 g of the isotope will remain after 10 years

#b. Mass of the isotope that will remain after 20 years

Remember the formula;

Remaining mass = Original mass × 0.5^n

n = Time ÷ half life

n = 20 years ÷ 10 years

= 2

Therefore;

Remaining mass = 100 g × 0.5^2

= 25 g

Thus, 25 g of the isotope will be left after 20 years

#3. Number of half lives in 40 years

1 half life = 10 years

But; n = time ÷ half life

= 40 years ÷ 10 years

= 4

Thus, the number of half lives in 40 years is 4.

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Antacids work to relieve stomach acids because of which of the following types of reactions?

Anastaziya [24]

Answer: The correct answer is Option B.

Explanation:

Decomposition is a type of chemical reaction in which larger compound breaks down into two or more smaller compounds.

$AB\rightarrow A+B$

Double displacement reactions is defined as the chemical reaction in which exchange of ions takes place.

$AB+CD\rightarrow AD+CB$

Synthesis reaction is a type of reaction in which two or more smaller compounds combines to form a single large compound.

$A+B\rightarrow AB$

Single displacement reaction is a type of reaction in which a more reactive metal displaces a less reactive metal from its chemical reaction.

$A+BC\rightarrow AC+B$

In stomach, an acid is present known as hydrochloric acid and to neutralize its effect, antacid is taken which has $CaCO_$ as a component.

The reaction between HCl and $CaCO_3$ is a type of neutralization reaction and it is a type of double displacement reaction.

The equation between the two follows:

$CaCO_3+HCl\rightarrow CaCl_2+H_2O+CO_2$

Hence, the correct answer is Option B.

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3 years ago

Which part of the water cycle would follow step C in the diagram shown?

castortr0y [4]

Evaporation

Explanation:

From C to D, the phase changes from liquid to gas and it is called evaporation.

Evaporation is a phase change process in which liquid water is converted to gaseous vapor.

During evaporation, the liquid is heated till the particles gain sufficient energy that allows the molecules to escape. The reverse process in which gas is turned to liquid is called condensation.

Learn more:

Evaporation brainly.com/question/10972073

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6 0

3 years ago

Ethanol (C2H5OH) is produced from the fermentation of sucrose in the presence of enzymes.

Klio2033 [76]

Explanation:

to calculate theoratical yield...first we must calculate how much Sucrose moles we have..in order to to do so...we must calculate molar mass of sucrose.

molar mass of sucrose is 342.3 g/mol.

now we can calculate how many sucrose moles we have by dividing the mass with molar mass of sucrose

735g/(342.3g/mol)=2.147mol

theoratically...according to stoichiometry..every 1 mole of sucrose yields 4 moles of ethanol...so...

2.147mole yield 2.147*4mol=8.588mol

now we must calculate weight of that much ethanol..molar mass of ethanol is 46.07 g/mol...

so we can multiple moles by molar mass to obtain the weight 8.588mol*46.07g/mol=395.649g

but we only obtained 310.5g...so percentage we have is

$\frac{310.5}{395.649} \times 100$

78.47%

if there is trouble with molar mases I used....use what you calculated...

If my explanation is good enough...please mark it as brainliest.thanks

5 0

3 years ago

You are asked to prepare 500 mL 0.200 M acetate buffer at pH 5.00 using only pure acetic acid ( MW=60.05 g/mol, pKa=4.76), 3.00

Vilka [71]

Answer:

You need to weight 6,005 g of acetic acid

Explanation:

Using Henderson-Hasselbalch formula you will obtain:

5,00 = 4,76 +log₁₀ $\frac{[Ac^-]}{[Acac]}$

Where Ac⁻ is the salt of acetic acid (Acac).

Solving:

1,738 = $\frac{[Ac^-]}{[Acac]}$ (1)

Also, yo know that:

0,200 M = [Ac⁻] + [Acac] (2)

Replacing (2) in (1):

[Acac] = 0,0730 M.

Thus:

[Ac⁻] = 0,127 M

The moles of each compound are:

Acac = 0,0730 M × 0,500 L = 0,0365 mol

Ac⁻ = 0,127 M × 0,500 L = 0,0635 mol

To prepare these moles it is necessary to use:

Acac + NaOH → AcNa + H₂O

The initial moles of Acac must be:

0,0365 moles + 0,0635 moles = 0,100 moles

To obtain 0,0635 moles of Ac⁻ you need to take this quantity of NaOH moles.

Thus, to obtain a acetate buffer of 5,00 you need to add 0,100 moles of acetic acid and 0,0635 moles of NaOH because This NaOH will react with acetic acid producing 0,0635 moles of Ac⁻ and surplus 0,0365 moles of acetic acid.

Now, to obtain 0,100 moles of acetic acid from pure acetic acid:

0,100 moles × $\frac{60,05 g}{1 mol}$ = 6,005 g

You need to weight 6,005 g of acetic acid

I hope it helps!

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4 years ago

How many grams are equivalent to 20 moles of water?

Eduardwww [97]

Answer: 360.3056 grams i think

Explanation:

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3 years ago