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3 years ago

Astatine-210 has a half-life of 8.08 days. What fraction of a sample of astatine-210 is left unchanged after 16.16 days?

Chemistry

1 answer:

Flauer [41]3 years ago

6 0

Answer:

Explanation:

Given parameters:

Half-life = 8.08days

Unknown:

What fraction is left unchanged after 16.16days = ?

Solution:

The half - life of a substance is the time taken for the half of a radioactive material to decay to half.

Day 0 Day 8.08 Day 16.16

100% 50% 0% Parent

0% 50% 100% Daughter

After 16.16 days, non of the original sample will remain unchanged.

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Write a balanced chemical equation, including physical state symbols, for the decomposition of solid sodium azide (NaN3) into so

LenaWriter [7]

Answer:

39g

Explanation:

Details of the solution is shown below. From the information provided regarding the N2 produced, we could calculate the amount of N2 produced and use that to find the mass of sodium azide reacted.

5 0

3 years ago

How many calories of heat are required to raise the temperature of 1.00kg of water from 10.2 degrees Celsius to 26.8 degrees Cel

Vika [28.1K]

Answer:

Explanation:

we know that specific heat is the amount of heat required to raise the temperature of substance by one degree mathmeticaly

Q=mcΔT

ΔT=T2-T1

ΔT=26.8-10.2=16.6

C for water is 4.184

therefore

Q=1.00*4.184*16.6

Q=69.4 j

now we have to covert joule into calorie

1 calorie =4.2 j

x calorie=69.4 j/2

so 69.4 j =34.7 calorie thats why 34.7 calorie heat is required to raise the temperature of water from 10.2 to 26.8 degree celsius

6 0

3 years ago

How much work (in J) is involved in a chemical reaction if the volume decreases from 4.35 to 1.20 L against a constant pressure

Marianna [84]

Answer:

W = -262 J.

Explanation:

Hello there!

In this case, according to the given information, we can recall the definition of work in terms of constant pressure and variable volume as follows:

$W=P(V_2-V_1)$

So we plug in the given pressure and volumes to obtain:

$W=0.822atm(1.20L-4.35L)\\\\W=-2.60atm*L$

Now, we convert this number to J (Pa*m³) by using the shown below conversion factor:

$W=-2.60atm*L*\frac{101325Pa}{1atm} *\frac{1m^3}{1000L}\\\\W=-262J$

Regards!

3 0

3 years ago

7. How many chlorine atoms are in a 34.2 g sample of dichlorine pentoxide?

Reptile [31]

Answer:

The answer to your question is 1.36 x 10²³ atoms

Explanation:

Data

number of atoms = ?

mass of the sample = 34.2 g

Molecule = Cl₂O₅

Process

1.- Calculate the molar mass of Cl₂O₅

Cl₂O₅ = (35.5 x 2) + (16 x 5) = 71 + 80 = 151 g

2.- Calculate the atoms of Cl₂O₅

151 g of Cl₂O₅ ---------------- 6 .023 x 10²³ atoms

34.2 g of Cl₂O₅ ------------ x

x = (34.2 x 6.023 x 10²³) / 151

x = 1.36 x 10²³ atoms

4 0

4 years ago

Is this a correct electron config?

lina2011 [118]

Yes, hope this helped!

3 0

3 years ago