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3 years ago

If you start with 100 grams of hydrogen-3, how many grams will you have after 24.6 years?

Chemistry

1 answer:

svet-max [94.6K]3 years ago

4 0

Answer:

The mass left after 24.6 years is 25.0563 grams

Explanation:

The given parameters are;

The mass of the hydrogen-3 = 100 grams

The half life of hydrogen-3 which is also known as = 12.32 years

The formula for calculating half-life is given as follows;

$N(t) = N_0 \times \left (\dfrac{1}{2} \right )^{\dfrac{t}{t_{\frac{1}{2} }} }$

Where;

N(t) = The mass left after t years

N₀ = The initial mass of the hydrogen-3 = 100 g

t = Time duration of the decay = 24.6 years

$t_{\frac{1}{2} }$ = Half-life = 12.32 years

$N(24.6) = 100 \times \left (\dfrac{1}{2} \right )^{\dfrac{24.6}{12.32}} } = 25.0563$

The mass left after 24.6 years = 25.0563 grams.

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When Z-4,5-dimethyloct-4-ene is treated with hydrogen chloride, HCl, the result is:________.

Vika [28.1K]

Answer:

The correct option is c

Explanation:

The chemical equation for the reaction of Z-4,5-dimethyloct-4-ene and HCl is shown on the first uploaded image

Now looking at the product we see that there are two who has four different groups attached to them this carbon are known as chiral carbons hence the product formed is a pair of diastereomers

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3 years ago

Titration of 0.824 g of potassium hydrogen phthalate required 38.314 g of naoh solution to reach the end point detected by pheno

melamori03 [73]

1.062 mol/kg.

Step 1. Write the balanced equation for the neutralization.

MM = 204.22 40.00

KHC8H4O4 + NaOH → KNaC8H4O4 + H2O

Step 2. Calculate the moles of potassium hydrogen phthalate (KHP)

Moles of KHP = 824 mg KHP × (1 mmol KHP/204.22 mg KHP)

= 4.035 mmol KHP

Step 3. Calculate the moles of NaOH

Moles of NaOH = 4.035 mmol KHP × (1 mmol NaOH/(1 mmol KHP)

= 4.035 mmol NaOH

Step 4. Calculate the mass of the NaOH

Mass of NaOH = 4.035 mmol NaOH × (40.00 mg NaOH/1 mmol NaOH)

= 161 mg NaOH

Step 5. Calculate the mass of the water

Mass of water = mass of solution – mass of NaOH = 38.134 g - 0.161 g

= 37.973 g

Step 6. Calculate the molal concentration of the NaOH

b = moles of NaOH/kg of water = 0.040 35 mol/0.037 973 kg = 1.062 mol/kg

3 0

3 years ago

Fires in forests can sometimes be stopped by cutting down trees to make a gap called a for break. Which part of the fire triangl

Ber [7]

It removes the fuel for the flame.

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3 years ago

Question 11 (3 points)

Ugo [173]

Answer:

The outer core of the earth is liquid.

Explanation:

Seismic waves travel through the Earth at different speeds and are reflected and refracted diffracted differently as they encounter layers of different densities.

By analyzing these waves, they have concluded that there is an inner core of solid iron surrounded by a liquid outer core consisting of nickel and iron.

A is wrong. The Earth consists of several layers

C is wrong. Seafloor spreading is caused by the movement of tectonic plates away from each other.

D is wrong. The mantle is mostly solid rock.

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3 years ago

The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151

vitfil [10]

Answer:- The natural abundance of $^1^5^1_E_u$ is 0.478 or 47.8% and $^1^5^3_E_u$ is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for $^1^5^1_E_u$ and $^1^5^3_E_u$ as 150.919860 and 152.921243 amu, respectively. Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of $^1^5^1_E_u$ as n then the abundance of $^1^5^3_E_u$ would be 1-n .

Let's plug in the values in the formula:

$151.964=150.919860(n)+152.921243(1-n)$