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Nadusha1986 [10]
2 years ago
15

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]2 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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Which of the following types of molecules always has a dipole moment? Linear molecules with two identical bonds. Trigonal pyrami
natali 33 [55]

Answer:

Trigonal pyramid molecules (three identical bonds)

Explanation:

In trigonal pyramidal molecule  like molecule of ammonia , the vector some of intra- molecular dipole moment is not zero because the bonds are not symmetrically oriented . In other molecules , bonds are symmetrically oriented in space so the vector sum of all the internal dipole moment  vectors cancel each other to make total dipole moment zero.

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2 years ago
A fox spots a rabbit in a field. The fox begins to chase the rabbit, and the rabbit runs away. Which statement best describes th
Ira Lisetskai [31]

Answer:

2 I Think

Explanation:

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2 years ago
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If a bottle of olive oil contains 1.2 kg of olive oil, what is the volume, in milliliters (mL), of the olive oil?
Paul [167]

Answer:

1.3 mL

Explanation:

First, get the density of the olive oil, which is 0.917 kg/mL. Then divide the mass by the density:

1.2kg/0.917kg/mL= 1.3086150491 mL. The kg cancel out, leaving us with mL.

It should have 2 significant figures, because 1.2kg has 2 and we are dividing.

7 0
2 years ago
What features of this model will help Armando answer the question?
Scrat [10]

Answer:

The adjustable legs and the table of sand.

<em>Note:The question is incomplete. The complete question is given below.</em>

Using Models to Answer Questions About Systems

Armando’s class was looking at images of rivers formed by flowing water. Most of the rivers were wide and shallow, but one river was narrow and deep. Armando’s class thinks that this river is narrow and deep because:

  • the hill that the water flowed down was very steep, or
  • the sand grains that the water flowed through were very small.

Armando designed the model below to try to answer the question: Why is this river so narrow and deep?

Explanation:

The model designed by Armando will be helpful to answer the question because of the following features it possesses:

1. An adjustable leg- since one of the hypotheses put forward by the class to explain why the river was narrow and deep was that the hill that the water flowed down was very steep, the adjustable legs can be lowered or raised in order to make the slope shallower or steeper so that their hypothesis can be tested.

2. A table of sand- the table of sand serves as the streambed. By adjusting the size of the sand grains to be larger or smaller, the students will be able to to test their second hypothesis that the small size sand grains that the water flowed through was the reason for the river to be narrow and deep.

The results of their experiments will enable them to come to a conclusion.

5 0
2 years ago
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A 157-mL sample of gas is collected over water at 22°C and 753 torr. What is the volume of the dry gas at STP? (The vapor pressu
galina1969 [7]

Answer:

Volume at STP = 0.1401 L

Explanation:

We are given:

Vapor pressure of water = 20 torr

Total vapor pressure = 753 torr

Vapor pressure of gas = Total vapor pressure - Vapor pressure of water = 753 torr - 20 torr = 733 torr

Using

PV=nRT

where,

P = pressure of the gas = 733 torr

V = Volume of the gas = 157 mL= 0.157 L

T = Temperature of the gas = 22^oC=[22+273]K=295K

R = Gas constant = 62.3637\text{ L.torr}mol^{-1}K^{-1}

n = number of moles of gas = ?

Putting values in above equation, we get:

733 torr\times 0.157L=n\times 62.3637\text{L.torr}mol^{-1}K^{-1}\times 295K\\\\n=\frac{733\times 0.157}{62.3637\times 295}=0.006255mol

At STP, one mole of the gas occupies a volume of 22.4 L

So, 0.006255 mole of the gas occupies a volume of 22.4*0.006255 L

<u>Volume at STP = 0.1401 L</u>

3 0
2 years ago
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