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3 years ago

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!! How much has CO2 in the atmosphere increased since the Industrial Revolution? In the 10,000 years before the Industrial Revol

ution in 1751, carbon dioxide levels rose less than 1 percent. Since then, they've risen by

1 answer:

sergeinik [125]3 years ago

3 0

Answer:

Im pretty sure its B) 43

Explanation:

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If the half-life of hydrogen-3 is 11.8 years, after two half-lives the radioactivity of a sample will be reduced to one-half of

maw [93]

Answer:

False

Explanation:

Half life is the time period at which the concentration of the radioactive substance in decay reduced to half.

<u>Thus, if the hydrogen-3 has gone 2 half lives, it means that it has first reduced to its half and then again the half of what it was, i.e. 1/4</u>

Thus, after two successive half-lives, the concentration must be 1/4 of the initial concentration and hence, the statement is false.

4 0

4 years ago

Which of the following are the requirements when creating a hypothesis for an experiment? Check ALL that apply. A: A hypothesis

kkurt [141]

Answer:

You're answer would be B.

Explanation:

5 0

3 years ago

Read 2 more answers

A student titrates 10.00 milliliters of hydrochloric acid of unknown molarity with 1.000 m naoh. it takes 21.17 milliliters of b

Dima020 [189]

Mole ratio for the reaction is 1:1
no of moles in NaOH that reacted= 1*21.17/1000=0.02117mols
molarity of HCl=0.02117*10/1000
=2.117M

4 0

3 years ago

At its critical point, ammonia has a density of 0.235 g cm23. You have a special thickwalled glass tube that has a 10.0mm outs

salantis [7]

Answer:

$\large \boxed{\text{69.3 mg}}$

Explanation:

1. Volume of sealed tube

Assume the sealed tube is a right circular cylinder in which the cap and the base are also 4.20 mm thick.

Its outside dimensions are 155 mm long × 10.0 mm diameter.

Its inside dimensions are

h = 155 mm - 2 × 4.20 mm = 146.6 mm

r = 5.0 mm - 4.20 mm = 0.8 mm

V = πr²h = π(0.8)²× 146.6 mm³ = 294.8 mm³ = 0.2948 cm³

2. Calculate the mass of NH₃

$\begin{array}{rcl}\text{Density} & = & \dfrac{\text{Mass}}{\text{Volume}}\\\\\rho & = &\dfrac{m}{V}\\\\0.235 \text{ g$\cdot$ cm}^{-3} & = & \dfrac{m}{\text{0.2948 cm}^{3}}\\\\m & = & \text{0.0693 g}\\& = & \textbf{69.3 mg}\\\end{array}\\\text{You must seal $\large \boxed{\textbf{69.3 mg}}$ of ammonia in the tube.}$

4 0

3 years ago

How many particles are present in 10.0 moles of table salt?

Dima020 [189]

1.01 x 10^24 molecules.

Explanation:

To calculate the number of molecules in a given number of mole, we can simply multiply by Avogadro's number which is equal to 6.022 x 10 ^23.

Therefore,

10 molecules = 1.68 mol x (6.022 x 10^23 molecules) / (1 mol = 1.01 x 10^24) molecules.

I hope this helps :)

7 0

3 years ago