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3 years ago

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Chemistry

1 answer:

skad [1K]3 years ago

8 0

Answer:

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Explanation:

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Since density depends on the mass and volume of an object, we need both of these values combined in the correct way to solve for

alukav5142 [94]

Answer:

7.86 g/cm³

Explanation:

11.0 kg = 11,000 g

The density in g/cm³ is ...

(11,000 g)/(1,400 cm³) = 7.86 g/cm³

7 0

3 years ago

Its gelatin and a fruit cup the same (its for a project)

gizmo_the_mogwai [7]

Answer:

Explanation:

Gelatin is a substance that makes jello, and a fruit cup is a cup full of fruit

6 0

2 years ago

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Is conversating a word? I am trying to prove a point to my friend that conversate is not a word

Andru [333]

Conversating and conversate are actual words

4 0

3 years ago

Determine the number of grams in each of the quantities<br><br> 1.39.0 x 1024 molecules Cl2

STatiana [176]

Mass of Cl₂ : 164.01 g

<h3>Further explanation</h3>

A mole is a number of particles(atoms, molecules, ions) in a substance

This refers to the atomic total of the 12 gr C-12 which is equal to 6.02.10²³, so 1 mole = 6.02.10²³ particles

Can be formulated :

N = n x No

N = number of particles

n = mol

No = 6.02.10²³ = Avogadro's number

mol Cl₂ :

$\tt n=\dfrac{N}{No}\\\\n=\dfrac{1.39.10^{24}}{6.02.10^{23}}\\\\n=2.31$

mass Cl₂(MW=71 g/mol) :

$\tt mass=mol\times MW\\\\mass=2.31\times 71=164.01$

8 0

3 years ago

In the early 1960s, radioactive strontium-90 was released during atmospheric testing of nuclear weapons and got into the bones o

Advocard [28]

Answer:

52.54 %

Explanation:

Half life = 29 years

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac {ln\ 2}{29.0}\ {years}^{-1}$

The rate constant, k = 0.023902 hour⁻¹

From 1964 to 1991:

Time = 27 years

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

So,

$\frac {[A_t]}{[A_0]}=e^{-0.023902\times 27}$

$\frac {[A_t]}{[A_0]}=0.5245$

<u>The strontium-90 remains in the bone = 52.54 %</u>

5 0

3 years ago