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tresset_1 [31]
3 years ago
10

The average density of a planet is 3.05 g/cm3. What is its density in kg/m3? (Enter your answer in scientific notation.)

Chemistry
1 answer:
jasenka [17]3 years ago
4 0

Answer:

3.05 ×10³ Kg/m³

Explanation:

Density:

Density is equal to the mass of substance divided by its volume.

Units:

SI unit of density is Kg/m3.

Other units are given below,

g/cm3, g/mL , kg/L

Formula:

D=m/v

D= density

m=mass

V=volume

Symbol:

The symbol used for density is called rho. It is represented by ρ. However letter D can also be used to represent the density.

Given data:

Density = 3.05 g/cm³

Density in Kg/m³ = ?

Solution:

In order to convert the g/cm³ to  Kg/m³ we will multiply the value with 1000.

3.05 g/cm³ × 1000 Kg/m³ / 1 g/cm³

3050 Kg/m³

In scientific notation:

3.05 ×10³ Kg/m³

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What is the pH of a solution with an H3O+ concentration of 5.67 x 10-4 M?
Valentin [98]

Answer:

<h2>3.25 </h2>

Explanation:

The pH of a solution can be found by using the formula

pH = - log [ { H_3O}^{+}]

From the question we have

ph =  -  log(5.67 \times  {10}^{ - 4} )  \\  = 3.2464

We have the final answer as

<h3>3.25 </h3>

Hope this helps you

3 0
3 years ago
Which is the best term to use when describing the energy of position?
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Kinetic energy is the energy due to the speed.

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3 years ago
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A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal
SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^-} = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻×\frac{2molCe^{4+}}{1molI_{3}^-} = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = <em>0,0625 g of Ce(IV)</em>

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

5 0
3 years ago
How do chemists explain on a molecular basis
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Explanation:

It happens because particles of gas are in constant random motion. Thus they can collide with the walls of the container causing pressure on the walls.

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