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Marat540 [252]
1 year ago
5

3. You measure a cube and determine that its sides are 0.65m. You place the cube on a mass scale and determine that this cube ha

s a mass of 10,500 grams. What is the density of this cube in units of kg/m3 and in units of g/mL?
Chemistry
1 answer:
amid [387]1 year ago
7 0

The density of the cube in-

Kg/m³= 38.234 kg/m³

g/mL= 0.038234 g/mL

Explanation:

We know that volume of the cube equals m ³

Where “m” equals side of the cube

Given data-

Side of the cube=0.65m

mass of the cube= 10,500 gm

We know that 1000gm= 1 kg

Hence, 10,500 gm= 10.5 kg

Volume of the cube= (0.65)³

∴ Volume= 0.274625 m ³

We know that density = mass/volume

⇒Substituting the value of mass and volume, we get-

⇒Density= 10.5/0.274625= 38.234 kg/m³

We know that 1 kg/m³= 0.001 g/mL

Hence 38.234 kg/m³ would equal 0.038234 g/mL

Hence the density of the cube is 38.234 kg/m³ and 0.038234 g/mL

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Amanda [17]

<u>Answer:</u> The value of K_c for the given reaction is 1.435

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Given mass of N_2O_4 = 9.2 g

Molar mass of N_2O_4 = 92 g/mol

Volume of solution = 0.50 L

Putting values in above equation, we get:

\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M

For the given chemical equation:

                 N_2O_4(g)\rightleftharpoons 2NO_2(g)

<u>Initial:</u>          0.20

<u>At eqllm:</u>     0.20-x        2x

We are given:

Equilibrium concentration of N_2O_4 = 0.057

Evaluating the value of 'x'

\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143

The expression of K_c for above equation follows:

K_c=\frac{[NO_2]^2}{[N_2O_4]}

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Putting values in above expression, we get:

K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435

Hence, the value of K_c for the given reaction is 1.435

6 0
2 years ago
2 red and 2 blue overlapping balls in the center are surrounded by a green, fuzzy, circular cloud with a white line running thro
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Answer:

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Explanation:

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4 0
2 years ago
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A solution contains cr3+ ions. the addition of 0.063 l of 1.50 m naf solution was needed to completely precipitate the chromium
steposvetlana [31]

Cr{3+} + 3 NaF → CrF3 + 3 Na{+} <span>

First calculate the total mols of NaF. 

(0.063 L) x (1.50 mol/L NaF) = 0.0945 mol NaF total </span>

 

Using stoichiometric ratio:

<span>0.0945 mol NaF * (1 mol Cr3+ / 3 mol NaF) * (51.9961 g Cr3+/mol) = 1.6379 g Cr3+</span>
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2 years ago
How many molecules are in a 189 g sample of carbon tetrabromide, CBr4 ?
anyanavicka [17]

Answer:

3.4 × 10²³ molecules of CBr₄

Explanation:

Given data:

Mass of CBr₄ = 189 g

Number of molecules = ?

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Number of moles = mass / molar mass

Number of moles = 189 g/ 331.63 g/mol

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Now the given problem will solve by using Avogadro number.

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The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

Foe 0.6 moles of CBr₄:

0.6 mol × 6.022 × 10²³ molecules of CBr₄ / 1 mol

3.4 × 10²³ molecules of CBr₄

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1 year ago
Consider the following data concerning the equation: H2O2 + 3I– + 2H+ → I3– + 2H2O [H2O2] [I–] [H+] rate I 0.100 M 5.00 × 10–4 M
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Answer:

See explaination

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