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3 years ago

If a 60-g object has a volume of 30 cm3, what is its density? 2 g/cm3 0.5 cm3/g 1800 g * cm3 none of the above

Chemistry

1 answer:

jasenka [17]3 years ago

6 0

Answer : The density of an object is, $2g/cm^3$

Solution : Given,

Mass of an object = 60 g

Volume of an object = $30cm^3$

Formula used :

$\text{Density of an object}=\frac{\text{Mass of an object}}{\text{Volume of an object}}$

Now put all the given values in this formula, we get the density of an object.

$\text{Density of an object}=\frac{60g}{30cm^3}=2g/cm^3$

Therefore, the density of an object is, $2g/cm^3$

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disa [49]

Answer:

11.33

Explanation:

-log(2.3x10^-3) = 2.67

14-2.67

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3 years ago

A beaker is filled to the 500 mL mark with alcohol. What increase in volume (in mL) does the beaker contain when the temperature

Aliun [14]

Answer:

"1.4 mL" is the appropriate solution.

Explanation:

According to the question,

$v_0=500$
$\alpha =1.12\times 10^{-4}$
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Now,

Increase in volume will be:

⇒ $\Delta V = \alpha\times v_0\times \Delta \epsilon$

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$=1.12\times 10^{-4}\times 500\times 25$

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3 years ago

Which of the above electron dot diagrams is wrong and why?

Elodia [21]

Answer:

Oxygen, it's supposed to have six valenge electrons.

Explanation:

Count the dots on the oxygen atom, you'll see seven, but there's supposed to be six.

5 0

3 years ago

Read 2 more answers

What is the equilibrium expression for the reaction below?

Mariana [72]

Answer:

A. $K=\frac{[N_2O]^2}{[N_2]^2[O_2]}$

Explanation:

Hello there!

In this case, for us to figure out the appropriate equilibrium expression, it will be firstly necessary for us to recall the law of conservation of mass which states that the equilibrium constant of an equilibrium chemical reaction is written by dividing the products and reactants and including the stoichiometric coefficients as exponents. In such a way, for the given reaction, we will have:

$K=\frac{[N_2O]^2}{[N_2]^2[O_2]}$

As N2O is the product whereas N2 and O2 are reactants; thus, the equilibrium expression will be A.

Regards!

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2 years ago

In the laboratory a student combines 26.2 mL of a 0.234 M chromium(III) acetate solution with 10.7 mL of a 0.461 M chromium(III)

Natalka [10]

Answer: The molarity of $Cr^{3+}$ ions in the solution is 0.299 M

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For chromium (III) acetate:

Molarity of chromium (III) acetate solution = 0.234 M

Volume of solution = 26.2 mL

Putting values in equation 1, we get:

$0.234=\frac{\text{Moles of chromium (III) acetate}\times 1000}{26.2}\\\\\text{Moles of chromium (III) acetate}=\frac{0.234\times 26.2}{1000}=0.00613mol$

1 mole of chromium (III) acetate $(Cr(CH_3COO)_3)$ produces 1 mole of chromium $(Cr^{3+})$ ions and 3 moles of acetate $(CH_3COO^-)$ ions

Moles of $Cr^{3+}\text{ ions}=(1\times 0.00613)=0.00613moles$

For chromium (III) nitrate:

Molarity of chromium (III) nitrate solution = 0.461 M

Volume of solution = 10.7 mL

Putting values in equation 1, we get:

$0.461=\frac{\text{Moles of chromium (III) nitrate}\times 1000}{10.7}\\\\\text{Moles of chromium (III) nitrate}=\frac{0.461\times 10.7}{1000}=0.00493mol$

1 mole of chromium (III) nitrate $(Cr(NO_3)_3)$ produces 1 mole of chromium $(Cr^{3+})$ ions and 3 moles of nitrate $(NO_3^-)$ ions

Moles of $Cr^{3+}\text{ ions}=(1\times 0.00493)=0.00493moles$

For chromium cation:

Total moles of chromium cations = [0.00613 + 0.00493] = 0.01106 moles

Total volume of solution = [26.2 + 10.7] = 36.9 mL

Putting values in equation 1, we get:

$\text{Molarity of }Cr^{3+}\text{ cations}=\frac{0.01106\times 1000}{36.9}\\\\\text{Molarity of }Cr^{3+}\text{ cations}=0.299M/tex]Hence, the molarity of [tex]Cr^{3+}$ ions in the solution is 0.299 M

5 0

3 years ago