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marishachu [46]
3 years ago
8

The density of gold is 19.3g/mL. If you have a 9.00 g piece of gold, what volume will it occupy (in milliliters)

Chemistry
1 answer:
kiruha [24]3 years ago
5 0

Answer:

The answer to your question is volume = 0.47 ml

Explanation:

To solve this problem use the formula of density, that relates the mass and volume.

Data

density = 19.3 g/ml

mass = 9.0 g

volume = ?

Formula

density = \frac{mass}{volume}

Solve for volume

volume = \frac{mass}{density}

Substitution

Volume = 9 / 19,3

Simplification and result

Volume = 0.47 ml

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What is the number of grams in 3.2 moles of CuSO4×5H2O​
son4ous [18]

Hey there!:

Molar mass CuSO4*5H2O = 249.68 g/mol

Therefore:

1 mole CuSO4*5H2O  ---------------- 249.68 g

3.2 moles --------------------------------- ?? ( mass of CuSO4*5H2O )

mass CuSO4*5H2O  = 249.68 * 3.2

mass CuSO4*5H2O = 798.97 g

Hope that helps!

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Using a case scenario describe how you will use the scientific method to conduct an investigation within your community
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Depends what kind of case we taking about
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1. Complete the reaction illustrating the hydration reaction of a strong electrolyte CaCl2​
makkiz [27]

Answer:

CaCl₂(s) ⟶ Ca²⁺(aq) + 2Cl⁻(aq)

Explanation:

When the calcium chloride dissolves. the calcium and chloride ions leave the surface of the solid and go into solution as hydrated ions.

8 0
4 years ago
If I have 4.5 liters of gas at a temperature of 33 0C and a pressure of 6.54 atm, what will be the pressure of the gas if I rais
Nikitich [7]

This is an exercise in the general or combined gas law.

To start solving this exercise, we must obtain the following data:

<h3>Data:</h3>
  • V₁ = 4.5 l
  • T₁ = 33 °C + 273 = 306 k
  • P₁ = 6.54 atm
  • T₂ = 94 °C + 273 = 367 k
  • V₂ = 2.3 l
  • P₂ = ¿?

We use the following formula:

  • P₁V₁T₂ = P₂V₂T₁ ⇒ General Formula

Where

  • P₁ = Initial pressure
  • V₁ = Initial volume
  • T₂ = Initial temperature
  • P₂ = Final pressure
  • V₂ = Final volume
  • T₁ = Initial temperature

We clear the general formula for the final pressure.

\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{P_{1}V_{1}T_{2} }{V_{2}T_{1}}  \ \to \ Clear \ formula \end{gathered}$}

We solve by substituting our data in the formula:

\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{(6.54 \ atm)(4.5 \not{l})(367 \not{K}) }{(2.3 \not{l})(306 \not{k})}  \end{gathered}$}

\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{10800.81}{ 703.8 } \ atm  \end{gathered}$}

\boxed{\large\displaystyle\text{$\begin{gathered}\sf P_{2}=15.346 \ atm \end{gathered}$} }

If I raise the temperature to 94°C and decrease the volume to 2.3 liters, the pressure of the gas will be 15,346 atm.

8 0
2 years ago
Read 2 more answers
If Sally has 250 grams of cement, how many kilograms does she have? How many milligrams?
MissTica

Answer:

0.250 kg

2500 g

Explanation:

1 kg = 1000g

1g = 10 mg

8 0
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