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kaheart [24]
3 years ago
13

The mass of the quartz crystal shown below is 10 grams, and its volume is 4 cm3. What is the density of quartz?

Chemistry
1 answer:
Tju [1.3M]3 years ago
8 0

Answer:

{\boxed{\text{b. 2.5 g/cm}^{3}}}

Explanation:

\text{Density} = \dfrac{m}{V} = \dfrac{\text{mass}}{\text{volume}}

Data:

m = 10 g

V = 4 cm³

Calculation:

\text{Density} = \dfrac{\text{10 g}}{\text{4 cm}^{3}} = \textbf{2.5 g/cm}^{3}\\\text{The density of the crystal is ${\boxed{\textbf{2.5 g/cm}^{3}}}$}

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Do all forms of radioactive decay change the identity of the original element?
hoa [83]

Explanation:

<h3>yes, Radioactive decay involves the emission of a particle and/or energy as one atom changes into another. In most instances, the atom changes its identity to become a new element.</h3>

5 0
2 years ago
Very Important!! Electron lender or borrower?
worty [1.4K]
Sodium lends 1 electron.
Phosphorus borrows 3 electrons.
Potassium lends one electron.
Oxygen borrows 2 electrons.
Iodine borrows one electron.
Cesium lends 1 electron.
Bromine borrows 1 electron.
Sulfur borrows 2 electrons.
And magnesium lends 2 electrons.
4 0
3 years ago
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Which of the following atoms would gain two electrons to fill its valence energy level? (2 points) Ca Br S B
Andrei [34K]
Sulfur is the answer according to the 2'8-8 rule
4 0
3 years ago
the formula of calcium oxide is CaO. What is the formula of the ionic compound containing calium abd sulfate ions
sineoko [7]

Answer:

CaS

Explanation:

Hello!

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Ca^{2+}S^{2-}

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CaS

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5 0
2 years ago
A 0.450 g sample of solid lead(II) nitrate is added to 250 mL of 0.250 M sodium iodide solution. Assume no change in volume of t
Verdich [7]

Pb(NO₃)₂ ⇒limiting reactant

moles PbI₂ = 1.36 x 10⁻³

% yield  = 87.72%

<h3>Further explanation</h3>

Given

Reaction(unbalanced)

Pb(NO₃)₂(s) + NaI(aq) → PbI₂(s) + NaNO₃(aq)

Required

  • moles of PbI₂
  • Limiting reactant
  • % yield

Solution

Balanced equation :

Pb(NO₃)₂(s) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)

mol Pb(NO₃)₂ :

= 0.45 : 331 g/mol

= 1.36 x 10⁻³

mol NaI :

= 250 ml x 0.25 M

= 0.0625

Limiting reactant (mol : coefficient)

Pb(NO₃)₂ : 1.36 x 10⁻³ : 1 = 1.36 x 10⁻³

NaI : 0.0625 : 2 = 0.03125

Pb(NO₃)₂ ⇒limiting reactant(smaller ratio)

moles PbI₂ = moles Pb(NO₃)₂ = 1.36 x 10⁻³(mol ratio 1 : 1)

Mass of PbI₂ :

= mol x MW

=  1.36 x 10⁻³ x 461,01 g/mol

= 0.627 g

% yield = 0.55/0.627 x 100% = 87.72%

7 0
3 years ago
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