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11Alexandr11 [23.1K]
3 years ago
13

Which of the following resources is a nonrenewable resource? (1 Point)

Chemistry
1 answer:
katrin2010 [14]3 years ago
5 0
I'm pretty sure the only renewable resource here is C.
You might be interested in
What is the volume of a 5.30g piece of aluminum? The density for aluminum is 2.70mL
Makovka662 [10]

Answer:

1.96mL

Explanation:

Density = mass/volume, and rearranged to solve for volume, volume = mass/density.

So:

volume = 5.30g/2.70g/mL = 1.96mL (assuming your unit was g/mL for density)

7 0
3 years ago
URGENT CHEMISTRY EXPERT!
vovangra [49]

Answer:

Part 1: 7.42 mL; Part 2: 3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ 2Cu₃(PO₄)₂(s)

Explanation:

Part 1. Volume of reactant

(a) Balanced chemical equation.

\rm 2Na_{3}PO_{4} + 3CuCl_{2} \longrightarrow Cu_{3}(PO_{4})_{2} + 6NaCl

(b) Moles of CuCl₂

\text{Moles of CuCl}_{2} =\text{ 16.7 mL CuCl}_{2} \times \dfrac{\text{0.200 mmol CCl}_{2}}{\text{1 mL CuCl}_{2}} =  \text{3.340 mmol CuCl}_{2}

(c) Moles of Na₃PO₄

The molar ratio is 2 mmol Na₃PO₄:3 mmol CuCl₂

\text{Moles of Na$_{3}$PO}_{4} =  \text{3.340 mmol CuCl}_{2} \times \dfrac{\text{2 mmol Na$_{3}$PO}_{4}}{\text{3 mmol CuCl}_{2}} =\text{2.227 mmol Na$_{3}$PO}_{4}

(d) Volume of Na₃PO₄

V = \text{2.227 mmol Na$_{3}$PO}_{4}\times \dfrac{\text{1 mL Na$_{3}$PO}_{4}}{\text{0.300 mmol Na$_{3}$PO}_{4}} = \text{7.42 mL Na$_{3}$PO}_{4} \\\\\text{The reaction requires $\large \boxed{\textbf{7.42 mL Na$_{3}$PO}_{4}}$}

Part 2. Net ionic equation

(a) Molecular equation

\rm 2Na_{3}PO_{4}(\text{aq}) + 3CuCl_{2}(\text{aq}) \longrightarrow Cu_{3}(PO_{4})_{2}(\text{s}) + 6NaCl(\text{aq})

(b) Ionic equation

You write molecular formulas for the solids, and you write the soluble ionic substances as ions.

According to the solubility rules, metal phosphates are insoluble.

6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)  

(c) Net ionic equation

To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.

<u>6Na⁺(aq)</u> + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + <u>6Cl⁻(aq)</u> ⟶ Cu₃(PO₄)₂(s) + <u>6Na⁺(aq)</u> + <u>6Cl⁻(aq)</u>  

The net ionic equation is

3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s)

7 0
3 years ago
Name:
Liono4ka [1.6K]

Answer:

1. 505g is the mass of the aluminium.

2. The answer is in the explanation

Explanation:

1. To solve this question we need to find the volume of the rectangle. With the volume and density we can find the mass of the solid:

Volume = 7.45cm*4.78cm*5.25cm

Volume = 187cm³

Mass:

187cm³ * (2.702g/cm³) = 505g is the mass of the aluminium

2. When the temperature of a liquid increases, the volume increases doing the density decreases because density is inversely proportional to volume. And works in the same way for gases because the temperature produce more collisions and the increasing in volume.

4 0
3 years ago
Reema took 5ml of Lead Nitrate solution in a beaker and added approximately 4ml of Potassium Iodide solution to it. What would s
Brrunno [24]

She would observe a yellowish  solid precipitate which is the lead iodide and a white solid precipitate which is the potassium nitrate.

This is because the lead nitrate solution which contains particles of lead will mix with the potassium iodide solution containing  particles of iodide. Upon mixing,the  lead particles  from the Lead nitrate solution combines with the  iodide particles from Potassium iodide and form two compounds, a yellowish solid precipitate called lead iodide and a white solid  precipitate called Potassium nitrate.

The formation of entirely two new compounds is known as the double displacement reaction and can be written in a chemical equation as  

2KI(aq.)+Pb(NO₃)₂(aq.)------>2KNO₃(aq.)+PbI ₂(s)

See similar answer here  :https://brainly.in/question/46262462

4 0
2 years ago
Read 2 more answers
5.0 g of copper was heated from 20°C to 80°C. How much energy was used to heat Cu?
ale4655 [162]

Answer:

100 J of energy are needed to heat the copper from 20∘C to 80∘C .

5 0
3 years ago
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