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3 years ago

Hydroelectricity and tidal energy are similar because

Chemistry

1 answer:

vovikov84 [41]3 years ago

3 0

C. both sources of energy rely on water.

Explanation:

Hydroelectricity and tidal energy are similar energy sources because they both rely on water.

They harness the power of moving water to produce their energy.

Tidal energy is the energy derived from moving water as crashing waves in an ocean.
Hydroelectricity is produce when water from a dam channel is used to drive heavy turbines.
They are both renewable energy sources.

learn more:

Non-renewable resources brainly.com/question/2948717

#learnwithBrainly

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NaOH (aq) + HCl(aq) H2O + NaCl (aq)

yuradex [85]

Answer:

The answer to your question is 0.005

Explanation:

Data

Volume of NaOH = 25 ml

[NaOH] = 0.2 M

moles of NaOH = ?

To solve this problem is not necessary to have the chemical reaction. Just use the formula of Molarity and solve it for moles.

Formula

Molarity = moles / volume

-Solve for moles

moles = Molarity x volume

-Convert volume to liters

1000 ml ---------------- 1 l

25 ml ---------------- x

x = (25 x 1) / 1000

x = 0.025 l

-Substitution

moles = 0.2 x 0.025

-Result

moles = 0.005

7 0

3 years ago

What is the specific heat of a substance if 1450 calories are required to raise the temperature of a 240g sample by 20℃?

bazaltina [42]

Answer:

6960 J/kg°C

Explanation:

specific heat= mass×specific heat capacity×increase in temperature

specific heat= 0.240×1450×20= 6960 J/kg°C

hope it helps!

5 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.

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

The net ionic equation is

3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s)

7 0

3 years ago

A 0.187 M weak acid solution has a pH of 3.99. Find Ka for the acid. Express your answer using two significant figures.

Anna35 [415]

Answer:

5.56 × 10⁻⁸

Explanation:

Step 1: Given data

Concentration of the weak acid (Ca): 0.187 M

pH of the solution: 3.99

Step 2: Calculate the concentration of H⁺

We will use the following expression.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -3.99 = 1.02 × 10⁻⁴ M

Step 3: Calculate the acid dissociation constant (Ka)

We will use the following expression.

$Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(1.02 \times 10^{-4})^{2} }{0.187} = 5.56 \times 10^{-8}$

3 0

3 years ago

A uranium atom has 92 protons in its nucleus. how many electrons does it have?

sweet [91]

In a balanced, non-charged atom, the number of protons (positive charge) is equal to the number of electrons (negative charge)

So there are 92 electrons

4 0

3 years ago

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