What is one thing that wind, solar, and hydropower all have in common?

Chemistry

2 answers:

vampirchik [111]2 years ago

7 0

Answer:

they are all renewable resources!

Explanation:

Hope this helps! ;-)

nordsb [41]2 years ago

4 0

It’s all can be used as the source of the energy

You might be interested in

Compare the rates of effusion for carbon monoxide and carbon dioxide gases

Airida [17]

According to Grahams law the rate of effusion of a gas is inversely proportional to the square root of it's molecular weight. The rate of diffusion is the measure of rate at which two gases mix. From this law we can say that for the two gases carbon monoxide and carbon dioxide, the rate of effusion of carbon monoxide is greater than that of carbon dioxide, this is because carbon monoxide is lighter (28 g) compared to carbon dioxide (44 g).

7 0

3 years ago

The attraction between particles give solids a definite

Harman [31]

Answer:Shape and volume

Explanation:

5 0

4 years ago

1 Na2CO3(aq) + 1 CaCl2(aq) → 1 CaCO3(s) + 2 NaCl(aq) 4. Use the balanced chemical equation from the last question to solve this

LenKa [72]

<h3>Answer:</h3>

0.6 g NaCl

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] Na₂CO₃ (aq) + CaCl₂ (aq) → CaCO₃ (s) + 2NaCl (aq)

[Given] 0.5 g Na₂CO₃ reacted with excess CaCl₂

<u>Step 2: Identify Conversions</u>

[RxN] Na₂CO₃ → 2NaCl

Molar Mass of Na - 22.99 g/mol

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of Na₂CO₃ - 2(22.99) + 12.01 + 3(16.00) = 105.99 g/mol

Molar Mass of NaCl - 22.99 + 35.45 = 58.44 g/mol

<u>Step 3: Stoichiometry</u>

Set up: $\displaystyle 0.5 \ g \ Na_2CO_3(\frac{1 \ mol \ Na_2CO_3}{105.99 \ g \ Na_2CO_3})(\frac{2 \ mol \ NaCl}{1 \ mol \ Na_2CO_3})(\frac{58.44 \ g \ NaCl}{1 \ mol \ NaCl})$
Multiply/Divide: $\displaystyle 0.551373 \ g \ NaCl$