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Effectus [21]
2 years ago
6

Chemistry gcse

Chemistry
1 answer:
mart [117]2 years ago
8 0
Yes, anything with carbonate, hydrogen carbonate (bicarbonate) at the end is a carbonate.

Examples:NaHCO3 (Sodium hydrogen carbonate or Sodium bicarbonate)Na2CO3 (Sodium carbonate)
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Sort the vocabulary word below. Make sure you have come to live session or watched the 5.03 recording. These words will be impor
Stells [14]

Answer:

Electric Current

This is a flow of electrons (such as the flow of electrons in a wire to light up a lamp).

Insulator

These stop the electricity from flowing (think rubber and glass).

Resistance

This is slowing the flow of electrons (the current) and where some of the electrons' energy gets converted into heat.

Conductor

These allow electricity to flow easily through them (think copper and aluminum).

5 0
3 years ago
If it takes 38.70cm of 1.90M NaOH to neutralize 10.30cm of H2SO4 in a battery, what is the molarity of H2SO4?
slamgirl [31]

Answer:

The molarity of the acid, H₂SO₄ is 3.57 M

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O

From the balanced equation above,

Mole ratio of the acid, H₂SO₄ (nₐ) = 1

Mole ratio of the base, NaOH (n₆) = 2

Finally, we shall determine the molarity of the acid, H₂SO₄. This can be obtained as follow:

Volume of base, NaOH (V₆) = 38.70 cm³

Molarity of base, NaOH (M₆) = 1.90M

Volume of acid, H₂SO₄ (Vₐ) = 10.30 cm³

Molarity of acid, H₂SO₄ (Mₐ) =?

MₐVₐ / M₆V₆ = nₐ/n₆

Mₐ × 10.3 / 1.9 × 38.70 = 1/2

Mₐ × 10.3 / 73.53 = 1/2

Cross multiply

Mₐ × 10.3 × 2 = 73.53 × 1

Mₐ × 20.6 = 73.53

Divide both side by 20.6

Mₐ = 73.53 / 20.6

Mₐ = 3.57 M

Thus, the molarity of the acid, H₂SO₄ is 3.57 M

8 0
2 years ago
Why do we have rain........................
VladimirAG [237]

Answer:

to watered our source of food

the "PLANTS & TREES"

6 0
2 years ago
8. A 15.0 gram unknown piece of metal at 99 oC is added to 75 grams of water at 23 oC. The final temperature of the metal and wa
Setler79 [48]

The specific heat of the unknown metal is 0.859 J/g⁰C.

<h3>Specific heat of the unknown metal</h3>

Heat lost by the unknown metal = Heat gained by water

m₁c₁Δθ₁ = m₂c₂Δθ₂

where;

  • c₁ is specific heat of the unknown metal
  • c₂ is specific heat of water

(15)(c₁)(99 - 26) = (75)(4.18)(26 - 23)

1095c₁ = 940.5

c₁ = 940.5/1095

c₁ = 0.859 J/g⁰C

Learn more about specific heat here: brainly.com/question/16559442

#SPJ1

3 0
2 years ago
Jorge conducted an experiment, and included the graph shown below as part of his lab report.
sergiy2304 [10]

Answer:physical change

Explanation:

8 0
2 years ago
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