There are 0.566 moles of carbonate in sodium carbonate.
<h3>CALCULATE MOLES:</h3>
- The number of moles of carbonate (CO3) in sodium carbonate (Na2CO3) can be calculated by dividing the mass of carbonate in the compound by the molar mass of the compound.
- no. of moles of CO3 = mass of CO3 ÷ molar mass of Na2CO3
- Molar mass of Na2CO3 = 23(2) + 12 + 16(3)
- = 46 + 12 + 48 = 106g/mol
- mass of CO3 = 12 + 48 = 60g
- no. of moles of CO3 = 60/106
- no. of moles of CO3 = 0.566mol
- Therefore, there are 0.566 moles of carbonate in sodium carbonate.
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Answer:
coral bleaching is when the temperature of the water changes (even slight changes in temperature effect coral) and the coral reacts negatively to it causing it to die. unfortunately corals cannot recover once they have been bleached.
Answer:
There are
grams contained in all the seawater in the world.
Explanation:
At first let is determinate the total mass of seawater (
), measured in grams, in the world by definition of density and considering that mass is distributed uniformly:

Where:
- Density of seawater, measured in grams per liters.
- Volume of seawater, measured in liters.
If
and
, then:


The total mass of sodium chloride is determined by the following ratio:


Given that
and
, the total mass of sodium chloride in all the seawater in the world is:

There are
grams contained in all the seawater in the world.
Answer:
The answer to your question is 34 g of H₂O₂
Explanation:
Data
mass of H₂O₂ = ?
mass of O₂ = 32 g
Process
1.- Write the balanced chemical reaction
H₂O₂ ⇒ H₂ + O₂
2.- Calculate the molar mass of H₂O₂ and O₂
Molar mass H₂O₂ = (1 x 2) + (16 x 2) = 34 g
Molar mass of O₂ = 2 x 16 = 32 g
3.- Use proportions to calculate the mass of O₂
34g of H₂O₂ -------------- 32g of O₂
x -------------- 32g of O₂
x = (32 x 34) / 32
x = 34 g of H₂O₂
Answer:
0.0613 L
Explanation:
Given data
- Initial pressure (P₁): 1.00 atm
- Initial volume (V₁): 1.84 L
- Final pressure (P₂): 30.0 atm
Since we are dealing with an ideal gas, we can calculate the final volume using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁ / P₂
V₂ = 1.00 atm × 1.84 L / 30.0 atm
V₂ = 0.0613 L