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kap26 [50]
3 years ago
8

what is the balanced equation of magnesium carbonate decomposes on heating to form magnesium oxide and carbon dioxide​

Chemistry
1 answer:
fenix001 [56]3 years ago
6 0

Answer:

ExplaAt high temperatures MgCO3 decomposes to magnesium oxide and carbon dioxide. This process is important in the production of magnesium oxide. This process is called calcining: MgCO3 → MgO + CO2 (ΔH = +118 kJ/mol)

nation:

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First 20 chemical elements in the periodic table​
kolbaska11 [484]
This should be able to help you

3 0
3 years ago
How much ice (in grams) would have to melt to lower the temperature of 353 mL of water from 26 ∘C to 6 ∘C? (Assume the density o
Rashid [163]

Answer:

The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg

Explanation:

Heat gain by ice = Heat lost by water

Thus,  

Heat of fusion + m_{ice}\times C_{ice}\times (T_f-T_i)=-m_{water}\times C_{water}\times (T_f-T_i)

Where, negative sign signifies heat loss

Or,  

Heat of fusion + m_{ice}\times C_{ice}\times (T_f-T_i)=m_{water}\times C_{water}\times (T_i-T_f)

Heat of fusion = 334 J/g

Heat of fusion of ice with mass x = 334x J/g

For ice:

Mass = x g

Initial temperature = 0 °C

Final temperature = 6 °C

Specific heat of ice = 1.996 J/g°C

For water:

Volume = 353 mL

Density (\rho)=\frac{Mass(m)}{Volume(V)}

Density of water = 1.0 g/mL

So, mass of water = 353 g

Initial temperature = 26 °C

Final temperature = 6 °C

Specific heat of water = 4.186 J/g°C

So,  

334x+x\times 1.996\times (6-0)=353\times 4.186\times (26-6)

334x+x\times 11.976=29553.16

345.976x = 29553.16

x = 85.4197 kg

Thus,  

<u>The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg</u>

7 0
4 years ago
Explain HOW this chemical reaction is NOT balanced correctly.
NeTakaya

Answer: there is more atoms of each element on the right.

Explanation:

On the left of the arrow you have:

Si-1

F-4

H-4

O-2

On the right of the arrow you have:

Si-3

F-6

H-8

O-4

A balanced equation always has the same amount of each element on the left of the arrow (the reactants) as the right (the products). This isn’t balanced because you have a lot more of each element on the right.

5 0
3 years ago
Read 2 more answers
Which option below describes a chemical property of a substance?
boyakko [2]

Answer:

C

Explanation:

Rust is a chemical change (reaction) all of the other options only change the state meaning that it is a physical reaction.

4 0
3 years ago
Read 2 more answers
12. How many grams of glucose (C6H1206) is produced if 17.3 mol of H20 is reacted according to this
iren2701 [21]

Answer:

1. 17.3 MOLES OF WATER WILL PRODUCE 518.4 G OF GLUCOSE.

2. 87.4 G OF HNO3 WILL PRODUCE 7.86 G OF NITROGLYCERIN WHEN REACTED WITH EXCESS GLYCEROL.

Explanation:

1. How many grams of glucose is produced from 17.3 mole of water?

Equation:

6CO2 + 6H20 ------> C6H12O6 + 6O2

From the reaction, 6 moles of carbon dioxide reacts with 6 moles of water to produce 1 mole of glucose

So therefore,

6 moles of water = 1 mole of glucose

Since 17.3 mole of water was used, we can first calculate the number of moles of glucose produced:

Then, we have:

6 moles of water = 1 mole of glucose

17.3 moles of water = ( 17.3 * 1/ 6) moles of glucose

= 2.883 moles of glucose

So we say 17.3 moles of water produces 2.883 moles of glucose

At standard conditions, 1 mole of a substance is its molar mass

Molar mass of water = 18 g/mol

Molar mass of glucose = 180 g/mol

From the reaction:

17.3 moles of water produces 2.883 moles of glucose

17.3 * 18 g of water produces 2.833 * 180 g of glucose

= 518.94 g of glucose.

From 17.3 mole of water, 518.4 g of glucose will be produced.

2.

C3H5(OH)3 + 3HNO3 -------> C3H5(ONO2)3 + 3H20

3 moles of HNO3 reacts to produce 1 mole of nitroglycerin

Molar mass of HNO3 = ( 1 + 14 + 16*3) = 63 g/mol

Molar mass of nitroglycerin = ( 12 *3 + 1 *5 + 16*6 + 14*3) = 179 g/mol

3 moles of HNO3 = 1 mole of nitroglycerin

3 * 63 g of HNO3 = 179 g of nitroglycerin

if 87.4 g of HNO3 were to be reacted, we have:

189 g of HNO3 = 179 g of nitroglycerin

87.4 g of HNO3 = ( 87.4 * 179 / 189) of nitroglycerin

= 7.86 g of nitroglycerin

So therefore, 87.4 g of HNO3 will produce 7.86 g of nitroglycerin when reacted with excess glycerol.

4 0
4 years ago
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