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Mumz [18]
3 years ago
6

When ammonium carbonate decomposes what gas is produced and will a glowing splint burn brighter in the presence?

Chemistry
1 answer:
mestny [16]3 years ago
7 0
When ammonium carbonate is heated it decomposes to give ammonia gas, carbon dioxide gas and water. The equation for the decomposition is;
(NH4)2CO3(s) = 2 NH3 (g) + CO2(g) + H2O(l)
A glowing splint would extinguish almost immediately because of the presence of carbon dioxide. Carbon dioxide does not support burning which is the property that makes it used a s a fighter extinguisher.
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Are produced along with a large quantitu of heat
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gThe mole fraction of potassium nitrate in an aqueous solution is 0.0194. The solution's density is 1.0627 g/mL. What is the mol
xxMikexx [17]

Answer:

0.595 M

Explanation:

The number of moles of water in 1L = 1000g/18g/mol = 55.6 moles of water.

Mole fraction = number of moles of KNO3/number of moles of KNO3 + number of moles of water

0.0194 = x/x + 55.6

0.0194(x + 55.6) = x

0.0194x + 1.08 = x

x - 0.0194x = 1.08

0.9806x= 1.08

x= 1.08/0.9806

x= 1.1 moles of KNO3

Mole fraction of water= 55.6/1.1 + 55.6 = 0.981

If

xA= mole fraction of solvent

xB= mole fraction of solute

nA= number of moles of solvent

nB = number of moles of solute

MA= molar mass of solvent

MB = molar mass of solute

d= density of solution

Molarity = xBd × 1000/xAMA ×xBMB

Molarity= 0.0194 × 1.0627 × 1000/0.981 × 18 × 0.0194×101

Molarity= 20.6/34.6

Molarity of KNO3= 0.595 M

8 0
3 years ago
Relative formula mass of glucose? (C6H12O6)
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To find the mass of glucose, you must multiply the atomic weight of each of the elements in the molecule by the subscripts in the formula:

C_{6}=6*(12.01g)=72.06g

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O_{6}=6*(15.999g)=95.994g

Then you add all of them together:

72.06g+12.096g+95.994g=180.15g

Therefore, the molar weight of glucose is 180.15 grams.

3 0
3 years ago
How many sulfur atoms are present in 25.6 g of Al2(S2O3)3
IgorC [24]

Given the mass of Al_{2}(S_{2}O_{3})_{3}=25.6 g

The molar mass of Al_{2}(S_{2}O_{3})_{3}=390.35g/mol

Converting mass of Al_{2}(S_{2}O_{3})_{3}to moles:

25.6 g Al_{2}(S_{2}O_{3})_{3}*\frac{1molAl_{2}(S_{2}O_{3}}{390.35 gAl_{2}(S_{2}O_{3}} =0.0656molAl_{2}(S_{2}O_{3}

Converting mol Al_{2}(S_{2}O_{3})_{3}to mol S:

0.0656mol Al_{2}(S_{2}O_{3})_{3}*\frac{6molS}{1mol Al_{2}(S_{2}O_{3})_{3}}=0.3936 molS

Converting mol S to atoms of S using Avogadro's number:

1 mol = 6.022*10^{23}atoms

0.3936mol S *\frac{6.022*10^{23}atoms S}{1 mol S}=2.37*10^{23} S atoms

5 0
3 years ago
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