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tatiyna
3 years ago
6

ilver bromide is used to coat ordinary black-and-white photographic film, while high-speed film uses silver iodide. (a) When 56.

6 mL of 5.00 g/L AgNO3 is added to a coffee-cup calorimeter containing 56.6 mL of 5.00 g/L NaI, with both solutions at 25°C, what mass of AgI forms?
Chemistry
1 answer:
olya-2409 [2.1K]3 years ago
8 0

Answer:

Mass of AgI formed=0.352g

Explanation:

AgNO3 + NaI ----‐>AgI + NaNO3

Volume of AgNO3 =5g/L

Mass of AgNO3 =C×V=0.25g

n(AgNO3)= m/Mr=0.0015mol

For NaI

V= 5g/l

Mass= C×v=0.25g

n=m/Mr=0.0017mol

n=0.0015mol

Mass=n× mr=0.352g

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5 0
3 years ago
When 91.5 g of isopropyl alcohol which has an empirical formula of C3H8O is burned in excess oxygen gas, how many grams of H2O a
Kobotan [32]

Answer:

109.7178g of H2O

Explanation:

First let us generate a balanced equation for the reaction. This is illustrated below:

2C3H8O + 9O2 —> 6CO2 + 8H2O

Next we will calculate the molar mass and masses of C3H8O and H20. This is illustrated below:

Molar Mass of C3H8O = (3x12.011) + (8x1.00794) + 15.9994 = 36.033 + 8.06352 + 15.9994 = 60.09592g/mol.

Mass of C3H8O from the balanced equation = 2 x 60.09592 = 120.19184g

Molar Mass of H2O = (2x1.00794) + 15.9994 = 2.01588 + 15.9994 = 18.01528g/mol

Mass of H2O from the balanced equation = 8 x 18.01528 = 144.12224g

From the equation,

120.19184g of C3H8O produced 144.12224g of H20.

Therefore, 91.5g of C3H8O will produce = (91.5 x 144.12224) /120.19184 = 109.7178g of H2O

7 0
3 years ago
A chemistry student adds a quantity of an unknown solid compound X to 5.00 L of distilled water at 15.° C . After 10 minutes of
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Answer:

34 g/100 mL

Explanation:

The solubility of a compound can be expressed in g/100mL, for this we must divide the mass of the compound that dissolves in the solute by the volume of the solvent.

The solvent, in this case, is water, and that mass of the solute X that dissolved is the mass that was recovered after the solvent was drained and evaporated. So the solubility of X (S) is:

S = 0.17 kg/5L

S = 170g/5000mL

S = 170g/(5*1000)mL

S = 34 g/100 mL

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