16.34 g of CuSO4 dissolved in water giving out 55.51 kJ and 25.17 g CuSO4•5H2O absorbs 95.31 kJ. From the following reaction cyc

The result is 324.3 grams. In this problem, we have 6.0 moles of nitrous acid and need to convert it into grams. This conversion requires multiplying the moles by the molar mass of the substance. Molar mass is computed from the formula mass, with atomic masses of each constituent atom multiplied by their subscripts. For instance, A single nitrogen atom and a single oxygen atom yield the molar mass of the formula calculated as: atomic mass of N + 2(atomic mass of O) = 14 + 2(16) = 14 + 32 = 46 gram per mol. This molar mass indicates that 1 mole of the compound weighs 46 grams. Now let’s compute the molar mass of nitrous acid similarly. Its molar mass would be 1 + 14 + 2(16) = 1 + 14 + 32 = 47 grams per mol. Subsequently, the grams equivalent of 6.9 moles of nitrous acid calculates to 324.3 g.

Answer:

Explanation:

Hello,

In this scenario, since 11 mg of cyanide per kilogram of body weight is lethal for 50% of domestic chickens, we can determine the lethal amount for a chicken weighing 3 kg using a proportion:

Thus, we find:

Which equates to grams as follows:

Regards.

When dealing with an unsaturated NaNO3(aq) solution at standard pressure, the boiling point increases when...

5.451 X 10³ kg of sodium carbonate is required to neutralize 5.04×10³ kg of sulfuric acid solution.

Explanation:

• Sodium carbonate neutralizes sulfuric acid (H₂SO₄). This compound is derived from a strong base (NaOH) and a weak acid (H₂CO₃). The chemical equation for this neutralization process is represented as:

Na₂CO₃ + H₂SO₄ ----> Na₂SO₄ + H₂CO₃

• The balanced equation indicates that one mole of Na₂CO₃ is needed to neutralize one mole of H₂SO₄.

• Molar mass of Na₂CO₃= 106 g/mol = 0.106 kg/mol, while Molar mass of H₂SO₄= 98 g/mol = 0.098 kg/mol.

• To neutralize 0.098 kg of H₂SO₄, the required Na₂CO₃ is 0.106 kg, thus, to neutralize 5.04×10³ kg of H₂SO₄, Na₂CO₃ needed is 5.451 X 10³ kg.

In the reaction: <span>caco3(s) → cao(s) + co2(g), it is evident that
1 mol (which is 100 g) of CaCO3 yields 1 mol (which is 44 g) of CO2
Now, the molarity of CaCO3 present in the reaction system is
</span>= 
=  = 0.45 mol

Thus, 0.45 mol of CaCO3 leads to the formation of 0.45 mol of CO2.

According to the ideal gas equation, we have PV = nRT
V = .
Considering P = 645 torr = 0.8487 atm (because 1 atm = 760 torr)
In that case, V = 

= 34.8 l