Determine the empirical formula of a compound that contains 30.45% nitrogen and 69.55% oxygen

The vapor pressure of a liquid __________.

mash [69]

Answer:

B

Explanation:

The tendency of a liquid to convert to vapour increases smoothly with increasing temperature. Vapour pressure shows the tendency of a liquid to convert to vapour. Increase In vapour pressure shows an increased tendency to convert to vapour. The higher the temperature, the higher the vapour pressure.

6 0

3 years ago

What functional group is present in all carbohydrates? chemistry?

Oxana [17]

Carbohydrates contain a carbonyl group =C=O and hydroxyl groups -OH.

7 0

3 years ago

Nitric acid, a major industrial and laboratory acid, is produced commercially by the multi-step Ostwald process, which begins wi

Marizza181 [45]

Answer: 3.178 Kg of ammonia must be used to produce 7.839 kg of nitric acid and The equations are not balanced. Explanation: Ostwald process is a multi-step for manufacturing Nitric Acid. First, We must check if our equations are balanced or not, by checking the amount of each element before and after the arrow. Step 1: NH3(g) + O2(g) ⟶ NO(g) + H2O(l) Step 2: NO(g) + O2(g) ⟶ NO2(g) Step 3: NO2(g) + H2O(l) ⟶ HNO3(l) + NO(g) For example in Step 1, the amount of Hydrogen atoms are different on both sides. On left side we can count 3 hydrogen atoms while on the right side we can count only 2 hydrogen atoms. In the Step 2, the amount of Oxygen atoms are different on both sides too. On the Left side we can count 3 oxygen atoms while on the right side we can count only 2 oxygen atoms and in step 3 the amount of Nitrogen atoms are different on both sides too. On the left side we can count 1 nitrogen atom while on the right side we can count 2 nitrogen atoms. So, Let´s balance equations by inspection, just adding coefficients to each compound, until the amount of atoms are equal on both sides. Step 1: 4NH3(g) + 5O2(g) ⟶4 NO(g) +6 H2O(l) Step 2: 2NO(g) + O2(g) ⟶ 2NO2(g) Step 3: 3NO2(g) + H2O(l) ⟶ 2HNO3(l) + NO(g) Second we gather the information what we are going to use in our calculations. Final Mass of HNO3 = 7,839Kg and Molecular Weigth = 63,01g/mol NH3 Molecular Weight= 17,031 g/mol Third we start discoverying the amount of NH3 that reacted completed to generate 7,839Kg of HNO3, using the giving equation and its respective molecular weights. 7,839Kg of HNO3 x 1000g / 1Kg x 1mol HNO3 / 63,01g HNO3 x 3mol NO2 / 2 mol HNO3 x 2 mol NO/ 2 mol NO2 x 4 mol NH3/4 mol NO x 17,031g NH3/1 mol NH3 x 1 Kg / 1000g= 3,178 Kg NH3 The amount of NH3 that is required to produce 7,839 Kg of HNO3 is 3,178 Kg NH3

7 0

3 years ago

Hi :)<br> take some points for the road

Maurinko [17]

Answer:

ty

Explanation:

6 0

2 years ago

Read 2 more answers

You want to determine ΔH o for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) To do so, you first determine the heat capacity

Assoli18 [71]

Answer:

(A) The heat capacity of the calorimeter is therefore = −2.1428KJ÷13.5°C

= −0.1587KJ/°C

(B) ΔHo for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) = –15.42KJ

Explanation:

Solution

Calculate the heat actually evolved.

q = mcΔt

Finding the mass of the reactants in grams we have.

Use density. (50 mL + 50 mL ) = 100 mL of solution.

100 mL X 1.04g/mL = 104 grams of solution. (mass = Volume X Density)

Find the temperature change.

Δt =tfinal - tinitial = 30.4°C – 16.9°C = 13.5°C

q = mcΔt

= 104grams × 3.93J/g°C × 13.5°C = 5.51772×103J

= 5.51772 × 103 J

This is the heat lost in the reaction between HCl and NaOH, therefore q = -5.52 × 103 J.

this is an exothermic heat producing reaction.

To calculate the total heat of the reaction or heat per mole we have

50.0 mL of HCl X 2.00 mol HCl /(1000 mL HCl ) = 0.100 mol HCl

The same quantity of base, 0.100 mole NaOH, was used.

The energy per unit mole is given by

i.e. molar enthalpy = J/mol = -5.52 × 103J / 0.100 mol

= -5.52 × 104 J/mol

= -55177.2 J/mol

= -55.177 kJ/mol

Therefore, the enthalpy change for the neutralization of HCl and NaOH, that is the enthalpy, heat, of reaction is ΔH = -55.177 kJ/mol

Heat absorbed by the calorimeter = −57.32kJ − 55.177 kJ = −2.1428KJ

The heat capacity of the calorimeter is therefore = −2.1428KJ÷13.5°C

= −0.1587KJ/°C

(B) For the ZnCl we have

Calculate the heat actually evolved.

q = mcΔt

Finding the mass of the reactants in grams we have.

Use density. 100 mL of solution of HCl

100 mL X 1.015g/mL = 101.5 grams of solution. (mass = Volume X Density)

Find the temperature change.

Δt =tfinal - tinitial = 20.5°C – 16.8°C = 3.7 °C

q = mcΔt

= 101.5grams × 3.95J/g°C × 3.7°C = 1483.422×103J

= -1483.422×103J

This is the heat lost in the reaction between HCl and NaOH, therefore q = -1.483 × 103 J.

this is an exothermic heat producing reaction.

To calculate the total heat of the reaction or heat per mole we have

100.0 mL of HCl X 1.00 mol HCl /(1000 mL HCl ) = 0.100 mol HCl

The energy per unit mole is given by

i.e. molar enthalpy = J/mol = -1.483 × 103J / 0.100 mol

= -1.483 × 104 J/mol

= -14834.22 J/mol

= -14.834 kJ/mol

Therefore, the enthalpy change for the neutralization of HCl and NaOH, that is the enthalpy, heat, of reaction is ΔH = -14.834 kJ/mol

ΔHo for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

= -14.834 kJ –(0.1587KJ/°C×3.7°C) = -15.42KJ

ΔHo for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) = –15.42KJ

5 0

2 years ago