As the mass number of the isotopes of hydrogen increases, the number of protons

a mixture of carbon and sulfur has a mass of 9.0 g. complete combustion with excess o2 gives 22.5 g of a mixture of co2 and so2.

photoshop1234 [79]

S + O2 → SO2
z / (32.0655 g S/mol) x (1 mol SO2 / 1 mol S) x (64.0638 g SO2/mol) = (1.9979 z) g SO2 

C + O2 → CO2 
(9.0-z) / (12.01078 g C/mol) x (1 mol CO2 / 1 mol C) x (44.00964 g CO2/mol) = (32.9776 - 3.66418 z) g CO2 

Add the two masses of SO2 and CO2 and set them equal to the amount given in the problem: 
(1.9979 z) + (32.9776 - 3.66418 z) = 27.9 
Solve for z algebraically: 
z = 3.0 g S

7 0

3 years ago

Which of the following statements concerning the density of a gas is true?

Naya [18.7K]

Answer:

can u give us the options

5 0

3 years ago

When determining the specific heat from a given metal in a calorimeter the mass of the water remains constant, but the mass of t

alexandr1967 [171]

The heat is hotter than cold

6 0

3 years ago

If the amount of copper is 6g and whole ratio to obtain copper sulfide is 4:1. How much sulfur is required for the reaction to o

Yuki888 [10]

Answer:

1.5g

Explanation:

According to this question, the amount of copper required to obtain copper sulfide in a 4:1 is 6g. This means that the ratio of copper to sulfur in the compound (copper sulfide) is 4:1.

Hence, to calculate the amount of sulfur required for the reaction to obtain copper sulfide using the above ratio, we say:

1/4 of the amount of copper required (6g)

= 1/4 × 6

= 6/4

= 1.5g of sulfur is required for the reaction to obtain carbon sulfide.

4 0

2 years ago

En un matraz, disponemos de 100 g de gas oxígeno que se encuentran a 1 at de presión y 273 K de temperatura. Calcular : a) el nú

Misha Larkins [42]

Answer:

Explanation:

Dado que:

masa de oxígeno gaseoso = 100 g

presión = 1 atm

temperatura = 273 K

(a)

número de moles de oxígeno contenidos en el matraz = masa de oxígeno / masa molar de oxígeno

= 100 g / 16 gmol⁻¹

= 6.25 moles

(b) El número de moléculas de oxígeno es el siguiente:

Dado que 1 mol de oxígeno gaseoso contiene 6.023 * 10²³ moléculas de oxígeno.

Entonces, 6.25 moles contendrán:

= (6.25 × 6.023 * 10²³) moléculas de oxígeno.

≅ 3.764 × 10²³ moléculas de oxígeno.

(c) El número de átomos de oxígeno es:

= 2 × 3.764 × 10²³

= 7.528 × 10²³ átomos de oxígeno

(d) Usando la ecuación de gas ideal

PV = nRT

El volumen ocupado por el oxígeno = $\dfrac{nRT}{P}$

Volumen ocupado por oxígeno = $\dfrac{ 6.25 * 8.314 *273}{1}$

Volumen ocupado por oxígeno= 14185.76 m³

3 0

2 years ago