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MaRussiya [10]
3 years ago
13

The density of a metal is 10.5 g/cm3. If the mass of themetal

Chemistry
1 answer:
vitfil [10]3 years ago
4 0

Answer:

11.7 mL

Explanation:

Density of a substance is given by the mass of the substance divided by the volume of the substance .

Hence , d = m / V

V = volume

m = mass ,  

d = density ,

From the question ,

The density of the metal = 10.5 g/cm³

The mass of the metal = 5.25 g

Hence , the volume can be calculated from the above formula , i.e. ,

d = m / V  

V = m / d

V = 5.25 g / 10.5 g/cm³

V = 0.5 cm³

Since , The unit 1 mL = 1 cm³

V = 0.5 mL

The vessel has 11.2 mL of water ,

The new volume becomes ,

11.2 mL +  0.5 mL = 11.7 mL

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7. How many chlorine atoms are in a 34.2 g sample of dichlorine pentoxide?
Reptile [31]

Answer:

The answer to your question is 1.36 x 10²³ atoms

Explanation:

Data

number of atoms = ?

mass of the sample = 34.2 g

Molecule = Cl₂O₅

Process

1.- Calculate the molar mass of Cl₂O₅

Cl₂O₅ = (35.5 x 2) + (16 x 5) = 71 + 80 = 151 g

2.- Calculate the atoms of Cl₂O₅

                     151 g of Cl₂O₅ ---------------- 6 .023 x 10²³ atoms

                       34.2 g of Cl₂O₅ ------------ x

                          x = (34.2 x 6.023 x 10²³) / 151

                          x = 1.36 x 10²³ atoms

4 0
4 years ago
Balance the following equations. Do not include the states of matter.<br><br> (a) C + O2 → CO
krek1111 [17]

Answer:

C + O2 → CO2

Explanation:

C + O2 → CO ----------------- (1)

from equ (1) on reactant side, C has 1 mole, O has 2 moles

from equ (1) on product side, C has 1 mole, O has 1 mole

Thus, to balance the equation, O should have 2 moles

C + O2 → CO2

7 0
3 years ago
Read 2 more answers
Hydrogen sulfide burns form sulfur dioxide:
Helga [31]

Answer: 404.04 kJ.

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

moles of H_2S

\text{Number of moles}=\frac{26.7g}{34.1g/mol}=0.78moles

2H_2S(g)+3O_2(g)\rightarrow 2SO_2(g)+2H_2O(g)    \Delta H=-1036kJ

According to stoichiometry :

2 moles of H_2S on burning produces = 1036 kJ

Thus 0.78 moles of H_2S on burning produces =\frac{1036kJ}{2}\times 0.78=404.04

Thus the enthalpy change when burning 26.7 g of hydrogen sulfide is 404.04 kJ.

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3 years ago
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Answer:

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