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3 years ago

What mass of lead (density 11.4 g/cm3) would have an identical volume to 28.5 g of mercury (density 13.6 g/cm3)?

1 answer:

6 0

The density of a material is its unique property which is different for different substances. The volume or mass of a compound can be determined from the other given mass of a compound, if the density of both the material is available. We know density of a material is equal to the mass per unit volume. From the given data 13.6 g of mercury has 1 cm³ volume. Thus 28.5g of mercury will have $\frac{28.5}{13.6}=2.095$ cm³ volume. Now as per the given density value of lead 11.4g of lead has 1 cm³ volume. Thus 2.095 cm³ volume of lead has (2.095×11.4)=23.889g.

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Helga [31]

Your answer would be D you welcome !

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3 years ago

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Campare and Conteast Physical and chemical properties.

I am Lyosha [343]

Answer:

Physical properties are changes in color, weight, mass, density, volume, whether it is freezing, tarnishing, boiling, etc. while chemical properties are changes in the compound of the object such as whether an object is/isn't flammable, affected by acid, it dissolves, ferments, rots, ignites in water, etc.

Explanation:

Physical property examples: a piece of steel is cut in half, a sugar cube is ground up into a powder, ice melts into water, clouds form in the sky, you mow the lawn, the water turned purple when you dropped red and blue dye in it, leaves grow in the spring, ethyl alcohol boils at 79° C, you cut apples up

Chemical property examples: pancakes cook on the griddle, fireworks explode in th sky, Alka-Seltzer gives off carbon dioxide when added to water, roasting a marshmellow, milk sours when left on the counter, burning wood in a bonfire, salt dissolves in water, food is digested in the stomach

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3 years ago

Q82. Liquid petrol does not ignite spontaneously when exposed to the air because

sashaice [31]

Liquid petrol does not ignite spontaneously when exposed to the air because there is not enough molecules that contains energy to reach the activation energy. As a result, the reaction cannot proceed and a supply of energy should be added for the reaction to occur.

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3 years ago

Liquid nitrogen has a density of 0.808 g/mL and boils at 77 K. Researchers often purchase liquid nitrogen in insulated 185 L tan

Serga [27]

Answer:

37.14 %

Explanation:

Using the equation, mass, M = D1 * V1

= D2 * V2

Where,

D1 = density of the liquid Nitrogen

D2 = density of gaseous Nitrogen

V1 = volume of the liquid Nitrogen

V2 = volume of the gaseous Nitrogen

Calculating V2,

0.808 * 185 = 1.15 * V2

Volume of Nitrogen after expansion = 129.98 m3.

Volume = L * b * h

= 10 * 10 * 3.5

Volume of the room = 350 m3.

Fraction of air = volume of Nitrogen after expansion/volume of the room * 100

= 129.98/350 *100

= 37.14 %

5 0

3 years ago

What is the percent composition of NaHCO3?

zhuklara [117]

Answer:

Option-C (27.36% Na, 1.20% H, 14.30% C, and 57.14% O)

Explanation:

Percent Composition is defined as the %age by mass of each element present in a compound. Therefore, it is a relative amount of each element present in a compound.

Calculating Percent Composition of NaHCO₃:

1: Calculating Molar Masses of all elements present in NaHCO₃:

a) Na = 22.99 g/mol

b) H = 1.01 g/mol

c) C = 12.01 g/mol

d) O₃ = 16.0 × 3 = 48 g/mol

2: Calculating Molecular Mass of NaHCO₃:

Na = 22.99 g/mol

H = 1.01 g/mol

C = 12.01 g/mol

O₃ = 48 g/mol

----------------------------------

Total 84.01 g/mol

3: Divide each element's molar mass by molar mass of NaHCO₃ and multiply it by 100:

For Na:

= 22.99 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

= 27.36 %

For H:

= 1.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

= 1.20 %

For C:

= 12.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

= 14.29 % ≈ 14.30 %

For O:

= 48.0 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

= 57.13 % ≈ 57.14 %

8 0

3 years ago