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Soloha48 [4]
2 years ago
7

Calculate the density of an object with a mass of 3.8 g, that when placed in a 10.0 mL graduated cylinder with an initial volume

of 4.5 mL rises to a final volume of 8.6 mL. *
Chemistry
1 answer:
Arturiano [62]2 years ago
3 0

Answer : The density of an object is 0.93 g/mL

Explanation : Given,

Mass of an object = 3.8 g

Initial volume = 4.5 mL

Final volume = 8.6 mL

First we have to calculate the volume of an object.

Volume of an object = Final volume - Initial volume

Volume of an object = 8.6 mL - 4.5 mL

Volume of an object = 4.1 mL

Now we have to calculate the density of an object.

Formula used:

\text{Density}=\frac{\text{Mass of an object}}{\text{Volume of an object}}

Now putting all the given values in this formula, we get:

\text{Density}=\frac{3.8g}{4.1mL}

\text{Density}=0.93g/mL

Therefore, the density of an object is 0.93 g/mL

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Suppose 50.0g of silver nitrate is reacted with 50g of hydrochloric acid producing silver chloride and a mixture of other produc
docker41 [41]

Answer:

53.6 grams of silver chloride was produced.

Explanation:

AgNO_3+HCl+\rightarrow AgCl+HNO_3

Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

Mass of silver nitrate = 50.0 g

Mass of hydrogen chloride = 50.0 g

Mass of silver chloride = x

Mass of  nitric acid = 46.4 g

Mass of silver nitrate + Mass of hydrogen chloride =

                             Mass of silver chloride + Mass of  nitric acid

[te]50.0 g+50.0 g=x+46.4 g[/tex]

x=50.0 g+50.0 g - 46.4 g = 53.6 g

53.6 grams of silver chloride was produced.

8 0
3 years ago
Materials expand when heated. Consider a metal rod of length L0 at temperature T0. If the temperature is changed by an amount ΔT
marysya [2.9K]

Answer:

(a) The length at temperature 180°C is 40.070 cm

(b) The length at temperature 90°C is 64.976 inches

(c) L(T, α) = 60·α·T - 9000·α + 60

Explanation:

(a) The given parameters are

The thermal expansion coefficient, α for steel = 1.24 × 10⁻⁵/°C

The initial length of the steel L₀ = 40 cm

The initial temperature, t₀ = 40°C

The length at temperature 180°C = L

Therefore, from the given relation, for change in length, ΔL, we have;

ΔL = α × L₀ × ΔT

The amount the temperature changed ΔT = 180°C - 40°C = 140°C

Therefore, the change in length, ΔL, is found as follows;

ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 40 × 140°C = 0.07 cm

Therefore, L =  L₀ + ΔL = 40 + 0.07 = 40.07 cm

The length at temperature 180°C = 40.07 cm

(b) Given that the length at T = 120°C is 65 in., we have;

The temperature at which the new length is sought = 90°C

The amount the temperature changed ΔT = 90°C - 120°C = -30°C

ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 65 × -30°C = -0.024375 inches

The length, L at 90°C is therefore, L = L₀ + ΔL = 65 - 0.024375 = 64.976 in.

The length at temperature 90°C = 64.976 inches

(c) L = L₀ + ΔL  = L₀ +  α × L₀ × ΔT = L₀ +  α × L₀ × (T - T₀)

Therefore;

L = 60 +  α × 60 × (T - 150°C)

L = 60 + α × 60 × T - 9000 × α

L(T, α) = 60·α·T - 9000·α + 60

7 0
3 years ago
n the periodic table, the atomic mass of oxygen is listed as 15.9994 amu. Based on this, one can deduce that the most common ___
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Isotope- variation of an element

Sixteen- atomic number of oxygen
4 0
3 years ago
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What is the abbreviation for the element with atomic number 11?
VikaD [51]

Answer:

Sodium (Na) has atomic number 11.

5 0
3 years ago
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What mass of H2SO4 is contained in 60.00 mL of a 5.85 M solution of sulfuric acid
strojnjashka [21]

First, we have to calculate the number of moles of H2SO4 in the solution:

V=60 mL = 0.06 L

c=5.85 mol/L

n=V×c=0.06×5.85=0.351 mol

Then we need to find the molar mass of H2SO4:

2×Ar(H) + Ar(S) + 4×Ar(O) =

=2 + 32 + 64 = 98 g/mol

Finally, we need to find the mass of H2SO4:

m=0.351 × 98 = 34.398 g

5 0
3 years ago
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