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3 years ago

calculate density of a rectangular solid, which has a mass of 25.71 g. It is 2.30 cm long, 4.01 cm wide and 1.82 cm high.

1 answer:

5 0

Density is a property of the substances that is obtained by dividing its mass by the volume. For a rectangular solid, the volume may be solved by the following equation,

V = L x W x H

Substituting the given values for the dimension,

V = (2.30 cm) x (4.01 cm) x (1.82 cm) = 16.78786 cm³

Calculating for the density,

Density = mass / volume

Density = 25.71 cm / 16.78786 cm³ = 1.53 grams per cm³

Thus, the density of the given solid is approximately 1.53 grams per cm³.

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2 years ago

Find the mass of an object on earth if its weight is 100N

Kaylis [27]

Answer:

0.19 kg

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Explanation:

w = Weight of an object = 100 N

g = Acceleration due to gravity of Earth = $9.81\ \text{m/s}^2$

m = Mass of an object

Weight is given by

$w=mg\\\Rightarrow m=\dfrac{w}{g}\\\Rightarrow m=\dfrac{100}{9.81}\\\Rightarrow m=10.19\ \text{kg}$

Mass of the object is 0.19 kg

g = $2\ \text{m/s}^2$

m = 4 kg

$w=mg\\\Rightarrow w=4\times 2\\\Rightarrow w=8\ \text{N}$

Weight of the object is 8 N.

m = 200 g

$w=mg\\\Rightarrow w=0.2\times 9.81\\\Rightarrow w=1.964\ \text{N}$

Weight of the object is 1.964 N.

Weight is the force the Earth exerts on an object which is on the surface of the Earth.

m = 6kg

$w=mg\\\Rightarrow w=6\times 9.81\\\Rightarrow w=58.86\ \text{N}$

The force of the object is 58.86 N.

w = 150 N

a = g = $10\ \text{m/s}^2$

$m=\dfrac{w}{g}\\\Rightarrow m=\dfrac{150}{10}\\\Rightarrow m=15\ \text{kg}$

Mass of the object is 15 kg.

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3 years ago

Rutherford proposed that atoms have positive nuclei. Which statement correctly describes the nucleus of an atom?

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3 years ago

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The reaction is first order.

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3 years ago

Given these reactions, X ( s ) + 1 2 O 2 ( g ) ⟶ XO ( s ) Δ H = − 668.5 k J / m o l XCO 3 ( s ) ⟶ XO ( s ) + CO 2 ( g ) Δ H = +

qwelly [4]

Answer: The $\Delta H^o_{rxn}$ for the reaction is -1052.8 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$X(s)+\frac{1}{2}O_2(g)+CO_2(g)\rightarrow XCO_3(s)$ $\Delta H^o_{rxn}=?$