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2 years ago

What is the volume occupied by 6.56 kg of Al?

2 answers:

8 0

Answer is:volume occupied by 6.56 kg of Al is 2430 dm³.
The density of aluminum is 2,7 kg/m³.
m(Al) = 6,56 kg.
V(Al) = m(Al) ÷ d(Al).
V(Al) = 6,56 kg ÷ 2,7 kg/m³.
V(Al) = 2,43 m³ = 2,43 · 1000 = 2430 dm³.
d(Al) - density of aluminium.
V(Al) - volume of aluminium.

Norma-Jean [14]2 years ago

3 0

If we are given the mass, the volume that the certain substance occupies is calculated by using the density in the equation,

V = m/D

where V is the volume, mi s the mass, and D is density. From the reliable sources, the value of the density of the substance is equal to 2,700 kg/m³.

Substituting the known values from to the equation,

V - (6.56 kg)/(2700 kg/m³)
V = 2.429 x 10^-3 m³

Answer: 2.429 x 10^-3 m³

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Phosphoric acid, which is commonly used as rust inhibitor, food additive and etching agent for dental and orthopedic use, can be

Sphinxa [80]

Answer:

$P_4_{(I)}+5O_2_{(g)}+6H_2O_{(l)}\rightarrow +4H_3PO_4_{(l)}$

Explanation:

The first step is:

$P_4_{(I)}+5O_2_{(g)}\rightarrow 2P_2O_5_{(g)}$

Second step is:

$P_2O_5_{(g)}+3H_2O_{(l)}\rightarrow 2H_3PO_4_{(l)}$

Multiplying second step by 2, and adding both the steps, we get that:

$P_4_{(I)}+5O_2_{(g)}+2P_2O_5_{(g)}+6H_2O_{(l)}\rightarrow 2P_2O_5_{(g)}+4H_3PO_4_{(l)}$

Cancelling common species, we get that:

$P_4_{(l)}+5O_2_{(g)}+6H_2O_{(l)}\rightarrow +4H_3PO_4_{(l)}$

6 0

2 years ago

A student obtained an unknown metal sample that weighed 65.3 g and at a temperature of 99.8oC, he placed it in a calorimeter con

lawyer [7]

Answer:

0.377 J/gºC

Explanation:

From the question given above, the following data were obtained:

Mass of metal (Mₘ) = 65.3 g

Initial temperature of metal (Tₘ) = 99.8 °C

Mass of water (Mᵥᵥ) = 43.7 g

Initial temperature of water (Tᵥᵥ) = 25.7 °C

Equilibrium temperature (Tₑ) = 34.5 °C

Specific heat capacity of water (Cᵥᵥ) = 4.18 J/gºC

Specific heat capacity of metal (Cₘ) =?

The specific heat capacity of metal can be obtained as illustrated below:

Heat lost by metal = heat gained by water.

MₘCₘ(Tₘ – Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Cᵥᵥ)

65.3 × Cₘ (99.8 – 34.5) = 43.7 × 4.18 (34.5 – 25.7)

65.3Cₘ × 65.3 = 182.666 × 8.8

4264.09Cₘ = 1607.4608

Divide both side by 4264.09

Cₘ = 1607.4608 / 4264.09

Cₘ = 0.377 J/gºC

Therefore the specific heat capacity of the metal is 0.377 J/gºC

3 0

2 years ago

If an equal number of moles of reactants are used, do the following equilibrium mixtures contain primarily reactants or products

puteri [66]

Answer:

For a: The equilibrium mixture contains primarily reactants.

For b: The equilibrium mixture contains primarily products.

Explanation:

There are 3 conditions:

When $K_{eq}>1$ ; the reaction is product favored.
When $K_{eq}$ ; the reaction is reactant favored.
When $K_{e}=1$ ; the reaction is in equilibrium.

For the given chemical reactions:

For a:

The chemical equation follows:

$HCN(aq.)+H_2O(l)\rightleftharpoons CN^-(aq.)+H_3O^+(aq.);K_{eq}=6.2\times 10^{-10}$

The expression of $K_{eq}$ for above reaction follows:

$K_{eq}=\frac{[CN^-][H_3O^+]}{[HCN][H_2O]}=6.2\times 10^{-10}$

As, $K_{eq}$ , the reaction will be favored on the reactant side.

Hence, the equilibrium mixture contains primarily reactants.

For b:

The chemical equation follows:

$H_2(g)+Cl_2(g)\rightleftharpoons 2HCl(g);K_{eq}=2.51\times 10^{4}$

The expression of $K_{eq}$ for above reaction follows:

$K_{eq}=\frac{[HCl]^2}{[H_2][Cl_2]}=2.51\times 10^{4}$

As, $K_{eq}>1$ , the reaction will be favored on the product side.

Hence, the equilibrium mixture contains primarily products.

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2 years ago

A sample of a compound containing only carbon and oxygen decomposes and produces 24.50g of carbon and 32.59g of oxygen. what is

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