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Trava [24]
3 years ago
15

A student experimentally determines the density of a metal cube. The edge length of the cube is measured using calipers. The cub

e's true density is 12.30 g/cm. Use the student's collected data below and calculate the percent error for the measurement.
Student's Collected Data:
Cube's Mass 551786
Edge Length 1.72 cm
Percent error= _______%
Chemistry
1 answer:
IgorC [24]3 years ago
8 0

Answer:

10.9%.

Explanation:

The first thing to do in order to solve this question is to Determine the value for the volume of the the cube. This can be done by taking the cube root of the length of the cube;

The volume of the cube = (length of the cube)^3 = length × length × length = 1.72 × 1.72 × 1.72 =( 1.72)^3 = 5.09cm^3.

The next thing you do is to Determine the exponential density, the can be done by using the formula below;

The exponential density = mass/ volume = 55. 786/ 5.09 = 10.96 g/cm^3.

Therefore, the percent error = (true density of the cube - exponential density of the cube)÷ true density of the cube × 100.

Hence, the percent error = 12.30 - 10.96/12.30 × 100 = 10.9%.

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The total pressure in a mixture of gases is equal to the partial pressures of
baherus [9]

Answer:

"The total pressure in a mixture of gases is equal to the sum of partial pressures of each gas"

Explanation:

Dalton's law of partial pressures state that, in a mixture of gases, the total pressure is equal to the sum of the partial pressure exerted by each gas of the mixture. The equation is:

Total pressure = Partial pressure Gas 1 + Partial pressure Gas 2 + .... + Partial pressure Gas n

To complete the sentence we can say:

"The total pressure in a mixture of gases is equal to the sum of partial pressures of each gas"

5 0
3 years ago
Manganese dioxide (MnO2(s), Hf = –520.0 kJ) reacts with aluminum to form aluminum oxide (AI2O3(s), Hf = –1699.8 kJ/mol) and mang
Temka [501]

Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

\Delta (H_{f})_{MnO_{2}} = -520.0 KJ/mole

\Delta (H_{f})_{Al_{2}O_{3}} = -1699.8 KJ/mole

The balanced chemical reaction is,

3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)

Formula used :

\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}

\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

Now, put all the values in above formula, we get

\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]

                        = (-3399.6) + (1560)

                        = -1839.6 KJ



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Answer:

Explanation:

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