Which of the following are physical properties of nonmentals

2. Calculate the pressure in Pa of a brick, which weighs 40 lbs (178N) sitting flat on the

Aloiza [94]

Answer:

P = 741.7 Pa

Explanation:

Pressure:

It is the force applied on object per unit area. The force is applied perpendicular to the surface.

Mathematical expression:

P = F/A

P = pressure

F = force

A = area

Given data:

Weight = 178 N

Area = 10 cm × 24 cm

Pressure = ?

Solution:

Area = 10 cm × 24 cm

Area = 240 cm²

Area = 240 /10000 = 0.024 m²

Formula:

P = F/A

P = 178 N / 0.024 m²

P = 741.7 Nm⁻²

Nm⁻² = Pa

P = 741.7 Pa

6 0

3 years ago

One orbit of the earth around the sun equals _____________ miles.

garri49 [273]

Answer:

it's about 580millon, but the closest you have is 595 million, so pick that one

4 0

3 years ago

Read 2 more answers

Calculate the cell potential for the reaction as written at 25.00 °C 25.00 °C , given that

Taya2010 [7]

Answer:

E = 2.02 V

Explanation:

In order to do this, we need to apply the Nernst equation which is:

E = E° - RT/nF lnQ

The value of RT/F can be simplified to just 0.059 because we are doing this experiment at 25 °C, and R and F are constants. so we need the value of Q which in this case is:

Q = [Mg²⁺] / [Ni²⁺]

We already have the concentrations, so, all we have left is the standard reduction potential, which are:

E° Mg = -2.38 V

E° Ni = -0.25 V

According to the overall reaction:

Mg(s) + Ni²⁺(aq) -------> Mg²⁺(aq) + Ni(s)

we can see that one element is reducting and the other is oxidizing, so we need to write the semi equation of reduction for each element:

Mg(s) ---------> Mg²⁺ + 2e⁻ E° = 2.38 V oxidizing (Value of E° inverted)

Ni²⁺ + 2e⁻ -----------> Ni(s) E° = -0.25 V reducting

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

Mg(s) + Ni²⁺(aq) -------> Mg²⁺(aq) + Ni(s) E° = 2.13 V

We have the value of the standard potential, now we need to replace all given data into the nernst equation to solve for the cell potential:

E = 2.13 - 0.059/2 ln(0.757/0.0160)

E = 2.13 - 0.0295 ln(47.3125)

E = 2.13 - 0.11

E = 2.02 V

This is the cell potential

3 0

3 years ago

What are some examples of physical changes?

Ludmilka [50]

Some examples of physical changes are:

Breaking a glass

Chopping wood

Tearing paper

Mixing sand and water

Melting an ice cube

These all are physical changes because the composition of the matter does not change.

3 0

3 years ago

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A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2

a_sh-v [17]

Answer: The empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=13.00g$

Mass of $H_2O=2.662g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 13.00 g of carbon dioxide, $\frac{12}{44}\times 13.00=3.54g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 2.662 g of water, $\frac{2}{18}\times 2.662=0.296g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.295 moles.

For Carbon = $\frac{0.295}{0.295}=1$

For Hydrogen = $\frac{0.296}{0.295}=1$

For Oxygen = $\frac{0.603}{0.295}=2.044\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is $CHO_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

$n=\frac{90.04g/mol}{45g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4$

Hence, the empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

3 0

4 years ago