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3 years ago

26. Which type of element has properties that are intermediate between metals and nonmetals?

Chemistry

1 answer:

Lesechka [4]3 years ago

7 0

Explanation:

.metallic has properties that are intermediate between mental and non metal

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How many atoms are in 3.5 grams of copper

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Avegadro's number = 6.02 x10^23 atoms
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We have air (21% O2 and 79% N2) at 23 bar and 30 C. 4. What is the ideal molar volume (m^3/kmol)? a. b. What is the Z factor? Wh

k0ka [10]

Answer:

The ideal molar volume is $\frac{V}{n} =V_z= 0.001095 \ m^3/mol$

The Z factor is $Z = 0.09997$

The real molar volume is $\frac{V_r}{n} = V_k= 0.0001095\ \frac{m^3}{mol}$

Explanation:

From the question we are told that

The pressure is $P = 23 \ bar = 23 *10^5 Pa$

The temperature is $T = 30 ^ oC = 303 \ K$

According to the ideal gas equation we have that

$PV = nRT$

=> $\frac{V}{n}=V_z= \frac{RT}{P}$

Where $\frac{V}{n }$ is the molar volume and R is the gas constant with value

$R = 8.314 \ m^3 \cdot Pa \cdot K^{-1}\cdot mol^{-1}$

substituting values

$\frac{V}{n} =V_z= \frac{ 8.314 * 303}{23 *10^{5}}$

$\frac{V}{n} =V_z= 0.001095 \ m^3/mol$

The compressibility factor of the gas is mathematically represented as

$Z = \frac{P * V_z}{RT}$

substituting values

$Z = \frac{23 *10^{5} * 0.001095}{8.314 * 303}$

$Z = 0.09997$

Now the real molar volume is evaluated as

$\frac{V_r}{n} = V_k= \frac{Z * RT }{P}$

substituting values

$\frac{V_r}{n} = V_k= \frac{0.09997 * 8.314 * 303}{23 *10^{5}}$

$\frac{V_r}{n} = V_k= 0.0001095\ \frac{m^3}{mol}$

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3 years ago

In the turbine of a nuclear power plant, the mechanical energy is converted

scoundrel [369]

It would be Thermal energy

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The density of benzene at 15 °c is 0.8787 g/ml. Calculate the mass of 0.1500 l of benzene at this temperature. Enter your answer

Otrada [13]

Answer:

131.8 g.

Explanation:

There is a relation that relates the density of the substance (d) to the mass (m) and the volume of the substance (V):

d = m / V.

d = 0.8787 g/ml.

V = 0.15 L x 1000 = 150 ml.

∴ the mass of 0.15 L of benzene = d x V = (0.8787 g/ml) (150 ml) = 131.8 g.

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