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2 years ago

Which of the following elements would form the smallest ionic radius and why?

Chemistry

1 answer:

AnnyKZ [126]2 years ago

5 0

Answer:

Sodium

Explanation:

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Which type of rock does B represent?

Tresset [83]

Answer:

A. because it came from a volcano

Explanation:

5 0

2 years ago

If 1.00 mol of argon is placed in a 0.500-L container at 27.0 degree C , what is the difference between the ideal pressure (as p

ElenaW [278]

Answer:

2.0 atm is the difference between the ideal pressure and the real pressure.

Explanation:

If 1.00 mole of argon is placed in a 0.500-L container at 27.0 °C

Moles of argon = n = 1.00 mol

Volume of the container,V = 0.500 L

Ideal pressure of the gas = P

Temperature of the gas,T = 27 °C = 300.15 K[/tex]

Using ideal gas equation:

$PV=nRT$

$P=\frac{1.00 mol\times 0.0821 L atm/mol K\times 300.15 K}{0.500 L}=49.28 atm$

Vander wall's of equation of gases:

The real pressure of the gas= $p_v$

For argon:

$a=1.345 L^2 atm/mol^2$

b=0.03219 L/mol.

$(p_v+(\frac{an^2}{V^2})(V-nb)=nRT$

$(p_v+(\frac{(1.345 L^2 atm/mol^2)\times (1.00 mol)^2}{(0.500 L)^2})(0.500 L-1.00 mol\times 0.03219L/mol)=1.00 mol\times 0.0821 L atm/mol K\times 300.15 K$

$p_v = 47.29 atm$

Difference : $p - p_v= 49.28 atm - 47.29 atm = 1.99 atm\approx 2.0 atm$

2.0 atm is the difference between the ideal pressure and the real pressure.

6 0

3 years ago

How do nuclear reactions (fusion & fission) compare to chemical reactions?

Zarrin [17]

Because chemicals they blow up

6 0

3 years ago

How many moles of Boron (B) are in 5.03 x 1024 B atoms?

goldfiish [28.3K]

Hey there!:

Number of moles = ( number of atoms / 6.023*10²³ atoms )

given number of atoms = 5.03*10²⁴

Therefore:

Number of moles B = 5.03*10²⁴ / 6.023*10²³

Number of moles B = 8.35 moles

Hope that helps!

3 0

3 years ago

What mass of NaC6H5COO should be added to 1.5 L of 0.40 M C6H5COOH solution at 25 °C to produce a solution with a pH of 3.87 giv

statuscvo [17]

Answer:

41 g

Explanation:

We have a buffer formed by a weak acid (C₆H₅COOH) and its conjugate base (C₆H₅COO⁻ coming from NaC₆H₅COO). We can find the concentration of C₆H₅COO⁻ (and therefore of NaC₆H₅COO) using the Henderson-Hasselbach equation.

pH = pKa + log [C₆H₅COO⁻]/[C₆H₅COOH]

pH - pKa = log [C₆H₅COO⁻] - log [C₆H₅COOH]

log [C₆H₅COO⁻] = pH - pKa + log [C₆H₅COOH]

log [C₆H₅COO⁻] = 3.87 - (-log 6.5 × 10⁻⁵) + log 0.40

[C₆H₅COO⁻] = [NaC₆H₅COO] = 0.19 M

We can find the mass of NaC₆H₅COO using the following expression.

M = mass NaC₆H₅COO / molar mass NaC₆H₅COO × liters of solution

mass NaC₆H₅COO = M × molar mass NaC₆H₅COO × liters of solution

mass NaC₆H₅COO = 0.19 mol/L × 144.1032 g/mol × 1.5 L

mass NaC₆H₅COO = 41 g

7 0

3 years ago