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Gennadij [26K]
3 years ago
10

Which substance would be expected to have the highest electrical conductivity? A. Al(s) B. Al₂O₃(s) C. MgCl₂(aq) D. HCl(g)

Chemistry
1 answer:
almond37 [142]3 years ago
3 0

Answer:

It's most likely gonna be C.

Explanation:

because ionic compounds can conduct electricity in aqueous solutions and covalent-which are the rest of the answers-don't conduct electricity at all.

You might be interested in
Which condition must be met in order for an equation to be balanced?
olga55 [171]

Answer:

The elements in the reactants are the same as the elements in the products.

Explanation:

You can't have more or less elements or atoms, but you also can't just have the same number of atoms because then it could be different elements, which can't happen.

5 0
3 years ago
Read 2 more answers
.an open flask contains 0.200 mol of air. atmospheric pressure is 745 mmhg and room temperature is 68˚f. how many moles are pres
aleksklad [387]

<u>0.219 moles </u><u>moles are present in the flask when the </u><u>pressure </u><u>is 1.10 atm and the temperature is 33˚c.</u>

What is ideal gas constant ?

  • The ideal gas constant is calculated to be 8.314J/K⋅ mol when the pressure is in kPa.
  • The ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas.
  • The combined gas law relates pressure, volume, and temperature of a gas.

We simple use this formula-

The basic formula is PV = nRT where. P = Pressure in atmospheres (atm) V = Volume in Liters (L) n =  of moles (mol) R = the Ideal Gas Law Constant.

68F = 298.15K

V = nRT/P = 0.2 * 0.08206 * 298.15K / (745/760) = 4.992Liters

n = PV/RT = 1.1atm*4.992L/(0.08206Latm/molK * 306K)

n = 0.219 moles

Therefore, 0.219 moles moles are present in the flask when the pressure is 1.10 atm and the temperature is 33˚c.

Learn more about ideal gas constant

brainly.com/question/3961783

#SPJ4

4 0
1 year ago
If 4.0 g of helium gas occupies a volume of 22.4 L at 0 o C and a pressure of 1.0 atm, what volume does 3.0 g of He occupy under
WINSTONCH [101]

Answer:

the volume occupied by 3.0 g of the gas is 16.8 L.

Explanation:

Given;

initial reacting mass of the helium gas, m₁ = 4.0 g

volume occupied by the helium gas, V = 22.4 L

pressure of the gas, P = 1 .0 atm

temperature of the gas, T = 0⁰C = 273 K

atomic mass of helium gas, M = 4.0 g/mol

initial number of moles of the gas is calculated as follows;

n_1 = \frac{m_1}{M} \\\\n_1 = \frac{4}{4} = 1

The number of moles of the gas when the reacting mass is 3.0 g;

m₂ = 3.0 g

n_2 = \frac{m_2}{M} \\\\n_2 = \frac{3}{4} \\\\n_2 = 0.75 \ mol

The volume of the gas at 0.75 mol is determined using ideal gas law;

PV = nRT

PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\since, \ \frac{RT}{P} \ is \ constant,\  then;\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1n_2}{n_1} \\\\V_2 = \frac{22.4 \times 0.75}{1} \\\\V_2 = 16.8 \ L

Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.

4 0
3 years ago
A sequence of thermonuclear fusion processes inside massive stars can continue to transform the nuclei of elements such as carbo
nlexa [21]

Answer:

Iron is the element that is produced at the limit of the reaction.

Explanation:

In nuclear fusion 2 lighter nuclei are combined together into a single nucleus releasing a tremendous amount of energy in the process.

Up to fusion of iron the reaction of fusion is exothermic but when iron atom pops out the reaction becomes endothermic and it requires very high amount of energy to fuse iron atoms which is not available thus marking an end to the fusion reaction.

5 0
3 years ago
The solubility of o2 at 20c is 1.38 x10^-3. the partial presure of o2 in the air at sea level is 0.27 atm. using henery;s law, c
netineya [11]

<u>Answer:</u> The solubility of oxygen at 682 torr is 4.58\times 10^{-3}M

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{A}=K_H\times p_{A}

Or,

\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}

where,

C_1\text{ and }p_1 are the initial concentration and partial pressure of oxygen gas

C_2\text{ and }p_2 are the final concentration and partial pressure of oxygen gas

We are given:

Conversion factor used:  1 atm = 760 torr

C_1=1.38\times 10^{-3}M\\p_1=0.27atm\\C_2=?\\p_2=682torr=0.897atm

Putting values in above equation, we get:

\frac{1.38\times 10^{-3}}{C_2}=\frac{0.27atm}{0.897atm}\\\\C_2=\frac{1.38\times 10^{-3}\times 0.897atm}{0.27atm}=4.58\times 10^{-3}M

Hence, the solubility of oxygen gas at 628 torr is 4.58\times 10^{-3}M

4 0
2 years ago
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