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3 years ago

Acids and bases are good conductors of electricity so they are also known as?

1 answer:

7 0

If Acid and Bases has the composition of being good conductors of electricity, then they could be known and considered as an electrolyte for these are substances or components that has the ability of being a good conductors of electricity. So acid and bases could be known as a electrolyte.

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What is sometimes described as the ability to do work?

diamong [38]

Kinetic and potential energy

6 0

3 years ago

which of the following solution is a buffer? a solution containing Hf and NaOH, NH3 and NH4+, HCN and HNO3?

Ivahew [28]

The answer is NH3 an NH4+. Buffer solutions contain a weak base and its conjugate acid or a weak acid and its conjugate base. NH3 is a weak base and NH4+ is its conjugate acid.

6 0

3 years ago

The standard molar enthalpy of formation of NH3(g) is -45.9 kJ/mol. What is the enthalpy change if 9.51 g N2(g) and 1.96 g H2(g)

Vladimir [108]

Answer:

$\Delta H=-29.7kJ$

Explanation:

Hello!

In this case, since the undergoing chemical reaction is:

$N_2+3H_2\rightarrow 2NH_3$

We first need to identify the limiting reactant given the masses of nitrogen and hydrogen:

$n_{NH_3}^{by\ H_2}=1.96gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.647molNH_3\\\\ n_{NH_3}^{by\ N_2}=9.51gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.679molNH_3$

It means that only 0.647 moles of ammonia are yielded, so the resulting enthalpy change is:

$\Delta H=0.647molNH_3*\frac{-45.9kJ}{1molNH_3}\\\\ \Delta H=-29.7kJ$

Best regards!

3 0

3 years ago

50cm3 of sodium hydroxide solution was titrated against a solution of sulfuric acid. The concentration of the sodium hydroxide s

miskamm [114]

Answer:

49 g/L is the concentration of the acid

Explanation:

Firstly, we proceed to write the equation of reaction.

2NaOH + H2SO4 ——-> Na2SO4 + 2H2O

We can see that 1 mole of the base reacted with two moles of the acid.

kindly note that dm^3 is same as liter

Firstly, we need to get the concentration of the reacted sulphuric acid in g/L

we use the simple titration equation below;

CaVa/CbVb = Na/Nb

From the question;

Ca = ?

Va = 25 cm^3

Cb = 20 g/L

we convert this to concentration in mol/L

Mathematically, that is concentration in g/L divided by molar mass in g/mole

molar mass of NaOH = 40 g/mol

so we have; 20g/L / 40 = 0.5 mol/L

Vb = 50 cm^3

Na = 1

Nb = 2

Where C represents concentrations, V volumes and N , number of moles

Now, substitute the values;

Ca * 25/0.5 * 50 = 1/2

25Ca/25 = 0.5

So Ca = 0.5 mol/L

Now to get the concentration of H2SO4 in g/L

What we do is to multiply the concentration in mol/L by molar mass in g/mol

That would be 0.5 * 98 = 49 g/L

4 0

4 years ago

0.785 moles of N2,fill a balloon at 1.5 atm and 301 K.<br> What is the volume of the balloon?

Tcecarenko [31]

Answer:

V = 12.93 L

Explanation:

Given data:

Number of moles = 0.785 mol

Pressure of balloon = 1.5 atm

Temperature = 301 K

Volume of balloon = ?

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

Now we will put the values.

V = nRT/P

V = 0.785 mol × 0.0821 atm.L/ mol.K × 301 K / 1.5 atm

V = 19.4 L /1.5

V = 12.93 L

7 0

3 years ago