Calculate the molar mass of each compound given below. c4h4

Answers

Answer 1

Answer: When asked to find molar mass, we first need to calculate the total number of atoms of each element in the compound.

In $C_(4) H_(4)$ , we see that there are 4 Carbons and 4 Hydrogens.

We then need to look up the atomic mass of each of these elements, which is found on the periodic table.

For carbon, the atomic mass is 12.01g
For hydrogen, the atomic mass is 1.008g

Then we multiply the number of atoms in the element by the atomic mass of the element:

C: $4*12.01g=48.04g$
H: $4*1.008g=4.032g$

And then we need to add these two values together to get the molar mass of the compound:

$48.04g+4.032g=52.072g$

So now we know that the molar mass of $C_(4) H_(4)$ is 52.072g.

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Chem help due in 2 hours. please help. problem 14.

Answers

14 )

a) 1 HBr + 1 NaHCO₃ = 1 CO₂ + 1 H₂O + 1 NaBr

b) 2 HNO₃ + 1 K₂SO₃ = 1 H₂O + 2 KNO₃ + 1 SO₂

c) 2 HI + 1 Na₂S = 1 H₂S + 2 NaI

d) 1 (NH₄)₂SO₄ + 1 Ca(OH)₂ = 1 CaSO₄ + 2 H₂O + 2 NH₃

1. Why the names such as sodium(l) chloride for NaCl andmagnesium(II) chloride for MgCl2 are not used?

Answers

Answer:

It's because the NaCl and MgCl2 is shorter and easier to remember

Study the image. Which image best describes what the arrows represent?

Answers

i’m pretty sure it’s d, please let me know !

50.00 mL of 0.10 M HNO 2 (nitrous acid, K a = 4.5 × 10 −4) is titrated with a 0.10 M KOH solution. After 25.00 mL of the KOH solution is added, the pH in the titration flask will be a. 2.17
b. 3.35
c. 2.41
d. 1.48
e. 7.00

Answers

Answer:

b. 3.35

Explanation:

To calculate the pH of a solution containing both acid and its salt (produced as a result of titration) we need to use Henderson’s equation i.e.

pH = pKa + log ([salt]/[acid]) (Eq. 01)

Where

pKa = -log(Ka) (Eq. 02)

[salt] = Molar concentration of salt produced as a result of titration

[acid] = Molar concentration of acid left in the solution after titration

Let’s now calculate the molar concentration of HNO2 and KOH considering following chemical reaction:

HNO2 + KOH ⇆ H2O + KNO2 (Eq. 03)

This shows that 01 mole of HNO2 and 01 mole of KOH are required to produce 01 mole of KNO2 (salt). And if any one of them (HNO2 and KOH) is present in lower amount then that will be considered the limiting reactant and amount of salt produced will be in accordance to that reactant.

Moles of HNO2 in 50 mL of 0.01 M HNO2 solution = 50/1000x0.01 = 0.005 Moles

Moles of KOH in 25 mL of 0.01 M KOH solution = 25/1000x0.01 = 0.0025 Moles

As it can be seen that we have 0.0025 Moles of KOH therefore considering Eq. 03 we can see that 0.0025 Moles of KOH will react with only 0.0025 Moles of HNO2 and will produce 0.0025 Moles of KNO2.

Therefore

Amount of salt produced i.e [salt] = 0.0025 moles (Eq. 04)

Amount of acid left in the solution [acid] = 0.005 - 0.0025 = 0.0025 moles (Eq.05)

Putting the values in (Eq. 01) from (Eq.02), (Eq. 04) and (Eq. 05) we will get the following expression:

pH= -log(4.5x10 -4) + log (0.0025/0.0025)

Solving above we get

pH = 3.35

Final answer:

The pH value in the titration flask after 25.00 mL of the 0.10 M KOH solution is added to 50.00 mL of 0.10 M HNO2 solution is 3.35.

Explanation:

The subject of this question is titration, which is a method used in chemistry to measure the concentration of an unknown solution. Given 50.00 mL of 0.10 M HNO2 (nitrous acid, Ka = 4.5 × 10-4), titrated with 0.10 M KOH (potassium hydroxide), we need to calculate the pH after 25.00 mL of the KOH solution is added.

First, we need to find the moles of the HNO2 and the KOH. Moles equals Molarity times Volume. So, for HNO2, it is 0.10 M * 0.050 L which equals 0.005 moles. For KOH, it is 0.10 M * 0.025 L which equals 0.0025 moles.

Then, subtract the moles of OH- from the moles of HNO2 to determine the concentration of HNO2 left, which is 0.005 moles - 0.0025 moles = 0.0025 moles. Divide this by the total volume of the solution (50.00 mL + 25.00 mL = 75.00 mL or 0.075 L to determine the new concentration of HNO2, 0.0025 moles / 0.075 L = 0.033 M. Then use the given Ka value with the equation [H+] = sqrt(Ka * [HNO2]) to get [H+].

To find acids' pH, we use the formula pH = -log[H+]. Use the calculated [H+] to find the pH.

Upon performing these calculations, the resulting pH value should be approximately 3.35 after 25.00 mL of the KOH solution is added, so the answer is (b) 3.35.

Learn more about titration here:

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If the molecule C6H12 does not contain a double bond, and there are no branches in it, what will its structure look like?

Answers

I have attached a photo of the structure.
You can get better at solving problems like this by practicing a lot!

The balanced combustion reaction for C 6 H 6 C6H6 is 2 C 6 H 6 ( l ) + 15 O 2 ( g ) ⟶ 12 CO 2 ( g ) + 6 H 2 O ( l ) + 6542 kJ 2C6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)+6542 kJ If 7.300 g C 6 H 6 7.300 g C6H6 is burned and the heat produced from the burning is added to 5691 g 5691 g of water at 21 ∘ 21 ∘ C, what is the final temperature of the water?