All categories

2 years ago

Hey! need help with chem, multiple choice questions.

Chemistry

1 answer:

ladessa [460]2 years ago

7 0

1. 6 carbon atoms and triple bonding between carbons 2 and 3
2. i think it’s 3 but i may be wrong

You might be interested in

What is the name of the compound made from Sn 4+ and PO4 3-?

givi [52]

Answer:

Sn₃(PO₄)₄ - tin(IV) phosphate.

Explanation:

Hope it helps! :)

8 0

2 years ago

Which answer tells the reason the earth’s climate is getting warmer?

Soloha48 [4]

Driving cars gives off gases that trap heat in the atmosphere

5 0

3 years ago

Read 2 more answers

A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0- L vessel at 300 K . The following equilibr

umka2103 [35]

Answer:

The concentration of N2 at the equilibrium will be 0.019 M

Explanation:

Step 1: Data given

Number of moles of NO = 0.10 mol

Number of moles of H2 = 0.050 mol

Number of moles of H2O = 0.10 mol

Volume = 1.0 L

Temperature = 300K

At equilibrium [NO]=0.062M

Step 2: The balanced equation

2NO(g) + 2H2(g) → N2(g) + 2H2O(g)

Step 3: Calculate the initial concentration

Concentration = Moles / volume

[NO] = 0.10 mol / 1L = 0.10 M

[H2] = 0.050 mol / 1L = 0.050 M

[H2O] = 0.10 mol / 1L = 0.10 M

[N2] = 0 M

Step 4: Calculate the concentration at the equilibrium

[NO] at the equilibrium is 0.062 M

This means there reacted 0.038 mol (0.038M) of NO

For 2 moles NO we need 2 moles of H2 to produce 1 mol N2 and 2 moles of H2O

This means there will also react 0.038 mol of H2

The concentration at the equilibrium is 0.050 - 0.038 = 0.012 M

There will be porduced 0.038 moles of H2O, this means the final concentration pf H2O at the equilibrium is 0.100 + 0.038 = 0.138 M

There will be produced 0.038/2 = 0.019 moles of N2

The concentration of N2 at the equilibrium will be 0.019 M

5 0

3 years ago

Pure ethylene glycol, , is added to 2.00 kg of water in the cooling system of a car. The vapor pressure of the water in the syst

spayn [35]

Answer:

1367.7 g of ethylene glycol was added to the solution

Explanation:

In order to find out the mass of glycol we added, we apply the colligative property of lowering vapor pressure: ΔP = P° . Xm

ΔP = Vapor pressure of pure solvent (P°) - Vapor pressure of solution(P')

525.8 mmHg - 451 mmHg = 451 mmHg . Xm

74.8 mmHg / 451 mmHg = Xm → 0.166 (mole fraction of solute)

Xm = Mole fraction of solute / Moles of solute + Moles of solvent

We can determine the moles of solvent → 2000 g . 1 mol/18 g = 111.1 mol

(Notice we converted the 2kg of water to g)

0.166 = Moles of solute / Moles of solute + 111.1 moles of solvent

0.166 (Moles of solute + 111.1 moles of solvent) = Moles of solute

18.4 moles = Moles of solute - 0.166 moles of solute

18.4 = 0.834 moles of solute → Moles of solute = 18.4/0.834 = 22.06 moles

Let's convert the moles to mass → 62 g/mol . 22.06 mol = 1367.7 g

6 0

3 years ago

Lab 27. stoichiometry and chemical reactions: which balanced chemical equation best represents the thermal decomposition of sodi

Cloud [144]

Answer:
Sodium Bicarbonate on decomposition produces Carbon dioxide gas and Water vapors.

 2 NaHCO₂ → Na₂CO₃ (s) + CO₂ (g) + H₂O (g)

Explanation:
Let suppose you burn 168 g ( 2 moles ) of NaHCO₃, a gas will produced and product is left behind. On measuring the product formed it will be almost equal to 105 g. This shows that the product is Na₂CO₃ and 1 mole of it is being produced after decomposition of sodium bicarbonate.

5 0

2 years ago