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qwelly [4]
3 years ago
8

Help me if you can please help me please​

Chemistry
2 answers:
Dovator [93]3 years ago
8 0

Answer: with? i can help:))

Explanation:

hmu love

olga2289 [7]3 years ago
6 0

Answer:

what you need help with ??

Explanation:

Il help

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How can you increase the energy of a glass of water?
balu736 [363]
B. Heat is a form of energy so boiling it would increase energy. (I guess)
8 0
3 years ago
Select the molecule where free rotation around one or more bonds is not possible. Group of answer choices N2H4 C2Cl4 NH3 C2H6 CC
Alexandra [31]

Answer:

C₂Cl₄

Explanation:

To know if free rotation around a bond in a compound is possible, we need to see the structure of the compound (picture in attachment).

In single bonds, which are formed by σ bonds, the atoms are not fixed in a single position, and free rotation is permitted.

Double and triple bonds are formed by a σ bond and one or two π bonds, respectively. These bonds do not allow rotation, since it is not possible to twist the ends without breaking the π bond.

The chloroethylene (C₂Cl₄) has two carbons with an sp2-sp2 hybridization, they are bonded together by a double bond. <u>Free rotation on this bond is not possible, because six atoms, including the carbon atoms, doubly bonded and the four chlorine atoms bonded to them, must be on the same plane. </u>

4 0
3 years ago
Please help:/ I can’t figure out the answer
jekas [21]
Because other people cannot see what you are sending to somebody
7 0
3 years ago
If 12.5 grams of strontium hydroxide is reacted with 150 mL of 3.5 M carbonic acid, identify the limiting reactant.
vesna_86 [32]

Answer:

Sr(OH)2

Explanation:

We'll begin by calculating the number of mole of carbonic acid in 150mL of 3.5 M carbonic acid solution. This is illustrated below:

Molarity = 3.5M

Volume = 150mL = 150/1000 = 0.15L

Mole of carbonic acid, H2CO3 =..?

Mole = Molarity x Volume

Mole of carbonic acid, H2CO3 = 3.5 x 0.15 = 0.525 mole.

Next, we shall convert 0.525 mole of carbonic acid, H2CO3 to grams.

Mole of H2CO3 = 0.525 mole

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 =..?

Mass = mole x molar mass

Mass of H2CO3 = 0.525 x 62 = 32.55g

Next, we shall write the balanced equation for the reaction. This is given below:

Sr(OH)2 + H2CO3 → SrCO3 + 2H2O

Next, we shall determine the mass of Sr(OH)2 and H2CO3 that reacted from the balanced equation. This is illustrated below:

Molar mass of Sr(OH)2 = 88 + 2(16 + 1) = 88 + 2(17) = 122g/mol

Mass of Sr(OH)2 from the balanced equation = 1 x 122 = 122g

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 from the balanced equation = 1 x 62 = 62g.

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Finally, we shall determine the limiting reactant as follow:

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Therefore, 12.5g of Sr(OH)2 will react with = (12.5 x 62)/122 = 6.35g.

We can see evidently from the calculations made above that it will take 6.35g out 32.55g of H2CO3 to react with 12.5g of Sr(OH)2. Therefore, Sr(OH)2 is the limiting reactant and H2CO3 is the excess reactant

5 0
3 years ago
Read 2 more answers
Determine whether you can swim in 1.00 x 10^27 molecules of water.​
zloy xaker [14]

Answer:

We can not swim in 1.00 × 10²⁷ molecules of water

Explanation:

The given number of molecules of water = 1.00 × 10²⁷ molecules

The Avogadro's number, N_A, gives the number of molecules in one mole of a substance

N_A ≈ 6.0221409 × 10²³ molecules/mol

Therefore

Therefore, we have;

The number of moles of water present in 1.00 × 10²⁷ molecules, n = (The number of molecules of water) ÷ N_A

∴ n = (1.00 × 10²⁷ molecules)/(6.0221409 × 10²³ molecules/mol) = 1,660.53902857 moles

The mass of one mole of water = The molar mass of water = 18.01528 g/mol

The mass, 'm', of water in 1,660.53902857 moles of water is given as follows;

Mass = (The number of moles of the substance) × (The molar mass of the substance)

∴ The mass of the water in the given quantity of water, m = 1,660.53902857 moles × 18.01528 g/mol ≈ 29.9150756 kg.

The density pf water, ρ = 997 kg/m³

Volume = Mass/Density

∴ The volume of the water present in the given quantity of water, v = 29.9150756 kg/(997 kg/m³) ≈ 30.0050909 liters

The volume of the water present in 1.00 × 10²⁷ molecules of water ≈ 30.0 liters

The average volume of a human body = 62 liters

Therefore, we can not swim in the given quantity of 1.00 × 10²⁷ molecules = 30.0 liters water

7 0
2 years ago
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