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2 years ago

In need of help very soon please.

Chemistry

1 answer:

NemiM [27]2 years ago

6 0

Answer:

hello, i hope this helps.

Explanation:

1 - group

2 - period

3 - periodic table

4 - family

5 - octet rule

6 - valence electrons

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A 1.25 g sample of aluminum is reacted with 3.28 g of copper (II) sulfate. What is the limiting reactant? 2Al(s) + 3CuSO4(aq) →

vova2212 [387]

Answer:

Copper (II) sulfate

Explanation:

Given reaction is

2Al(s) + 3CuSO4(aq) → Al2(SO4)3(aq) + 3Cu(s)

Amount of aluminum = 1·25 g

Amount of copper (II) sulfate = 3·28 g

Atomic weight of Al = 26 g

Molecular weight of CuSO4 ≈ 159·5

Number of moles of Al = 1·25 ÷ 26 = 0·048

Number of moles of CuSO4 = 3·28 ÷ 159·5 = 0·021

From the above balanced chemical equation for every 2 moles of aluminum, 3 moles of copper (ll) sulfate will be required

So for 1 mole of Al, 1·5 moles of copper (ll) sulfate will be required

For 0·048 moles of Al, 1.5 × 0·048 moles of copper (ll) sulfate will be required

∴ Number of moles of copper (ll) sulfate required = 0·072

But we have only 0·021 moles of copper (ll) sulfate

As copper (ll) sulfate is not there in required amount, the limiting reactant will be copper (ll) sulfate

∴ The limiting reactant is copper (ll) sulfate

7 0

3 years ago

What are the contents of a electron cloud?

PilotLPTM [1.2K]

Weast to east , east to west, nort to south, south to nort.

7 0

3 years ago

How many atoms are in ErAg

Semmy [17]

6.022 * 10^23 atoms

3 0

3 years ago

What molarity of nitric acid (HNO3) was used if 2.00 L must be used to prepare 4.5 L of a 0.25 M HNO3 solution?

Ymorist [56]

The molarity of (HNO₃) that was used if 2.00 L must be used to prepare 4.5 L of a 0.25M HNO₃ solution is 0.563 M

calculation

This is calculated usind M₁V₁=M₂V₂ formula

where,

M₁( molarity ₁) = ?

V₁( volume ₁) = 2.00 L

M₁ (molarity ₂) = 0.25M

V₂( volume₂) = 4.5 L

make M₁ the subject of the formula by diving both side of the formula by V₁

M₁ is therefore = M₂V₂/V₁

M₁ =[ (0.25 M x 4.5 L) / 2.00 L ] =0.563 M

5 0

2 years ago

Compound A is an alkene that was treated with ozone to yield only (CH3CH2CH2)2C=O. Draw the major product that is expected when

DochEvi [55]

Answer:

2,2,3,3-tetrapropyloxirane

Explanation:

In this case, we have to know first the alkene that will react with the peroxyacid. So:

What do we know about the unknown alkene?

We know the product of the ozonolysis reaction (see figure 1). This reaction is an oxidative rupture reaction. Therefore, the double bond will be broken and we have to replace the carbons on each side of the double bond by oxygens. If $(CH_3CH_2CH_2)_2C=O$ is the only product we will have a symmetric molecule in this case 4,5-dipropyloct-4-ene.

What is the product with the peroxyacid?

This compound in the presence of alkenes will produce peroxides. Therefore we have to put a peroxide group in the carbons where the double bond was placed. So, we will have as product 2,2,3,3-tetrapropyloxirane. (see figure 2)

3 0

3 years ago