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3 years ago

How many electrons are drawn for a electron dot structure for carbon?

Chemistry

1 answer:

koban [17]3 years ago

5 0

Answer:

6

Explanation:

yeah i’m pretty sure 2 in the first shell and 4 in the second

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1. BaBr2(aq) + H2SO4(aq) →?<br> please balance the equation and predict the products

Andrew [12]

Answer:

BaBr2 (aq) + H2SO4 (aq) → BaSO4 (s) + 2 HBr (aq)

Explanation:

This is a precipitation reaction: BaSO4 is the formed precipitate.

4 0

3 years ago

(ill give brainliest)

gizmo_the_mogwai [7]

Answer:

It is because the object that is charged attracts the uncharged because it is giving electrons and passing it on to the object. As shown in the image the balloon is attracting the paper that is not charged. Actually telling,the charged object will cause the uncharged object to become charged, with a positive charge on one side and a negative charge on the other side. This process is called induction.

Explanation: Pls Mark Brainliest

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3 years ago

Why is water classified as a renewable resource?

mrs_skeptik [129]

The answer is a because the water we use it comes back again

6 0

3 years ago

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How many atoms are present in 179.0 g of iridium?

posledela

160 atoms are present

4 0

4 years ago

Read 2 more answers

When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H

Debora [2.8K]

Answer:

$Y=58.15\%$

Explanation:

Hello,

For the given chemical reaction:

$Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)$

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

$n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3$

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

$n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol Al_2S_3$

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

$m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3$

Finally, we compute the percent yield with the obtained 2.10 g:

$Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%$

Best regards.

7 0

3 years ago