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3 years ago

Which element would be expected to have chemical and physical properties closest to those of rubidium?

Chemistry

1 answer:

DIA [1.3K]3 years ago

7 0

Rubidium is an element that belongs to Group 1. As such it will have physical properties similar to the other Group 1 elements. Rubidium is below Potassium in the periodic table but above Cesium. As such it would be most like one of those two elements.

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Electromagnetic forces are the predominant force in holding what subatomic particles together?

vova2212 [387]

The strong forces oppose the electromagnetic force of repulsion between protons. Like ”glue” the strong force keeps the protons together to form the nucleus. · The strong forces and electromagnetic forces both hold the atom together.

4 0

3 years ago

3) What are the long-term effects of the Chernobyl disaster? How many people were displaced?

I am Lyosha [343]

Answer: The issue of long-term effects of the Chernobyl disaster on civilians is very controversial. The number of people whose lives were affected by the disaster is sizable. Over 300,000 people were resettled because of the disaster; millions lived and continue to live in the contaminated area.

Explanation:

5 0

2 years ago

A chemical reaction produces 11.8 moles of tin atoms how many grams of tin are made

valina [46]

Convert mols to grams by multiplying grams of tin by the number of mols.

There are 119 grams per mol

119 x 11.8 = 1404 grams

7 0

2 years ago

What are the three different types of selective breeding

balandron [24]

first of all there is only two types of selective breeding and they are hybridization and inbreeding.

3 0

2 years ago

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

2 years ago