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4 years ago

Explain why CaCl2 is likely to have properties similar to those of CaBr2

Chemistry

1 answer:

qaws [65]4 years ago

4 0

Answer:

Because both CaCl2 and CaBr2 both contain elements (Chlorine and Bromine) from the same group (group 7)

Explanation:

Elements are placed into different groups in the periodic table. Elements in the same group are those that have the same number of valence electrons in their outermost shell and as a result will behave similar chemically i.e. will react with other elements in the same manner.

Chlorine and Bromine are two elements belonging to group 7 of the periodic table. They are called HALOGENS and they have seven valence electrons in their outermost shell. Hence, when they form a compound with Calcium, a group two element, these compounds (CaCl2 and CaBr2) will possess similar properties because they have elements that are from the same group (halogen group).

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To make a table of the elements, dimitri Mendeleev sorted the elements according to their?

mixer [17]

Answer: He sorted on the basis of average atomic weights.

Explanation:

Dmitri Mendeleev in 1863 arranged chemical elements in the groups according to their average atomic weights.

He said that elements which have similar chemical properties either have similar atomic weights or their atomic weights increases regularly.

Hence, he collaborated a great work in sorting the elements according to their average atomic weights.

8 0

3 years ago

Read 2 more answers

A certain alcohol contains only three elements, carbon, hydrogen, and oxygen. Combustion of a 60.00 gram sample of the alcohol p

marin [14]

Answer:

$C_2H_6O$

Explanation:

The first step is the calculation of the moles of $H_2O$ and $CO_2$ , so:

$114.6~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=2.6~mol~of~CO_2$

$70.44~g~H_2O\frac{1~mol~H_2O}{18~g~H_2O}=~3.9~mol~H_2O$

Now, in 1 mol of CO2 we have 1 mol of C and in 1 mol of $H_2O$ we have 1 mol of H. Additionally, if we want to calculate the moles of oxygen we need to calculate the grams of C and O and then do the substraction form the initial amount, so:

$2.6~mol~CO_2\frac{1~mol~C}{1~mol~CO_2}\frac{12~g~C}{1~mol~C}=31.25~g~of~C$

$3.9~mol~H_2O\frac{2~mol~H}{1~mol~H_2O}\frac{1~g~H}{1~mol~H}=7.82~g~of~H$

$Total~grams=~31.25~+~7.82=39.08~g$

$grams~of~O=60.00~g-~39.08~g=20.92~g~of~O$

Now we can convert the grams of O to moles, so:

$20.92~g~of~O\frac{1~mol~O}{16~g~O}=1.30~mol~O$

The next step is to divide all the mol values by the smallest one:

$O=\frac{1.30~mol~O}{1.30~mol~O}=~1$

$C=\frac{2.6~mol~C}{1.30~mol~O}=~2$

$H=\frac{7.82~mol~H}{1.30~mol~O}=6$

Therefore the formula is $C_2H_6O$

6 0

3 years ago

ACTUAL YIELD VS THEORETICAL YIELD?

lawyer [7]

Actual yield over theoretical yield, then multiply by 100

6 0

3 years ago

What is the law of reciprocal proportion

irina [24]

If two different elements combine separately with a fixed mass of a third element, the ratio of the masses in which they do so are either the same as or a simple multiple of the ratio of the masses in which they combine with each other.

7 0

3 years ago

2 points

erica [24]

9 grams of hydrogen gas (H2) will SC Johnson need to react in order to make 1 bottle of Windex.

Explanation:

Balance equation for the formation of ammonia from H2 gas.

N2 + 3H2 ⇒ 2 $NH_{3}$