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Papessa [141]
3 years ago
8

How many moles of cacl2 are in 10.5g of cacl2?

Chemistry
1 answer:
Arte-miy333 [17]3 years ago
5 0

Answer:

0.09460823181719888

Explanation:

Round to whatever decimal point is needed for your answer.

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EastWind [94]

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what kind of class u in like dam

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3 years ago
PLEASE ANSWER I GIVE BRAINLIEST AND MAX POINTS
d1i1m1o1n [39]

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I believe the answer is A (changes in air temperature  result in differences in air pressure which causes wind). Hope this helps

Explanation:

7 0
3 years ago
How is electronegativity related to covalent bonding?
Zolol [24]

Answer:

Atoms must have similar electronegativities in order to share electrons in a covalent bond.

Explanation:

Covalent bonding is one of the bondings that occurs between the atoms of elements. It is the bonding in which atoms share their valence electrons with one another. However, the ELECTRONEGATIVITY, which is the ability of an atom to be attracted to electrons play a major role in the formation of covalent bonds.

When atoms of different electronegativities combine, the more electronegative atom pulls more electrons towards itself, hence, an IONIC bond is formed. However, when the electronegativities of the atoms are similar, the sharing of their electrons becomes stronger. Hence, ATOMS MUST HAVE SIMILAR ELECTRONEGATIVITIES in order to share electrons in a covalent bond.

5 0
3 years ago
Read 2 more answers
Which of the following do water, rust, and salt have in common?
Cloud [144]
The answer is "elements" :)
6 0
2 years ago
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A stock solution of ammonia in water is 28 wt% NH3. The Formal Weight (FW) of NH3 is 17.03 g/mol). The solution density at room
weqwewe [10]

Answer:

[NH₃] = 14.7 mol/L

Explanation:

28 wt% is a type of concentration that indicates that 28 g of ammonia is contained in 100 g of solution.

Let's determine the amount of ammonia:

28 g . 1 mol / 17.03g = 1.64 moles of NH₃

You need to consider that, when you have density's data it is always referred to solution:

Mass of solution is 100 g, let's find out the volume

0.90 g/mL = 100 g /V

V = 100 g / 0.90mL/g → 111.1 mL

We convert the volume to L → 111.1 mL . 1 L/1000mL = 0.1111 L

mol/L = 1.64 mol/0.1111L → 14.7 M

mol/L = M → molarity a sort of concentration that indicates the moles of solute in 1L of solution

4 0
2 years ago
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