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Katyanochek1 [597]
3 years ago
9

The symbol Ca-40 denotes that

Chemistry
1 answer:
Roman55 [17]3 years ago
8 0
D. A calcium atom has mass of 40
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What binary compound would be formed from barium ions and fluoride ions?
LenaWriter [7]
Barium fluoride (BaF2) - Also known as Barium(II) fluoride - because it's a combination of two different kinds of ions (binary = two).
4 0
3 years ago
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How many grams of chlorine gas can be produced when 50.0 grams of aluminum chloride decompose? 2AlCl3→ 2Al + 3Cl2
lianna [129]

Answer:

Explanation:

Approx.

425

⋅

g

Explanation:

2

A

l

(

s

)

+

3

C

l

2

(

g

)

→

2

A

l

C

l

3

(

s

)

You have given a stoichiometrically balanced equation, so bravo.

The equation explicitly tells us that

54

⋅

g

of aluminum metal reacts with

6

×

35.45

⋅

g

C

l

2

gas to give

266.7

⋅

g

of

aluminum trichloride

hope this helps

6 0
3 years ago
What is liquor ammonia fortis?​
erik [133]

Answer:

Ammonia fortis liquor is a saturated solution of ammonia in water. It is also called 880 ammonia. Its relative density is 0.880. It is stored in tightly sealed bottles in a cold place. (Sorry if I'm wrong)

Explanation:

6 0
3 years ago
Read 2 more answers
Calculate the number of moles of magnesium, chlorine, and oxygen atoms in 3.50 moles of magnesium
Firdavs [7]

Answer:

n_{Mg}=3.50molMg\\\\ n_{Cl}=7.00molCl\\\\n_O=28.0molO

Explanation:

Hello there!

In this case, according to the given information it turns out possible for us to realize that one mole of the given compound, Mg(ClO₄)₂, has one mole of Mg, two moles of Cl and eight moles of O; thus, we proceed as follows:

n_{Mg}=3.50molMg(ClO_4)_2*\frac{1molMg}{1molMg(ClO_4)_2}=3.50molMg\\\\ n_{Cl}=3.50molMg(ClO_4)_2*\frac{2molCl}{1molMg(ClO_4)_2}=7.00molCl\\\\n_O=3.50molMg(ClO_4)_2*\frac{8molO}{1molMg(ClO_4)_2}=28.0molO

Best regards!

3 0
3 years ago
When 7.80 mL of 0.500 M AgNO3 is added to 6.25 mL of 0.300 M NH4Cl, how many grams of AgCl are formed?
irina1246 [14]

Answer:

The answer to your question is 0.269 grams of AgCl

Explanation:

Data

[AgNO₃] = 0.50 M

Vol AgNO₃ = 7.80 ml

[NH₄Cl] = 0.30 M

Vol NH₄Cl = 6.25 ml

mass of AgCL

Balanced reaction

                 AgNO₃(aq)  +  NH₄Cl(aq)   ⇒   AgCl (s) + NH₄NO₃ (aq)

Process

1.- Calculate the moles of AgNO₃

Molarity = moles / volume

moles = Molarity x volume

moles = 0.50 x 0.0078

moles = 0.0039

2.- Calculate the moles of NH₄Cl

moles = 0.30 x 0.0063

moles = 0.00188

3.- Calculate the limiting reactant

The proportion of     AgNO₃(aq)  to  NH₄Cl(aq) is 1 :1, then, we conclude that the limiting reactant is NH₄Cl(aq), because there are less amount of this reactant in the experiment.

4.- Calculate the moles of AgCl

                     1 mol of NH₄Cl  ---------------- 1 mol of AgCl

              0.00188 mol of NH₄Cl ------------- x

                     x = (0.00188 x 1) /1

                     x = 0.00188 moles of AgCl

5.- Calculate the grams of AgCl

molecular mass of AgCl = 108 + 35.5 = 143.5 g

                         143.5 grams of AgCl -------------- 1 mol

                         x -------------------------------------------0.00188 moles of AgCl

                          x = (0.00188 x 143.5) / 1

                          x = 0.269 grams of AgCl

8 0
3 years ago
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