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Fofino [41]
3 years ago
5

Find the mass of 1.112 moles HF?

Chemistry
1 answer:
Alexeev081 [22]3 years ago
7 0

Answer:

22.25 g

Explanation:

To find the mass, you need to convert moles to grams and get 22.25 g.

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lilavasa [31]

Answer:

I believe the answer is C. Stopping an object from moving

4 0
3 years ago
How does distillation work.
Usimov [2.4K]

Answer:

Distillation is an important commercial process that is used in the purification of a large variety of materials. However, before we begin a discussion of distillation, it would probably be beneficial to define the terms that describe the process and related properties. ... Elevation of the boiling point with an increase in external pressure, while important in cooking and sterilizing food or utensils, is less important in distillation.

Explanation:

8 0
2 years ago
In calculating the concentration of [Cu(NH3)4]2+[Cu(NH3)4]2+ from [Cu(H2O)4]2+[Cu(H2O)4]2+, the stepwise formation constants are
tamaranim1 [39]

Answer:

11.1×10^12

Explanation:

β4= K1× K2 × K3 × K4

β4= 1.90×10^4 × 3.90×10^3 × 1.00×10^3 ×1.50×102

β4=11.1×10^12

The overall formation (stability) constant β4= K1× K2 × K3 × K4. Hence the answer.

3 0
4 years ago
(d) If you have 9.44 g of NazPO4, how many grams of AgNO, will be needed for complete
AlladinOne [14]

30.6 g of AgNO₃

Explanation:

We have the following chemical reaction:

Na₃PO₄ + 3 AgNO₃ → Ag₃PO₄ +  3 NaNO₃

Now we calculate the number of moles of  Na₃PO₄:

number of moles = mass / molecular weight

number of moles of Na₃PO₄ = 9.44 / 164 = 0.06 moles

Now we formulate the following reasoning:

if        1 mole of Na₃PO₄ will react with 3 moles of AgNO₃

then  0.06 moles of Na₃PO₄ will react with X moles of AgNO₃

X = (0.06 × 3) / 1 = 0.18 moles of AgNO₃

mass = number of moles × molecular weight

mass of AgNO₃ = 0.18 × 170 = 30.6 g

Learn more about:

mole concept

brainly.com/question/1445383

brainly.com/question/10165629

#learnwithBrainly

3 0
3 years ago
Correct?
erica [24]
Correct answer is C

Na2O + H2O ----> 2 NaOH

:-) ;-)
4 0
4 years ago
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