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Rudik [331]
3 years ago
7

Which equations represent precipitation reactions?

Chemistry
1 answer:
viktelen [127]3 years ago
3 0

In a precipitation reaction, one of the product formed is an insoluble precipitate.

The given reactions can be written with phases included as:

A. Na_{2}S(aq)+FeBr_{2}(aq)-->2NaBr(aq)+FeS(s): In this reaction one of the product, FeS is insoluble. Therefore, this is a precipitation reaction.

B. MgSO_{4}(aq)+CaCl_{2}(aq)-->MgCl_{2}(aq)+CaSO_{4}(s): In this reaction, the product CaSO_{4}is a solid(insoluble). So, this is a precipitation reaction too.

C.LiOH(aq)+NH_{4}I(aq)-->LiI(aq)+NH_{4}OH(aq): In this reaction, both the products are soluble. So this is not a precipitation reaction.

D.2NaCl(aq)+K_{2}S(aq)-->Na_{2}S(aq)+2KCl(aq): In this reaction, both the products are soluble. So this is not a precipitation reaction.

E. AgNO_{3}(aq)+NaCl(aq)-->AgCl(s)+NaNO_{3}(aq): In this reaction, the product AgCl is a precipitate. So, it is a precipitation reaction.


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Answer:

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7 0
3 years ago
Do all substances absorb heat the same way?
miv72 [106K]

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5 0
3 years ago
A solution has a pH of 2.5. Answer the following questions.
olga_2 [115]
PH + pOH = 14  ⇒  pOH = 14 - pH
pOH = 14 - 2.5
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[H⁺] = 10^(-pH) = 10^(-2.5)
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[OH⁻] = 3 × 10⁻¹² M


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6 0
3 years ago
How much heat is required to raise the temperature of 65.8 grams of water from 31.5ºC to 46.9ºC?
vodomira [7]

Answer: 1,013.32 cal × 4.18 J/cal = 4,235.68 J

Explanation:

1) Data:

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m = 65.8 g

Ti = 31.5°C

Tf = 36.9°C

Heat, Q = ?

2) Formula:

Q = mCΔT

3) Calculations:

Q = 65.8g × 1 cal/g°C × (46.9°C - 31.5°C) = 1,013.2 cal

4) You can convert from calories to Joules using the conversion factor:

1 cal = 4.18 J

⇒ 1,013.32 cal × 4.18 J/cal = 4,235.68 J

3 0
3 years ago
What about 50 g of water?<br> I need help what this
ELEN [110]

Answer:

3.38  Tablespoons

10.14 Teaspoons

0.21 U.S. Cups

0.18 Imperial Cups

0.20 Metric Cups

50.00 Milliliters

Explanation:

3 0
3 years ago
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