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marta [7]
2 years ago
5

Phosphorus trichloride, PCl3, decomposes to form elemental phosphorus and chlorine. The equation is: 4PCl3 → P4 + 6Cl2. Balance

the equation and determine the mass in grams of chlorine that would be formed if 25 grams of PCl3 (molecular mass = 137.32 g/mol) decompose. grams of Cl2
Chemistry
2 answers:
Scilla [17]2 years ago
8 0

answer- 19 grams of CI2

TiliK225 [7]2 years ago
5 0
19 grams of Cl2 is the correct answer                <span>

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What are the chemical processes used to make chloroquine?
4vir4ik [10]
You spell Chlorine wrong but the process is called electrolysis  
6 0
3 years ago
You have 100 mL of a 12M solution of HCl, and you need to dilute it to 1.5M for an experiment. How many liters will your new sol
sashaice [31]

Answer:

800.0 mL.

Explanation:

  • To solve this problem; we must mention the rule states the no. of millimoles of a substance before and after dilution is the same.

<em>(MV)before dilution of HCl = (MV)after dilution of HCl</em>

M before dilution = 12.0 M, V before dilution = 100.0 mL.

M after dilution = 1.5 M, V after dilution = ??? mL.

∵ (MV)before dilution of HCl = (MV)after dilution of HCl

∴ (12.0 M)(100.0 mL) = (1.5 M)(V after dilution of HCl)

<em>∴ V after dilution of HCl = (12.0 M)(100.0 mL)/(1`.5 M) = 800.0 mL.</em>

8 0
2 years ago
How much heat is required to convert 90. g of ice at -4.0 C to water vapor (steam) at 110 C
lakkis [162]

Answer:  How much heat is required to convert 29 g of ice at -4.0 ˚C to water vapor at 105 ˚C (report your answer to three significant figures)? Answer all five parts, showing your detailed calculations: i) Heat to warm ice from -4.0 to 0 oC: ii) Heat to convert ice to liquid: iii) Heat to warm liquid to boiling: iv) Heat to convert liquid to gas

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Explanation:

e

6 0
2 years ago
Calculate the ph of a solution containing 0.0451 m potassium hydrogen tartrate and 0.028 m dipotassium tartrate. The ka values f
Marina86 [1]

Given buffer:

potassium hydrogen tartrate/dipotassium tartrate (KHC4H4O6/K2C4H4O6 )

[KHC4H4O6] = 0.0451 M

[K2C4H4O6] = 0.028 M

Ka1 = 9.2 *10^-4

Ka2 = 4.31*10^-5

Based on Henderson-Hasselbalch equation;

pH = pKa + log [conjugate base]/[acid]

where pka = -logKa

In this case we will use the ka corresponding to the deprotonation of the second proton i.e. ka2

pH = -log Ka2 + log [K2C4H4O6]/[KHC4H4O6]

     = -log (4.31*10^-5) + log [0.0451]/[0.028]

pH = 4.15



4 0
2 years ago
Give five examples of exothermic reactions in everyday life
DedPeter [7]

Answer:

Ice Cubes. When water freezes into ice cubes, the energy is released in the form of heat. ...

Formation Of Snow In Clouds. Sponsored. ...

Hot Packs. ...

Rusting Of Iron. ...

Burning Of Candles. ...

Lightning Of Match. ...

Setting Cement And Concrete.

Explanation:

7 0
3 years ago
Read 2 more answers
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