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Svet_ta [14]
2 years ago
5

If 45.0 mL of a 4.00 M sodium sulfate (Na2SO4) solution is used for a reaction, how many moles of solute were present in the sol

ution?
____ moles
Chemistry
1 answer:
mixas84 [53]2 years ago
6 0
797 jenehsbsehshehehbehshehhebwbbss
You might be interested in
ammonium nitrate, which is prepared from nitric acid, is used as a nitrogen fertilizer. determine the percent composition of amm
Alina [70]

Answer:

Ammonium nitrate, (NH4NO3), a salt of ammonia and nitric acid, used widely in fertilizers and explosives. The commercial grade contains about 33.5 percent nitrogen, all of which is in forms utilizable by plants; it is the most common nitrogenous component of artificial fertilizers.

5 0
2 years ago
Can someone pls help me with this question
artcher [175]

Answer:

The answer to your question is 24.325

Explanation:

Data

Magnesium-24  Abundance = 78.70%

Magnesium-25  Abundance = 10.13%

Magnesium-26  Abundance = 11.17%

Process

1.- Convert the abundance to decimals

Magnesium-24  Abundance = 78.70/100 = 0.787

Magnesium-25  Abundance = 10.13/100 = 0.1013

Magnesium-26  Abundance = 11.17/100 = 0.1117

2.- Write an equation

Average atomic mass = (Atomic mass-1 x Abundance 1) + (Atomic mass 2 x

                                       Abundance-2) + (Atomic mass 3 x Abundance 3)

3.- Substitution

Average atomic mass = (24 x 0.787) + (25 x 0.1013) + (26 x 0.1117)

4.- Simplification

Average atomic mass = 18.888 + 2.533 + 2.904

5.- Result

Average atomic mass = 24.325

5 0
3 years ago
How many liters is in a gallon
solong [7]
1 Gallon = 3.7854118 Liters
5 0
3 years ago
Read 2 more answers
Use standard reduction potentials to calculate the equilibrium constant for the reaction: 2Cr3+(aq) + Pb(s)2Cr2+(aq) + Pb2+(aq)
inn [45]

Answer:

3.47 ×10^-10

Explanation:

The equation of the reaction is 2Cr3+(aq) + Pb(s)------->2Cr2+(aq) + Pb2+(aq)

A total of two moles of electrons were transferred in the process. The chromium was reduced while the lead was oxidized. Hence the lead species will constitute the oxidation half equation and the chromium will constitute the reduction half equation.

E°cell = E°cathode - E°anode

E°cathode = -0.41 V

E°anode = -0.13 V

E°cell = -0.41 -(-0.13) = -0.28 V

From

E°cell = 0.0592/n log K

n= 2, K= the unknown

-0.28 = 0.0592/2 log K

log K = -0.28/0.0296

log K = -9.4595

K = Antilog ( -9.4595)

K= 3.47 ×10^-10

4 0
3 years ago
Helppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp
vlabodo [156]

Answer:

1. Delete lion, bison, reindeer, and giraffe

2. Delete cactus, oak, evergreen,

Explanation:

1. Lions live in the African savanna, bison live in open fields with lots of grass, reindeer live in Antarctica and other very cold places, and giraffes also live in the areas of the African savanna where there are trees.

2. Cacti live in the desert, oaks and evergreens live in open grassy areas.

Hope this helps

5 0
3 years ago
Read 2 more answers
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