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Consider the reaction between 15.0 mL of a 1.00 M aqueous solution of AgNO3 and 10.0 mL of a 1.00 M aqueous solution of K2CrO4.

Neporo4naja [7]

Answer:

Species present in the solution after complete reaction: $K^+ ,CrO_4^- , Ag_2CrO_4 , K^+ ,NO_3^-$

Explanation:

First balance the chemical equation,

$2AgNO_3 + K_2CrO_4 =Ag_2CrO_4 + 2KNO_3$

calculation of mili moles of each:

$mili mole= volume in ml \times molarity$

mili mole of AgNO3= 15mili mole;

mili mole of K2CrO4=10 mili mole;

From balance equation,

2 mili mole of AgNO3 reacts with 1 mili mole of K2CrO4

hence 1 mili mole of AgNO3 reacts with 0.5 mili mole of K2CrO4

15 mili mole of AgNO3 reacts with 7.5 mili mole of K2CrO4

so only 7.5 milimole will be used in the reaction and 2.5 mili mole left in solution.

species present in the solution after complete reaction:

$K^+ ,CrO_4^- , Ag_2CrO_4 , K^+ ,NO_3^-$

$K_2CrO4 , KNO_3$ are soluble in water so it will dissociate in the solution

7 0

4 years ago

Suppose that one of millikan's oil drops has a charge of −4.8×10−19

svlad2 [7]

The best and correct option is c

8 0

3 years ago

Read 2 more answers

You are a NASCAR pit crew member. Your employer is leading the race with 20 laps to go. He just finished a pit stop and has 5.0

jek_recluse [69]

Answer:

Since there are 3500 g of fuel left in the tank, and he needs only 1687.5 g to complete 20 laps, he has enough fuel to complete the race. I will tell the driver that he does not need to make another pit stop as he has enough fuel to complete the race.

Explanation:

Density = mass / volume

Density of fuel = 700 g/ 1 gal

Therefore, the mass of fuel in 1 gallon = 700 g

The driver has 5.0 gallons of fuel in the tank.

The mass of 5.0 gallons of fuel = 5 × 700 = 3500 g of fuel

Equation of the combustion of fuel, C₅H₁₂ is given below:

C₅H₁₂ + 8 O₂ ---> 6 H₂O + 5 CO₂

1 mole C₅H₁₂ requires 8 moles of O₂

1 mole of C₅H₁₂ has a mass = 72 g

8 moles of O₂ has a mass = 256 g

Therefore, 300 g of O₂ will require 300 × (72/256) g of C₅H₁₂ = 84.375 g of C₅H₁₂

84.375 g of fuel is used by the car per lap;

20 laps will require 20 × 84.375 g of fuel = 1687.5 g of fuel.

Since there are 3500 g of fuel left in the tank, and he needs only 1687.5 g to complete 20 laps, he has enough fuel to complete the race. I will tell the driver that he does not need to make another pit stop as he has enough fuel to complete the race.

7 0

3 years ago

State one function of Emitter??

stellarik [79]

Answer:

To emit

Explanation:

yea

7 0

3 years ago

Please help me vote you brainiest

IceJOKER [234]

Answer:

If two forces act on an object in the same direction, the net force is equal to the sum of the two forces. This always results in a stronger force than either of the individual forces alone

4 0

3 years ago