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Marizza181 [45]
2 years ago
11

Describe the three steps involved in producing crystals of the salt copper sulfate.

Chemistry
1 answer:
denpristay [2]2 years ago
7 0

Answer:

Add copper (II) oxide (insoluble base), a little at a time to the warm dilute sulfuric acid and stir until the copper (II) oxide is in excess (stops disappearing) Filter the mixture into an evaporating basin to remove the excess copper (II) oxide. Leave the filtrate in a warm place to dry and crystallize.

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Which one of Newton's Laws fits this statement: More mass means more force needed to accelerate?
ivolga24 [154]

Answer:

Newton's Second Law

Explanation:

Newton's second law basically states that the acceleration of a body which is produced by a net force is directly proportional to the magnitude of net force applied in the same direction.

This tells us that

     F   is directly proportional to a

⇒ F= ma

So we can also state from the above equation, that when we have more mass, we need more net force to accelerate it. Here, we are keeping the acceleration constant so we can surely say that force and mass varies directly.

Therefore, we have made good use of Newton's Second Law of motion to arrive at this conclusion.

4 0
3 years ago
HELP Which of the following fractions can be used in the conversion of 32 m3 to the unit mm3?
poizon [28]
We know that each millimeter contains 10⁻³ meters. Writing this as a ratio:
1 mm : 10⁻³ m

We require a conversion from m³ to mm³, so we must take the cube of the ratio we have made:
1 mm³ = (10⁻³)³ m³

Therefore, the conversion used will be:
(1 mm / 10⁻³ m)³

When we multiply by this conversion, we will get:
32 m³ = 32 x 10⁹ mm³
7 0
3 years ago
If a small pig runs 46 meters in 6 seconds, what is the pig's average speed?
Tamiku [17]

Answer:

the pig's average speed is 7 m/s

7 0
2 years ago
PLEASEEEE HELP MEE!
Ugo [173]

Answer:

2nd option

Explanation:

7 0
3 years ago
A city continuously disposes of effluent from a wastewater treatment plant into a river. The minimum flow in the river is 130 m3
Vera_Pavlovna [14]

Answer:

2.54\ \text{mg/L}

Explanation:

C = Allowable concentration = 1.1 mg/L

Q_1 = Flow rate of river = 130\ \text{m}^/\text{s}

Q_2 = Discharge from plant = 37\ \text{m}^3/\text{s}

C_1 = Background concentration = 0.69 mg/L

C_2 = Maximum concentration that of the pollutant

The concentration of the mixture will be

C=\dfrac{Q_1C_1+Q_2C_2}{Q_1+Q_2}\\\Rightarrow C_2=\dfrac{C(Q_1+Q_2)-Q_1C_1}{Q_2}\\\Rightarrow C_2=\dfrac{1.1(130+37)-130\times 0.69}{37}\\\Rightarrow C_2=2.54\ \text{mg/L}

The maximum concentration that of the pollutant (in mg/L) that can be safely discharged from the wastewater treatment plant is 2.54\ \text{mg/L}.

6 0
2 years ago
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