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tatiyna
2 years ago
11

During WW2, battleships would shoot their 16-inch main guns to project an 800km projectile at the enemy as far as 22 miles aways

. If the projectile leaves the muzzle moving at +850m/s, what is the momentum of that projectile?
Chemistry
1 answer:
Nesterboy [21]2 years ago
3 0

The momentum of the projectile is 6.8 * 10⁵ kgm/s

<h3>Momentum and projectiles</h3>

Momentum of an object is the product of the mass of that object and its velocity.

  • Momentum, p = m*v

where m is mass of object

v is velocity of object

A projectile is an object launched into space and allowed to fall freely under the force of gravity.

The momentum of a projectile is conserved in the horizontal direction since no net force acts on the projectile in the horizontal direction.

using the formula for momentum;

p = m*v

Assuming the mass of the projectile to be 800 kg

Velocity of the projectile = 850 m/s

p = 800 kg * 850 m/s

p = 6.8 * 10⁵ kgm/s

Therefore, the momentum of the projectile is 6.8 * 10⁵ kgm/s

Learn more about momentum and projectiles at: https://brainly.in/question/201038

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What mass of Fe(OH)3 is produced when 35 mL of 0.250 M Fe(NO3)3 solution is mixed with 55 mL of a 0.180 M
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Answer:

0.35 g.

Explanation:

We'll begin by calculating the number of mole of Fe(NO3)3 in 35 mL of 0.250 M Fe(NO3)3 solution.

This is illustrated below:

Molarity of Fe(NO3)3 = 0.250 M

Volume = 35 mL = 35/1000 = 0.035 L

Mole of Fe(NO3)3 =?

Molarity = mole /Volume

0.250 = mole of Fe(NO3)3 / 0.035

Cross multiply

Mole of Fe(NO3)3 = 0.25 x 0.035

Mole of Fe(NO3)3 = 8.75×10¯³ mole

Next, we shall determine the number of mole of KOH in 55 mL of 0.180 M

KOH solution. This is illustrated below:

Molarity of KOH = 0.180 M

Volume = 55 mL = 55/1000 = 0.055 L

Mole of KOH =.?

Molarity = mole /Volume

0.180 = mole of KOH /0.055

Cross multiply

Mole of KOH = 0.180 x 0.055

Mole of KOH = 9.9×10¯³ mole.

Next, we shall write the balanced equation for the reaction. This is given below:

3KOH + Fe(NO3)3 —> Fe(OH)3 + 3KNO3

From the balanced equation above,

3 moles of KOH reacted with 1 mole of Fe(NO3)3 to produce 1 mole of Fe(OH)3.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

3 moles of KOH reacted with 1 mole of Fe(NO3)3.

Therefore, 9.9×10¯³ mole of KOH will react with = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(NO3)3.

From the above illustration, we can see that only 3.3×10¯³ mole out of 8.75×10¯³ mole of Fe(NO3)3 given is needed to react completely with 9.9×10¯³ mole of KOH.

Therefore, KOH is the limiting reactant and Fe(NO3)3 is the excess reactant.

Next, we shall determine the number of mole of Fe(OH)3 produced from the reaction.

In this case, we shall use the limiting reactant because it will give the maximum yield of Fe(OH)3 as all of it is consumed in the reaction.

The limiting reactant is KOH and the mole of Fe(OH)3 produce can be obtained as follow:

From the balanced equation above,

3 moles of KOH reacted to produce 1 mole of Fe(OH)3.

Therefore, 9.9×10¯³ mole of KOH will react to produce = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(OH)3.

Finally, we shall convert 3.3×10¯³ mole of Fe(OH)3 to grams. This can be obtained as follow:

Molar mass of Fe(OH)3 = 56 + 3(16 + 1) = 56 + 3(17) = 107 g/mol

Mole of Fe(OH)3 = 3.3×10¯³ mole

Mass of Fe(OH)3 =?

Mole = mass /Molar mass

3.3×10¯³ = Mass of Fe(OH)3 / 107

Cross multiply

Mass of Fe(OH)3 = 3.3×10¯³ x 107

Mass of Fe(OH)3 = 0.3531 ≈ 0.35 g.

Therefore, 0.35 g of Fe(OH)3 was produced from the reaction.

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