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3 years ago

An aircraft is taking a group of skydivers up into the air. Evan is dressed in

Chemistry

1 answer:

meriva3 years ago

7 0

4,410 kJ

Explanation:

Gravitational Potential Enegry (GPE) is calculated as;

GPE = <em>m*g*h</em> where;

m = mass (kg)

g = gravity (m/s²)

h = height (meters)

= 90 * 9.8 * 5000

= 4,410,000‬ joules

= 4,410 kJ

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Sav [38]

Explanation:

what is the end of the question???

4 0

3 years ago

How many grams of ammonia (NH3) can be produced by the synthesis of excess hydrogen gas (H2) and 253.8 grams of nitrogen gas (N2

kogti [31]

Answer:

308.2 g of NH₃.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

3H₂ + N₂ —> 2NH₃

Next, we shall determine the mass of N₂ that reacted and the mass of NH₃ produced from the balanced equation. This can be obtained as follow:

Molar mass of N₂ = 2 × 14 = 28 g/mol

Mass of N₂ from the balanced equation = 1 × 28 = 28 g

Molar mass of NH₃ = 14 + (3×1)

= 14 + 3 = 17 g/mol

Mass of NH₃ from the balanced equation = 2 × 17 = 34 g

Summary:

From the balanced equation above,

28 g of N₂ reacted to produce 34 g of NH₃.

Finally, we shall determine the mass of NH₃ produced by the reaction of 253.8 g of N₂. This can be obtained as illustrated below:

From the balanced equation above,

28 g of N₂ reacted to produce 34 g of NH₃.

Therefore, 253.8 g of N₂ will react to produce = (253.8 × 34)/28 = 308.2 g of NH₃.

Thus, 308.2 g of NH₃ were obtained from the reaction.

8 0

3 years ago

How to tell the formula for elements compounds and mixtures?

Kay [80]

Search it up on google

3 0

4 years ago

Under which class of substitution reaction does this reaction appear to fall

liberstina [14]

This reaction is most likely to fall under SN2 because the thing called carbonication does not occur in SN1. The carbon forms a partial bond with the nucleophile during the intermediate phase and the leaving group. So for this question the reaction will fall under SN2.

3 0

3 years ago

Consider the reaction 2CO * O2 —> 2 CO2 what is the percent yield of carbon dioxide (MW= 44g/mol) of the reaction of 10g of c

Arturiano [62]

Answer:

Y = 62.5%

Explanation:

Hello there!

In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:

$m_{CO_2}^{theoretical}=10gCO*\frac{1molCO}{28gCO}*\frac{2molCO_2}{2molCO} *\frac{44gCO_2}{1molCO_2}\\\\ m_{CO_2}^{theoretical}=16gCO_2$

Finally, given the actual yield of the CO2-product, we can calculate the percent yield as shown below:

$Y=\frac{10g}{16g} *100\%\\\\Y=62.5\%$

Best regards!

8 0

3 years ago