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Hydrazine (N2H4), a rocket fuel , reacts with oxygen to form nitrogen gas and water vapor. The reaction is represented with the

Serggg [28]

85.33 grams of hydrazine is needed for the reaction.

Rocket fuel contains the reaction between hydrazine and oxygen to break and form nitrogen and water vapour along with steam as it is an exothermic reaction.

<h3>How to calculate the hydrazine amount?</h3>

The reaction can be shown as:

$\rm N_{2}H_{4}(l) + O_{2}(g) \rightarrow N_{2}(g) + 2H_{2}O(g)$

From the periodic chart, it can be said that:

Mass of nitrogen = 14 gm
Mass of oxygen = 16 gm
Mass of hydrogen = 1 gm

The molar mass of hydrazine will be:

$\begin{aligned} \rm Molar \;mass \;of \rm \;N_{2}H_{4}&= 2(14) + 4(1) \\\\&= 32 \;\rm grams\end{aligned}$

Mass of water:

$\begin{aligned}&= 2(1) + 16 \\\\&= 18 \;\rm grams\end{aligned}$

From the reaction, it can be said that,

1 mole of $\rm N_{2}H_{4}$ = 2 moles of $\rm H_{2}O$

32 grams of $\rm N_{2}H_{4}$ = ? grams of water

$\begin{aligned}\text{Gram of water} &= 18 \times 2\\\\&= 36 \;\rm gm\end{aligned}$

So, for 96 gm of water the hydrazine produced will be,

$\begin{aligned}\text{mass of }\rm N_{2}H_{4} &= \dfrac{96 \times 32}{36}\\\\&= 85.3334 \;\rm grams \end{aligned}$

Therefore, 85.33 grams of hydrazine is needed for the reaction.

Learn more about rocket fuel here:

brainly.com/question/10179270

7 0

2 years ago

What sort of nucleophile does NADH supply?

Elden [556K]

Answer:

Hydride ion

Explanation:

You often see the reduction with NADH written as

NADH ⟶ NAD⁺ + H⁺ + 2e⁻

If you think about it, H⁺ + 2e⁻ is equivalent to H:⁻, so we could write the reaction as

NADH ⟶ NAD⁺ + H:⁻

In terms of a mechanism, the dihydropyridine ring of NADH transfers a hydrogen atom with its pair of electrons (a hydride ion) to the substrate and becomes the more stable, aromatic pyridinium ion in NAD⁺.

7 0

4 years ago

What is the concentration of hydrogen ions in a solution with a pH of 2.36?

Viefleur [7K]

Answer:

b)

Explanation:

$pH = - log[H {}^{ + } ] \\ 2.36 = - log[H {}^{ + } ] \\ [H {}^{ + } ] = {10}^{ - 2.36} \\ [H {}^{ + } ] = 4.37 \times {10}^{ - 3} \: M$

8 0

3 years ago

A gas sample is held at constant pressure. The gas occupies 2.97 L of volume when the temperature is 21.6°C. Determine the tempe

STALIN [3.7K]

Answer:

339.2K

Explanation:

Using Charles law equation;

V1/T1 = V2/T2

Where;

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question,

V1 = 2.97 L

V2 = 3.42 L

T1 = 21.6°C = 21.6 + 273 = 294.6K

T2 = ?

Using V1/T1 = V2/T2

2.97/294.6 = 3.42/T2

Cross multiply

2.97 × T2 = 294.6 × 3.42

2.97T2 = 1007.532

T2 = 1007.532 ÷ 2.97

T2 = 339.236

The final temperature is 339.2K

5 0

3 years ago

Give chemical equation representing ionisation of carbonic acid

boyakko [2]

Answer:

HCO3− + OH− ⇌ CO32− + H2O

hope it helps!!!!

6 0

3 years ago

There is a close relation between forestand wildlife.​

There is a close relation between forestand wildlife.