What can the shape of a molecule affect?

Over [174]

In the given reaction potassium metal is placed in water and when a reactive metal is placed in water, it reacts with water to form metal hydroxide and hydrogen gas, as a result, potassium hydroxide and hydrogen gas are produced in the reaction. A redox reaction is a reaction in which a substance is oxidized during the reaction whereas some other substance is reduced during the reaction, simultaneously.

The given chemical reaction is:

2K(s) + 2H20(l) arrow 2KOH(aq) + H2(g)

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8 0

3 years ago

Pure nitrogen (N2) and pure hydrogen (H2) are fed to a mixer. The product stream has 40.0% mole nitrogen and the balance hydroge

LuckyWell [14K]

Explanation:

The given data is as follows.

Mass flow rate of mixture = 1368 kg/hr

$N_{2}$ in feed = 40 mole%

This means that $H_{2}$ in feed = (100 - 40)% = 60%

We assume that there are 100 total moles/hr of gas $(N_{2} + H_{2})$ in feed stream.

Hence, calculate the total mass flow rate as follows.

40 moles/hr of N_{2}/hr (28 g/mol of $N_{2}$ ) + 60 moles/hr of $H_{2}/hr$ (2 g/mol of $H_{2}$ )

$40 \times 28 g/hr + 60 \times 2 g/hr$

= 1120 g/hr + 120 g/hr

= 1240 g/hr

= $\frac{1240}{1000}$ (as 1 kg = 1000 g)

= 1.240 kg/hr

Now, we will calculate mol/hr in the actual feed stream as follows.

$\frac{100 mol/hr}{1.240 kg/hr} \times 1368 kg/hr$

= 110322.58 moles/hr

It is given that amount of nitrogen present in the feed stream is 40%. Hence, calculate the flow of $N_{2}$ into the reactor as follows.

$0.4 \times 110322.58 moles/hr$

= 44129.03 mol/hr

As 1 mole of nitrogen has 28 g/mol of mass or 0.028 kg.

Therefore, calculate the rate flow of $N_{2}$ into the reactor as follows.

$0.028 kg \times 44129.03 mol/hr$

= 1235.612 kg/hr

Thus, we can conclude that the the feed rate of pure nitrogen to the mixer is 1235.612 kg/hr.

3 0

3 years ago

PLEASE HELP!!!! show work and also balance equation plus stoic

siniylev [52]

Answer:

:)

Explanation:

3 0

3 years ago

What do scientists use molecular clocks to estimate?

astra-53 [7]

I think the answers B

4 0

3 years ago

Determine the concentrations of BaBr2, Ba2 , and Br– in a solution prepared by dissolving 1.18 × 10–4 g BaBr2 in 1.00 L of water

Anika [276]

Supposing complete ionization:
BaBr2 → Ba{2+} + 2 Br{-} 

(2.23 × 10^–4 g BaBr2) / (297.135 g BaBr2/mol) / (2.00 L) = 3.75 × 10^-7 mol/L BaBr2 

(3.75 × 10^-7 mol/L BaBr2) x (1 mol Ba{2+} / 1 mol BaBr2) = 3.75 × 10^-7 mol/L Ba{2+} 

(3.75 × 10^-7 mol/L BaBr2) x (2 mol Br(-} / 1 mol BaBr2) = 7.50 × 10^-7 mol/L Br{-}

5 0

3 years ago