Answer:
2,137... gallons
Explanation:
(160+140+14x16x2)/350=2,137...
Answer:
and 
Explanation:
Our goal for this question is the calculation of the number of moles of the molecules produced by the reaction of hydrazine (
) and <u>oxygen</u> (
). So, we can start with the <u>reaction</u> between these compounds:
Now we can <u>balance the reaction</u>:
In the problem, we have the values for both reagents. Therefore we have to <u>calculate the limiting reagent</u>. Our first step, is to calculate the moles of each compound using the <u>molar masses values</u> (32.04 g/mol for
and 31.99 g/mol for
):


In the balanced reaction we have 1 mol for each reagent (the numbers in front of
and
are 1). Therefore the <u>smallest value would be the limiting reagent</u>, in this case, the limiting reagent is
.
With this in mind, we can calculate the number of moles for each product. In the case of
we have a <u>1:1 molar ratio</u> (1 mol of
is produced by 1 mol of
), so:

We can follow the same logic for the other compound. In the case of
we have a <u>1:2 molar ratio</u> (2 mol of
is produced by 1 mol of
), so:

I hope it helps!
Ans: grinding one of the reactants into a powder.
this increases the surface area and hence there would be more chances of collisions happening. therefore reaction rate would increase
47 Kpa would bw the pressure needes
= 6.022 × 1020
Explanation<em>;</em>
Mole of aluminium oxide (Al2O3) is
⇒ 2 x 27 + 3 x 16
Mole of aluminium oxide = 102 g
i.e., 102 g of Al2O3= 6.022 x 1023 molecules of Al2O3
Then, 0.051 g of Al2O3 contains = 6.022 x 1023 / (102 x 0.051 molecules)
= 3.011 x 1020 molecules of Al2O3
The number of aluminium ions (Al3+) present in one molecule of aluminium oxide is 2.
Therefore, the number of aluminium ions (Al3+) present in 3.11 × 1020 molecules (0.051g) of aluminium oxide (Al2O3)
= 2 × 3.011 × 1020
=<em> 6.022 × 1020</em>
<em>hope </em><em>it </em><em>helps</em><em>_</em>