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Sedaia [141]
2 years ago
11

(need help asap)

Chemistry
1 answer:
stellarik [79]2 years ago
7 0

Explanation:

150 grams is .15 liters

1 mole of gas at STP occupies 22.414 liters

.15 liter of hydrogen represents .15/22.414 =

0.0066922 moles of H2

equation is

Zn + 2HCl → ZnCl2 + H2

1 mole of Zn will make 1 mole of H2

0.0066922 moles of Zn will make 0.0066922 moles of H2

To convert this molar quantity of zinc into grams we simply multiply by the atomic mass of zinc, which is

65.38 g/mol;-

Mass of Zn required = 0.0066922 moles * 65.38 g/mol = 0.4375390381 grams

https://www.quora.com/How-many-grams-of-Zinc-would-you-need-to-react-with-Hydrochloric-acid-to-produce-1-L-of-H2-gas

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The answers please to this
Vlad1618 [11]
I know for number 4 the answer is c, sorry I can't help with the others.
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3 years ago
For the reaction below , label each reactant as an electron pair acceptor or electron pair donor and as a Lewis acid or a Lewis
JulsSmile [24]

Since the reaction shown in the question is an acid - base reaction in the Lewis sense; the Lewis acid here is AlCl3  while the Lewis base here is Cl^- .

<h3>What is  a Lewis acid?</h3>

A Lewis acid is a substance that accepts electron pair while a Lewis base donates an electron pair.

Now consider the given reaction; AlCl3 +Cl^- ------> AlCl 4 ^-. The Lewis acid here is AlCl3  while the Lewis base here is Cl^- .

Learn more about acid - base reaction: brainly.com/question/14356798

6 0
2 years ago
How many grams of carbon dioxide will form if 5.5 g of C3H8 burns in 15 g of O2?
mr Goodwill [35]
C3H8+3O2--->3CO2+8H
Therefore for every 1:3 there are 3 Carbon dioxides that form. That means find the limiting reactant from the two reactants.
5.5g(1mole C3H8/44.03g of C3H8)=0.1249 moled of C3H8 and if for every one C3H8 we can form three CO2. We can assume 0.3747 miles of CO2 will be produced.
15g of O2(1 mole O2/32g of O2)=0.4685moles O2 and if for every three O2 we can produce three CO2 we may assume a 1:1 ratio.
This means C3H8 will be your limiting reactant. Therefore 0.3747 moles of CO2 will be produced.
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5 0
3 years ago
Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel ro
melamori03 [73]

<u>Answer:</u> The mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

  • <u>For nickel:</u>

Given mass of nickel = 14.8 g

Molar mass of nickel = 58.7 g/mol

Putting values in equation 1, we get:

\text{Moles of nickel}=\frac{14.8g}{58.7g/mol}=0.252mol

For the given chemical reaction:

3NiO(s)+2Al(s)\rightarrow 3Ni(l)+Al_2O_3(s)

  • <u>For nickel (II) oxide:</u>

By Stoichiometry of the reaction:

3 moles of nickel are produced from 3 moles of nickel (II) oxide

So, 0.252 moles of nickel will be produced from \frac{3}{3}\times 0.252=0.252mol of nickel (II) oxide

Now, calculating the mass of nickel (II) oxide by using equation 1:

Molar mass of nickel (II) oxide = 74.7 g/mol

Moles of nickel (II) oxide = 0.252 moles

Putting values in equation 1, we get:

0.252mol=\frac{\text{Mass of nickel (II) oxide}}{74.7g/mol}\\\\\text{Mass of nickel (II) oxide}=(0.252mol\times 74.7g/mol)=18.8g

  • <u>For aluminium:</u>

By Stoichiometry of the reaction:

3 moles of nickel are produced from 2 moles of aluminium

So, 0.252 moles of nickel will be produced from \frac{2}{3}\times 0.252=0.168mol of aluminium

Now, calculating the mass of aluminium by using equation 1:

Molar mass of aluminium = 27 g/mol

Moles of aluminium = 0.168 moles

Putting values in equation 1, we get:

0.168mol=\frac{\text{Mass of aluminium}}{27g/mol}\\\\\text{Mass of aluminium}=(0.168mol\times 27g/mol)=4.54g

Hence, the mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

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ELEN [110]
The unit of heat is Joule
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