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Kryger [21]
3 years ago
9

___ may be rinsed over materials to was off explosive residue. A) water B) bleach C) acid D) acetone

Chemistry
2 answers:
Sergio [31]3 years ago
4 0

Answer: D.

Explanation: D. Acetone, Acetone may be rinsed over materials to wash off the explosive residue.

Hope this helps :)

inysia [295]3 years ago
3 0
D ) acetone bc the others don’t make the cut /.
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To calculate the number of moles, simply take the ratio of mass and molar mass. The molar mass of Iron (Fe) is equal to 55.85 g/mol.Therefore the total number of moles is:
moles Fe = 25.7 g / (55.85 g / mol)

moles Fe = 0.46 mole
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How many moles of helium gas are contained in a 4.0-L flask at STP?
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1.0 mole of gas at STP occupies 22.4 Litres.
4/0L / 22.4 L/mole = 0.179 moles He.
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Which ion with a charge of -2 has an electron configuration of 1s22s22p6 ?
tester [92]

Answer:

The answer is option B.

<h3>O2-</h3>

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3 years ago
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You throw a barbeque for your friends. Before you start the grill, your propane tank has a pressure of 78 psi. When you’re finis
Dvinal [7]

Answer:

b. 485 kPa

Explanation:

Gay-Lussac's law express that the pressure of a gas under constant volume is directly proportional to the absolute temperature. The equation is:

P1T2 = P2T1

<em>P is pressure and T absolute temperature of 1, initial state and 2, final state of the gas</em>

<em>Where P1 = 74psi</em>

<em>T2 = 20°C + 273.15 = 293.15K</em>

<em>P2 = ?</em>

<em>T1 = (95°F -32) * 5/9 + 273.15 = 308.15K</em>

<em />

Replacing:

74psi*293.15K = P2*308.15K

70.4psi

In kPa:

70.4psi * (6.895kPa / 1psi) =

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3 years ago
100 POINTS PLEASE HELP!! Honors Stoichiometry Activity Worksheet Instructions: In this laboratory activity, you will taste test
Shtirlitz [24]

Answer:

2 water + sugar + lemon juice → 4 lemonade

Moles of water present in 946.36 g of water=\frac{946.36 g}{236.59 g/mol}=4 mol=

236.59g/mol

946.36g

=4mol

Moles of sugar present in 196.86 g of water=\frac{196.86 g}{225 g/mol}=0.8749 mol=

225g/mol

196.86g

=0.8749mol

Moles of lemon juice present in 193.37 g of water=\frac{193.37 g}{257.83 g/mol}=0.7499 mol=

257.83g/mol

193.37g

=0.7499mol

Moles of lemonade in 2050.25 g of water=\frac{2050.25 g}{719.42 g/mol}=2.8498 mol=

719.42g/mol

2050.25g

=2.8498mol

As we can see that number of moles of lemon juice are limited.

So, we will consider the reaction will complete in accordance with moles of lemon juice.

1 mole lemon juice reacts with 2 mol of water,then 0.7499 mol of lemon juice will react with:

\frac{2}{1}\times 0.7499 mol = 1.4998 mol

1

2

×0.7499mol=1.4998mol of water

Mass of water used = 1.4998 mol × 236.59 g/mol=354.8376 g

Water remained unused = 946.36 g - 354.8376 g =591.5223 g

1 mole lemon juice reacts with mol of sugar,then 0.7499 mol of lemon juice will react with:

\frac{1}{1}\times 0.7499 mol = 0.7499 mol

1

1

×0.7499mol=0.7499mol of water

Mass of sugar used = 0.7499 mol × 225 g/mol = 168.7275 g

Sugar remained unused = 196.86 g - 28.1325 g

1 mole of lemon juice gives 4 moles of lemonade.

Then 0.7499 mol of lemon juice will give:

\frac{4}{1}\times 0.7499 mol=2.996 mol

1

4

×0.7499mol=2.996mol of lemonade

Mass of lemonade obtained = 2.996 mol × 719.42 g/mol = 2157.9722 g

Theoretical yield of lemonade = 2157.9722 g

Experimental yield of lemonade = 2050.25 g

Percentage yield of lemonade:

\frac{\text{Experimental yield}}{\text{theoretical yield}}\times 100

theoretical yield

Experimental yield

×100

\frac{2050.25 g}{2157.9722 g}\times 100=95.00\%

2157.9722g

2050.25g

×100=95.00%

6 0
3 years ago
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