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Irina-Kira [14]
3 years ago
15

I want to ask a girl out what do i say please help

Chemistry
2 answers:
alexandr1967 [171]3 years ago
3 0

Answer:

just tell her that you like her and would like to take her out some time and if she says yes just be your self

Explanation:

taurus [48]3 years ago
3 0

Answer:

Compliment her and then go in for the i was thinking maybe we can go out sometime and it goes from there depending on her answer

Explanation:

i hope i helped =)

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Answer: 02 라

Explanation:

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3 years ago
PLS HELP I WILL REWARD MOST BRAINLIEST
kvv77 [185]

Answer:

Actual yield = 138.94 g

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2Na + Cl₂ —> 2NaCl

Next, we shall determine the mass of Na that reacted and the mass of NaCl produced from the balanced equation. This can be obtained as follow:

Molar mass of Na = 23 g/mol

Mass of Na from the balanced equation = 2 × 23 = 46 g

Molar mass of NaCl = 23 + 35.5

= 58.5 g/mol

Mass of NaCl from the balanced equation = 2 × 58.5 = 117 g

SUMMARY:

From the balanced equation above,

46 gof Na reacted to produce 117 g of NaCl.

Next, we shall determine the theoretical yield NaCl. This can be obtained as follow:

From the balanced equation above,

46 gof Na reacted to produce 117 g of NaCl.

Therefore, 57.50 g of Na will react to produce = (57.50 × 117)/46 = 146.25 g of NaCl.

Thus, the theoretical yield of NaCl is 146.25 g

Finally, we shall determine the actual yield of NaCl. This can be obtained as follow:

Percentage yield = 95%

Theoretical yield = 146.25 g

Actual yield =?

Percentage yield = Actual yield / Theoretical yield × 100

95% = Actual yield / 146.25

Cross multiply

Actual yield = 95% × 146.25

Actual yield = 95/100 × 146.25

Actual yield = 0.95 × 146.25

Actual yield = 138.94 g

4 0
3 years ago
g A sample of unknown gas has a mass of 1.95 g and occupies 3.00 L at 1.25 atm and 20 C. What is the molar mass of the unknown g
Viktor [21]

Answer:

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V = 3L

P = 1.25 atm

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m = 1.95g

Molar mass = ?

The variables are related by the follwing equation;

pV = nRT

where r = gas constant = 0.0821 L atm K−1

Solving for n, we have;

n = pV / RT

n = (1.25 * 3 ) / (0.0821 * 293)

n = 3.75 / 24.0553 = 0.1559 mol

The relationship between number of moles, n and molar mass is given as;

n = mass / molar mass

Molar mass = Mass / n = 1.95 / 0.1559

Molar mass = 12.51 g/mol

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