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Nutka1998 [239]
3 years ago
8

Help please :< I just need the answer for this as quickly as possible.

Chemistry
2 answers:
11Alexandr11 [23.1K]3 years ago
8 0

Answer:

I think its the 3th one

Explanation:

Anni [7]3 years ago
8 0
It’s the 3rd one or as you can say C
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How many grams would be in 4.45 x 10^23 molecules of N203
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Answer:

56.24g

Explanation:

To find the mass of N2O3 in 4.45 x 10^23 molecules, it must first be converted to moles by dividing the number of molecules in N2O3 by Avagadro's number (6.02 × 10^23).

number of moles in N2O3 = 4.45 x 10^23 ÷ 6.02 × 10^23

n = 4.45/6.02 × 10^(23 - 23)

n = 0.74 × 10^0

n = 0.74moles.

Using the formula below to find the mass of N2O3;

mole = mass ÷ molar mass

Molar mass of N2O3 = 14(2) + 16(3)

= 28 + 48

= 76g/mol

mass = mole × molar mass

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Questioning, creative, open-minded, objective

Explanation:

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4 years ago
If you have any two elements, how do you tell which one conducts heat and electricity better?
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4. A piece of metal weighing 0.0713 g was placed in a eudiometer containing dilute aqueous HCl. After the metal fully dissolved,
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Complete Question

4. A piece of metal weighing 0.0713 g was placed in a eudiometer containing dilute aqueous HCl. After the metal fully dissolved, 23.5 mL of hydrogen gas was collected by displace-ment of water and a 400 mm column of water was observed. The water temperature was 258C and the barometric pressure was 758.8 mm Hg (torr). Refer to the Introduction and data sheet to solve the following problems.

a) What is the vapor pressure of the water vapor in the column? (Consult Appendix E.)

b) What is the pressure of the water column expressed in mm Hg (torr)? The density of mercury is 13.6 g/mL.

c) Calculate the pressure of the hydrogen gas above the water in the column.

d) Calculate the volume occupied by the hydrogen gas at STP.

Answer:

a)  25\textdegree C=23.8 torrs

b) P_w=758.8

c)  P_w=758.8=735torr

d)  V_2=20.82mL

Explanation:

From the question we are told that:

Metal weight M_m=0.0713g

Volume Hydrogen V_h=23.5mL

Displace-ment Column of water 400 mm column of water

Temperature T =258\textdegree C

Barometric Pressure p=758.8mmHg

Vapour Pressure of water at 25^oC

Generally from (Consult Appendix E.)

a)

Va-pour Pressure of water at

25\textdegree C=23.8 torrs

b)

Pressure of Water column

P_w=758.8

c) Pressure of Water column (Consult Appendix E.)

P_w=758.8=735torr

d)

Generally the equation for ideal gas is mathematically given by

 \frac{p_1v_1}{T_1}=\frac{p_2v_2}{T_2}

Therefore

 V_2=\frac{p_1V_1T_2}{T_1p_2}

 V_2=\frac{735*23.5*273}{298*760}

 V_2=20.82mL

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