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rjkz [21]
3 years ago
11

How many molecules are there in 840 grams of na2SO4

Chemistry
1 answer:
beks73 [17]3 years ago
4 0

Answer:

3.56x10^24 molecules

Explanation:

The following data were obtained from the question:

Mass of Na2SO4 = 840g

Molar Mass of Na2SO4 = (23 x2) + 32 + (16x4) =46 + 12 + 64 = 142g/mol

From Avogadro's hypothesis, 1mole of any substance contains 6.02x10^23 molecules. This also gives that 1mole of Na2SO4 contains 6.02x10^23 molecules.

If 1 mole (i.e 142g) contains 6.02x10^23 molecules,

Therefore, 840g of Na2SO4 will contain = (840 x 6.02x10^23)/142 = 3.56x10^24 molecules

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If you weighed out 0.38 g of calcium and reacted it with an excess of hcl, how many moles of h2(g) would you expect to produce?
Hitman42 [59]
Answer is: 0,0095 mol of hydrogen gas will be produced in reaction.
Chemical reaction: Ca + 2HCl → CaCl₂ + H₂.
m(Ca) = 0,38 g.
n(H₂) = ?
n(Ca) = m(Ca) ÷ M(Ca).
n(Ca) = 0,38 g ÷ 40 g/mol
n(Ca) = 0,0095 mol.
from reaction: n(Ca) : n(H₂) = 1 : 1.
n(H₂) = n(Ca) = 0,0095 mol.
n - amount of substance.
4 0
3 years ago
Given that cao(s) + h2o(l) → ca(oh)2(s), δh°rxn = –64.8 kj/mol, how many grams of cao must react in order to liberate 525 kj of
USPshnik [31]

Answer:

454.3 g.

Explanation:

  • From the given data:

1.0 mol of CaO liberates → – 64.8 kJ.

??? mol of CaO liberates → - 525  kJ.

∴ The no. of moles needed = (1.0 mol)(- 525 kJ)/(- 64.8 kJ) = 8.1 mol.

<em>∴ The no. of grams of CaO needed = no. of moles x molar mass</em> = (8.1 mol)(56.077 g/mol) = <em>454.3 g.</em>

8 0
2 years ago
A student with an unknown code of Q3A conducts a qualitative analysis of her unknown. Her results indicated the formation of a y
insens350 [35]

Answer:

Grey precipitate implies the presence of silver ions

Yellow precipitate implies the presence of lead II ions

Explanation:

Qualitative analysis provides us a quick method of identifying ions present in a sample by chemical reactions involving simple reagents. Precipitates having a unique colour is formed. The identity of ions in the sample is deduced from the colour of precipitate obtained when particular reagents are added.

In the question, a precipitate containing silver ions upon standing turn into grey colour. Similarly, lead II ions give a yellow precipitate.

6 0
2 years ago
In a science lab, a student heats up a chemical from 10 °C to 25 °C which requires thermal energy of 30000 J. If mass of the obj
olganol [36]

Answer:

The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)

Explanation:

Step 1: Data given

Initial temperature = 10.0 °C

Final temperature = 25.0 °C

Energy required = 30000 J

Mass of the object = 40.0 grams

Step 2: Calculate the specific heat capacity of the object

Q = m* c * ΔT

⇒With Q = the heat required = 30000 J

⇒with m = the mass of the object = 40.0 grams

⇒with c = the specific heat capacity of the object = TO BE DETERMINED

⇒with ΔT = The change in temperature = T2 - T2 = 25.0 °C - 10.0°C = 15.0 °C

30000 J = 40.0 g * c * 15.0 °C

c = 30000 J / (40.0 g * 15.0 °C)

c = 50 J/g°C

The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)

3 0
3 years ago
Please answer all of the questions
anzhelika [568]

Here's your ans

Explanation:

1-1.50

2-d

3-c

4-b

5-d

Tq

3 0
2 years ago
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