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pochemuha
2 years ago
13

2 Na + Cl ---> 2NaCl

Chemistry
1 answer:
Greeley [361]2 years ago
5 0

Answer:

1. d. 84g

2. c. 77.4%

Explanation:

1. Based on the reaction 2 moles of Na produce 2 moles of NaCl.

The moles of Na (22.99g/mol) in 33g are:

33g * (1mol / 22.99g) = 1.435 moles Na = 1.435 moles NaCl must be produced.

The mass of NaCl is (Molar mass: 58.44g/mol):

1.435moles * (58.44g / mol) = 84g

d. 84g

2. Percent yield is defined as 100 times the ratio between actual yield (65g) and theoretical yield (84g). The percent yield is:

65g / 84g * 100 =

c. 77.4%

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3 years ago
The analysis of compound only magnesium, phosphorus and oxygen showed 36.23% MgO and 63.77 % P2O5. set up the simplest formula
Rzqust [24]

Answer:

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Explanation:

4 0
2 years ago
Calculate the amount of heat required to raise the temperature of a 32g sample of water from 8°C to 22°C.
qwelly [4]

Answer:

The amount of heat required to raise the temperature of a 32g sample of water from 8°C to 22°C is 1,874.432 J

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

Between heat and temperature there is a direct proportional relationship. The constant of proportionality depends on the substance that constitutes the body and its mass, and is the product of the specific heat and the mass of the body. So, the equation that allows to calculate heat exchanges is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature.

In this case:

  • c= 4.184 \frac{J}{g*C}
  • m= 32 g
  • ΔT= Tfinal - Tinitial= 22°C - 8°C= 14°C

Replacing:

Q= 32 g* 4.184 \frac{J}{g*C} *14 °C

Solving:

Q= 1,874.432 J

<u><em>The amount of heat required to raise the temperature of a 32g sample of water from 8°C to 22°C is 1,874.432 J</em></u>

7 0
2 years ago
What volume of chlorine gas at 27 °C, 812 mmHg, is required to react with an excess of carbon disulfide so that 5.00kJ of heat i
Kamila [148]

Answer:

The correct answer is 1.21 L.

Explanation:

Based on the given information, the reaction will be,

CS2 (l) + 3Cl2 (g) ⇒ CCl4 (l) + S2Cl2 (l)

By using the standard values of the substances, the standard enthalpy of the reaction is,

ΔH° = [(-139.5) + (-58.5) – 0 – (87.3)] kJ/mol

= -285.3 kJ/mol

The amount of heat evolved for 3 moles of chlorine reacted us 285.3 kJ.

Now the number of moles of chlorine needed to react to produce 5.00 kJ is,

= 5.00 kJ × 3 mol Cl2/285.3 kJ

= 0.0526 mol Cl2

Now the volume of chlorine gas at 27degree C and 812 mmHg will be,

Volume = 0.0526 mol Cl2 × 0.0821 Latm/mol K × 300 K/ 1.07 atm

= 1.21 L

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