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Murrr4er [49]
4 years ago
7

What is the mass of 1 mole of MgSO4?

Chemistry
1 answer:
White raven [17]4 years ago
7 0

Answer:

120.37 g

Explanation:

Mg = 24.31 g

S = 32.06 g

4 O = 16 X 4 = 64 g

Therefore, MgSO4 is (24.31 + 32.06 + 64) g = 120.37 g

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Determine the temperature of 1.42 mol of a gas contained in a 3.00-L container at a pressure of 123 kPa.
Schach [20]

Answer:

31.2K

Explanation:

The following data were obtained from the question:

n = 1.42 mol

V = 3L

P = 123 kPa = 123000Pa

Recall: 101325Pa = 1atm

123000Pa = 123000/101325 = 1.21atm

R = 0.082atm.L/Kmol

T =?

Using the ideal gas equation PV = nRT, we can obtain the temperature as follows:

PV = nRT

T = PV/nR

T = (1.21 x 3)/(1.42 x 0.082)

T = 31.2K

8 0
4 years ago
Give two reasons why a luminous flame is not used for heating purposes. (2mks) a​
Natali [406]

Answer:

The Two reasons are :-

1. Combustion efficiency

2. Standardization of test conditions as non-luminous flame.

Explanation:

A luminous flame isn't suitable for heating as it gives out soot (A black powdery or flaky substance consisting largely of amorphous carbon, produced by the incomplete burning of organic matter)

8 0
3 years ago
How much aluminum oxide are produced when 46.5g of Al react with 165.37g of MnO?
solong [7]

Aluminum oxide produced : = 79.152 g

<h3>Further explanation</h3>

Given

46.5g of Al

165.37g of MnO

Required

Aluminum oxide produced

Solution

Reaction

2 Al (s) + 3 MnO (s) → 3 Mn (s) + Al₂O₃ (s)

  • mol Al(Ar = 27 g/mol) :

mol = mass : Ar

mol = 46.5 : 27

mol = 1.722

  • mol MnO(Ar=71 g/mol) :

mol = 165.37 : 71

mol = 2.329

mol : coefficient ratio Al : MnO = 1.722/2 : 2.329/3 = 0.861 : 0.776

MnO as a limiting reactant(smaller ratio)

So mol Al₂O₃ based on MnO as a limiting reactant

From equation , mol Al₂O₃ :

= 1/3 x mol MnO

= 1/3 x 2.329

= 0.776

Mass Al₂O₃ (MW=102 g/mol) :

= 0.776 x 102

= 79.152 g

7 0
3 years ago
MARKING BRAINLIEST! please help asap, i need a and b, thank you
defon

The time required to reduce the concentration from 0.00757 M to 0.00180 M is equal to 1.52 × 10⁻⁴ s. The half-life period of the reaction is 9.98× 10⁻⁵s.

<h3>What is the rate of reaction?</h3>

The rate of reaction is described as the speed at which reactants are converted into products. A catalyst increases the rate of the reaction without going under any change in the chemical reaction.

Given the initial concentration of the reactant, C₀= 0.00757 M

The concentration of reactant after time t is C₁= 0.00180 M

The rate constant of the reaction, k = 37.9 M⁻¹s⁻¹

For the first-order reaction: C_t =C_0-kt

0.00180 = 0.00757 - (37.9) t

t =  1.52 × 10⁻⁴ s

The half-life period of the reaction: t_{\frac{1}{2} } =\frac{C_0}{2k}

t_{\frac{1}{2} } =\frac{0.00757}{2\times 37.9}

Half-life of the reaction = 9.98 × 10⁻⁵s

Learn more about the rate of reaction, here:

brainly.com/question/13571877

#SPJ1

5 0
1 year ago
A welding torch produces a flame by burning acetylene fuel in the presence of oxygen. This flame is used to melt a metal. Which
NISA [10]

Answer:

chemical energy into thermal energy

Explanation:

The reaction taking place is as follows

2C₂H₂ + 5O₂ = 4CO₂ + 2H₂O + Heat

In this reaction bonds present in acetylene is broken and new bond present in water and carbon dioxide are formed . In the whole process of bond breaking and bond formation , there is net loss of energy and that energy is released as heat energy .

Thus we can say that in the whole process , chemical energy is converted into heat energy .

5 0
3 years ago
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