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givi [52]
3 years ago
12

How much heat must be used to raise the

Chemistry
1 answer:
Artist 52 [7]3 years ago
8 0

Answer:

Q =  57934.8 J

Explanation:

Given data:

Mass of water = 180.0 g

Initial temperature = 19°C

Fina temperature = 96°C

Specific heat capacity of water = 4.18 J/g.°C

Heat absorbed = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 96°C - 19°C

ΔT = 77°C

Q = 180.0 g×4.18 J/g.°C×77°C

Q =  57934.8 J

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Read 2 more answers
Determine the concentrations of MgCl2, Mg2+, and Cl− in a solution prepared by dissolving 2.52 × 10−4 g MgCl2 in 2.00 L of water
dybincka [34]

Answer:

Explanation:

Given parameters:

Mass of MgCl₂ = 2.52 x 10⁻⁴g

Volume of water = 2L

Unkown:

Concentration of MgCl₂ =?

Concentration of Mg²⁺ = ?

Concentration of Cl⁻ =?

Solution:

  Concentration is defined as the number of moles of a solute contained in a solution.

   Concentration = \frac{number of moles }{volume}

 To find the number of moles"

           number of moles = \frac{mass}{molar mass}

   Molar mass of MgCl₂ = 24.3 + (2 x 35.5) = 95.3g/mol

       number of moles = \frac{0.000}{molar mass} = 0.000252moles

 Concentration of MgCl₂ = \frac{0.000252}{95.3} = 2.64 x 10⁻⁶moldm⁻³

 

from the formula of the compound;

    1 mole of MgCl₂ contains 1 mole of Mg²⁺

Therefore, 2.64 x 10⁻⁶moldm⁻³ of MgCl₂ will contains 2.64 x 10⁻⁶moldm⁻³ of Mg

Also;

   1 mole of MgCl₂ contains 2 mole of Cl⁻

  2.64 x 10⁻⁶moldm⁻³ contains 2 x 2.64 x 10⁻⁶moldm⁻³; 5.29 x 10⁻⁶moldm⁻³

 

Expressing in ppm;

               1ppm = 1mg/L

   2.64 x 10⁻⁶moldm⁻³ to mg/L for Mg²⁺

2.64 x 10⁻⁶moldm⁻³  = 2.64 x 10⁻⁶moldm⁻³  x molar mass(g/mol)

                                     = 2.64 x 10⁻⁶moldm⁻³ x 24.3 = 6.43 x 10⁻⁵g/L

    g/L to mg/L; 6.43 x 10⁻⁵g/L x 1000 =   6.43 x 10⁻²mg/L = 6.43 x 10⁻²ppm

5.29 x 10⁻⁶moldm⁻³ to mg/L for Cl⁻;

   5.29 x 10⁻⁶moldm⁻³ = 5.29 x 10⁻⁶moldm⁻³ x 35.5 = 1.88 x 10⁻⁴g/L

   g/L to mg/L; 1.88 x 10⁻⁴g/L x 1000 = 1.88 x 10⁻¹mg/L = 1.88 x 10⁻¹ppm

7 0
3 years ago
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