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Harlamova29_29 [7]
3 years ago
10

Calculate the heat absorbed by a sample of water that has a mass of 45.00 g when the temperature increases from 21.0oC to 38.5 o

C. (s=4.184 J/g.o C)
Chemistry
1 answer:
Elenna [48]3 years ago
8 0

Answer:

The heat absorbed by the sample of water is 3,294.9 J

Explanation:

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

The sensible heat of a body is the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous). Its mathematical expression is:

Q = c * m * ΔT

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

In this case:

  • Q=?
  • m= 45 g
  • c= 4.184 \frac{J}{g*C}
  • ΔT= Tfinal - Tinitial= 38.5 C - 21 C= 17.5 C

Replacing:

Q= 4.184 \frac{J}{g*C} * 45 g* 17.5 C

Solving:

Q=3,294.9 J

<u><em>The heat absorbed by the sample of water is 3,294.9 J</em></u>

<u><em></em></u>

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A 2.350×10−2 M solution of NaCl in water is at 20.0∘C. The sample was created by dissolving a sample of NaCl in water and then b
sveta [45]

Answer:

  • Part A: m = 0.02356 mol/kg = 0.02356 m
  • Part B: Xsolute = 4.243×10⁻⁴
  • Part C: % m/m = 0.1376%
  • Part D: ppm = 1,376 ppm

Explanation:

<u>1. Data:</u>

a) M = 2.350×10⁻² M

b) V sol = 1.000 L

c) V H₂O = 994.4 mL = 0.9944 L

d) d H₂O = 0.9982 g/mL

<u>2. Formulae</u>

  • M = n solute / V sol (L)
  • m = n solute / Kg solvent
  • X solute = n solute / N total
  • % m/m = (mass of solute / mass of solution) × 100
  • ppm = (mass of solute / mass of solution) × 1,000,000
  • density = mass in grams / volume in mL

<u>3. Solution</u>

<u>Part A: Calculate the molality of the salt solution. </u>

<u />

            m = n solute / Kg solvent

i) M = n solute / V sol (L) ⇒ n solute = M × V sol (L)

⇒ n solute = M = 2.350×10⁻² M × 1.000 L = M = 2.350×10⁻² mol

ii) density H₂O = mass H₂O / volume H₂O

⇒ mass H₂O = density H₂O × volume H₂O

⇒ mass H₂O = 0.9982 g/mL × 999.4 mL = 997.6 g

iii) kg  H₂O = 997.6 g / (1,000 g/Kg) = 0.9976 kg

iv) m = 2.350×10⁻² mol / 0.9976 kg = 0.02356 mol/kg = 0.02356 m

<u>Part B: Calculate the mole fraction of salt in this solution</u>.

          X solute = n solute / N total

i) n solute =  2.350×10⁻² mol

ii) n solvent = n H₂O = mass H₂O in grams/ molar mass H₂O

⇒ 997.6 g / 18.015 g/mol = 55.38 mol

iii) X solute = 2.350×10⁻² mol / 55.38 mol = 4.243×10⁻⁴

<u>Part C: Calculate the concentration of the salt solution in percent by mass</u>.

         % m/m = (mass of solute / mass of solution) × 100

i) molar mass = mass in grams / molar mass

⇒ mass of solute = mass of NaCl = n solute × molar mass NaCl

⇒ mass of solute = 2.350×10⁻² mol × 58.44 g/mol = 1.373 g

ii) % m/m = (1.373 g / 997.6 g) × 100 = 0.1376%

Part D: Calculate the concentration of the salt solution in parts per million.

       ppm = (mass of solute / mass of solution) × 1,000,000

i) ppm = ( (1.373 g / 997.6 g) × 1,000,000 = 1,376 ppm

5 0
3 years ago
What structures are contained in the nucleus?
zzz [600]

Answer:

nuclear envelope, nuclear la Nina, nucleolus, chromosomes, and nucleoplasm. Hope this helped?

4 0
3 years ago
Jade has a density of 3.3 g/cm3. If a piece of jade is placed into a graduated cylinder as shown below, the volume increases fro
just olya [345]

The volume of a sample is its unique property, which can be defined as the mass of the material per unit volume of that material. Now as per Archimedes law if a material is submerge in a liquid the increased volume of the liquid will be equivalent to the volume of the material.

Here due to addition of the Jade the volume of the liquid increases from 50 to 60.5 mL. Thus the change of the volume is (60.5-50) = 10.5 mL, which will be equivalent to the volume of the Jade. Thus the volume of the Jade is 10.5 mL and its density 3.3 g/cm³. Now as 1 mL = 1 cm³ thus the density can be rewritten as 3.3 g/mL. So 1 mL of the Jade has the mass 3.3 g, thus 10.5 mL of the Jade will have (10.5×3.3) = 34.65 g of mass.

The mass of the piece of Jade is 34.65 g.    

8 0
3 years ago
How many HCl molecules and moles are there in 3.65 grams of
ipn [44]

Answer:

0.6022×10²³ molecules

Number of moles = 0.1 mol

Explanation:

Given data:

Mass of HCl = 3.65 g

Number of molecules = ?

Number of moles = ?

Solution:

Number of moles:

Number of moles = mass/molar mass

Number of moles = 3.65 g/36.45 g/mol

Number of moles = 0.1 mol

Number of molecules:

1 mole of any substance contain 6.022×10²³ molecules

0.1 mol ×  6.022×10²³ molecules/ 1 mol

0.6022×10²³ molecules

4 0
2 years ago
The reaction described in part a required 3.77 l of magnesium chloride. what is the concentration of this magnesium chloride sol
katrin [286]
Assuming you are talking about the atomic mass of magnesium chloride
M = n/V
M = 1/3.77
M = .265


6 0
3 years ago
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