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Diano4ka-milaya [45]
3 years ago
7

Beaker A contains 100.0 g of water initially at 10EC. Beaker B contains 100.0 g of water initially at 20°C. The contents of the

two beakers are mixed. What is the final temperature of the contents? Assume that there are no heat losses and that the mass of the beakers is negligible. Useful data: The specific heat of water is 4.18 J/(g°C) Question 2 options: A. 15.0°C B. 17.5°C C.12.5°C D.10.0°C
Chemistry
2 answers:
Korvikt [17]3 years ago
7 0

Answer:

(A) 15.0 °C

Explanation:

The water in beaker A gains heat because its initial temperature (10 °C) is less than the initial temperature of the water in beaker B (20 °C) which loses heat.

Let T3 be the final temperature

Heat gained by beaker A = heat loss by beaker B

mc(T3 - T1) = mc(T2 - T3)

The mass and specific heat of water in both beakers are the same. Therefore, (T3 - T1) = (T2 - T3)

T1 is initial temperature of beaker A = 10 °C

T2 is initial temperature of beaker B = 20 °C

T3 - 10 = 20 - T3

T3 + T3 = 20 + 10

2T3 = 30

T3 = 30/2 = 15 °C

Masteriza [31]3 years ago
5 0

Answer:

Explanation:

Given:

Cp of water = 4.18 J/g.°C

Mass of beaker 1 and 2, M1 = M2

= 100 g

Initial temperature of beaker 1, Ti1 = 10°C

Initial temperature of beaker 2, Ti2 = 20°C

Delta H1 = -delta H2

The heat lost by one besjef is gained by the other.

Delta H = M × Cp × delta T

Delta T = tempf - Ti

100 × 4.18 × (10 - x) = -100 × 4.18 × (20 - x)

4180 - 418x = -8360 + 418x

836x = 12540

x = 15°C

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PLEASE HELP!!! PLEASE.
adell [148]

Answer:

Q₁: [HCl] = 0.075 N = 0.075 M.

Q₂: [KOH] = 7.675 mN = 7.675 mM.

Q₃: [H₂SO₄] = 0.2115 N = 0.105 M.

Q₄:  The equivalence point is the point at which the added titrant is chemically equivalent completely to the analyte in the sample whereas the endpoint is the point where the indicator changes its color.

Explanation:

<u><em>Q₁: If it takes 67 mL of 0.15 M NaOH to neutralize 134 mL of an HCl solution, what is the concentration of the HCl? </em></u>

  • As acid neutralizes the base, the no. of gram equivalent of the acid is equal to that of the base.
  • The normality of the NaOH and HCl = Their molarity.

<em>∵ (NV)NaOH = (NV)HCl</em>

<em>∴ N of HCl = (NV)NaOH / (V)HC</em>l = (0.15 N)(67 mL) / (134 mL) = 0.075 N.

∴ The concentration of HCl = 0.075 N = 0.075 M.

<u><em>Q₂: If it takes 27.4 mL of 0.050 M H₂SO₄ to neutralize 357 mL of KOH solution, what is the concentration of the KOH solution?</em></u>

  • As mentioned in Q1, the no. of gram equivalent of the acid is equal to that of the base at neutralization.
  • <em>The normality of H₂SO₄ = Molarity of H₂SO₄ x 2 = 0.050 M x 2 = 0.1 N.</em>

<em>∵  (NV)H₂SO₄ = (NV)KOH</em>

∴ N of KOH = (NV)H₂SO₄ / (V)KOH = (0.1 N)(27.4 mL) / (357 mL) = 7.675 x 10⁻³ N = 7.675 mN.

<em>∴ The concentration of KOH = 7.675 mN = 7.675 mM.</em>

<em></em>

<u><em>Q₃:If it takes 55 mL of 0.5 M NaOH solution to completely neutralize 130 mL of sulfuric acid solution (H₂SO₄), what is the concentration of the H₂SO₄ solution?</em></u>

  • As mentioned in Q1 and 2, the no. of gram equivalent of the acid is equal to that of the base at neutralization.

<em>The normality of NaOH = Molarity of NaOH = 0.5 N.</em>

<em>∵ (NV)H₂SO₄ = (NV)NaOH</em>

<em>∴ N of H₂SO₄ = (NV)NaOH / (V)H₂SO₄</em> = (0.5 N)(55 mL) / (130 mL) = 0.2115 N.

<em>∴ The concentration of H₂SO₄ = 0.2115 N = 0.105 M.</em>

<em></em>

<u><em>Q₄: Explain the difference between an endpoint and equivalence point in a titration.</em></u>

  • The equivalence point is the point at which the added titrant is chemically equivalent completely to the analyte in the sample whereas the endpoint is the point where the indicator changes its color.
  • The equivalence point in a titration is the point at which the added titrant is chemically equivalent completely to the analyte in the sample. It comes before the end point. At the equivalence point, the millimoles of acid are chemically equivalent to the millimoles of base.
  • End point is the point where the indicator changes its color. It is the point of completion of the reaction between two solutions.
  • The effectiveness of the titration is measure by the close matching between equivalent point and the end point. pH of the indicator should match the pH at the equivalence to get the same equivalent point as the end point.
6 0
3 years ago
A container of gas is initially at 0.25 atm and 0 ˚C. What will the pressure be at 125 ˚C?
Pani-rosa [81]

Answer:

0.37atm

Explanation:

Given parameters:

Initial pressure  = 0.25atm

Initial temperature  = 0°C  = 273K

Final temperature  = 125°C  = 125 + 273  = 398K

Unknown:

Final pressure  = ?

Solution:

To solve this problem, we use a derivative of the combined gas law;

           \frac{P1}{T1}  = \frac{P2}{T2}

  P and T are pressure and temperature

  1 and 2 are initial and final values

        \frac{0.25}{273}   = \frac{P2}{398}  

         P2  = 0.37atm

3 0
2 years ago
2. How many grams of water can be heated from 20.0 oC to 75oC using 12500.0 Joules?
kenny6666 [7]

Answer;

  = 0.054 kg or 54 g

Explanation;

Using the equation; Q = mcΔT where Q is the quantity of heat transferred, m is the mass, c is specific heat of the substance, ΔT is delta T, the change in temperature.  

ΔT = 75 - 20 = 55 C.  

Solve the equation for m  

m = Q/ cΔT

Mass = 12500 / (55 × 4200)

        = 0.054 kg or 54 g


4 0
3 years ago
Read 2 more answers
Pb(SO4)2 + 4 LiNO3 → Pb(NO3)4 + 2 Li2SO4
Anvisha [2.4K]

Answer:

4.5 moles of lithium sulfate are produced.

Explanation:

Given data:

Number of moles of lead sulfate = 2.25 mol

Number of moles of lithium nitrate = 9.62 mol

Number of moles of lithium sulfate = ?

Solution:

Chemical equation:

Pb(SO₄)₂ + 4LiNO₃      →     Pb(NO₃)₄ + 2Li₂SO₄

Now we will compare the moles of lithium sulfate with lead sulfate and lithium nitrate.

                       Pb(SO₄)₂        :         Li₂SO₄

                            1                :             2

                          2.25           :          2/1×2.25 = 4.5 mol

                       LiNO₃            :             Li₂SO₄

                           4                :                2

                           9.62           :             2/4×9.62 = 4.81 mol

Pb(SO₄)₂  produces less number of moles of Li₂SO₄ thus it will act as limiting reactant and limit the yield of  Li₂SO₄.      

7 0
3 years ago
How many atoms of carbon are combined to with 2.81 moles of hydrogen atoms in a sample of the compound ethane, C2H6?
puteri [66]

Answer:

5.64×10²³ atoms C

Explanation:

Convert moles of H to moles of C:

2.81 mol H × (2 mol C / 6 mol H) = 0.937 mol C

Convert moles of C to atoms of C:

0.937 mol C × (6.02×10²³ atoms C / mol C) = 5.64×10²³ atoms C

5 0
4 years ago
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