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Cerrena [4.2K]
2 years ago
9

use the gizmos to mix 200 G of copper at 100 C with 1000 g of water at 20 C whatis the final temperature

Chemistry
1 answer:
ziro4ka [17]2 years ago
5 0

Answer:

Explanation:

mass of copper = 200 g at temperature 100°C

mass of water = 1000 g at 20⁰C

Let the final temperature be T .

specific heat of copper = .385 J per gram

specific heat of water = 4.182 J per gram

heat lost = mass x specific heat x fall of temperature

heat lost by copper

= 200 x .385 x ( 100 - T )

heat gained by water

= 1000 x 4.182 x ( T - 20 )

heat lost by copper = heat gained by water

200 x .385 x ( 100 - T )  = 1000 x 4.182 x ( T - 20 )

7700 - 77 T = 4182 T - 83640

4259 T  = 91340

T = 21.44⁰C .

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Calculate the freezing point of a solution containing 5. 0 grams of kcl and 550. 0 grams of water. the molal-freezing-point-depr
lutik1710 [3]

The freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is  - 0.45°C

Using the equation,

ΔT_{f} = iK_{f}m

where:

ΔT_{f} = change in freezing point (unknown)

i = Van't Hoff factor

K_{f} = freezing point depression constant

m = molal concentration of the solution

Molality is expressed as the number of moles of the solute per kilogram of the solvent.

Molal concentration is as follows;

MM KCl = 74.55 g/mol

molal concentration = \frac{5.0g*\frac{1mol}{74.55g} }{550.0g*\frac{1kg}{1000g} }

molal concentration = 0.1219m

Now, putting in the values to the equtaion ΔT_{f} = iK_{f}m we get,

ΔT_{f} = 2 × 1.86 × 0.1219

ΔT_{f} = 0.4536°C

So, ΔT_{f} of solution is,

ΔT_{f_{solution} } = 0.00°C - 0.45°C

ΔT_{f_{solution} } =  - 0.45°C

Therefore,freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is  - 0.45°C

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7 0
1 year ago
What are the three different types of selective breeding
balandron [24]

first of all there is only two types of selective breeding and they are hybridization and inbreeding.

3 0
2 years ago
A city water district wants to encourage local businesses and homeowners to landscape with drought-tolerant plants. After disapp
tino4ka555 [31]

Answer:

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Explanation:

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3 0
2 years ago
Calculate the work (kJ) done during a reaction in which the internal volume expands from 20 L to 43 L against an outside pressur
Alenkinab [10]

Answer:

\large \boxed{\text{-10.0 kJ}}

Explanation:

1. Calculate the work

w = - pΔV = -4.3 atm × (43 L - 20 L) = -4.3 × 23 L·atm = -98.9 L·atm

2. Convert litre-atmospheres to joules

w = \text{-98.9 L\cdot$atm } \times \dfrac{\text{101.3 J}}{\text{1 L$\cdot$atm }} = \text{-10000 J} = \textbf{-10.0 kJ}\\\\\text{The work done is $\large \boxed{\textbf{-10.0 kJ}}$}

The negative sign indicates that the work was done against the surroundings.

4 0
3 years ago
In the "Méthode Champenoise," grape juice is fermented in a wine bottle to produce sparkling wine. The reaction is the following
prohojiy [21]

Answer:

The pressure inside the wine bottle at 21 °C is 4.8 · 10² atm

Explanation:

Hi there!

We know that 1 mol of CO₂ is produced per mol of produced ethanol.

If the final concentration of ethanol is 13%, let´s calculate how many moles of ethanol are present at that concentration.

A concentration of 13% means that in 100 ml of solution, 13 ml is dissolved ethanol. We have 754 ml of solution, then, the volume of ethanol will be:

754 ml solution · (13 ml ethanol/100 ml solution) = 98 ml ethanol

With the density, we can calculate the mass of ethanol present:

density = mass/ volume

0.79 g/ml = mass / 98 ml

mass = 0.79 g/ml · 98 ml

mass = 77 g

The molar mass of ethanol is 46.07 g/mol, then 77 g of ethanol is equal to:

77 g · (1 mol/46.07 g) = 1.7 mol

Then, the number of moles of CO₂ produced will be 1.7 mol.

Using the equation of the ideal gas law, we can calculate the pressure of CO₂:

P = nRT/V

Where:

P = pressure

n = number of moles

R = ideal gas constant

T = temperature

V = volume

The volume will be the headspace of the bottle (840 ml - 754 ml) 86 ml = 0.086 l.

The temperature in kelvin will be: 21 + 273 = 294 K

The gas constant is 0.082 l atm / K mol

Then:

P = (1.7 mol · 0.082 l atm/K mol · 294 K)/ 0.086 l

P = 4.8 · 10² atm

The pressure inside the wine bottle at 21 °C is 4.8 · 10² atm

5 0
2 years ago
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