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Cloud [144]
4 years ago
7

Convert the following to Celsius

Chemistry
1 answer:
Naya [18.7K]4 years ago
8 0

9. (80-32)5/9

48×5/9

240/9

26.66°C

9. (90-32)5/9

58×5/9

32.22°C

10 (212-32)5/9

180×5/9

20×5

100°C

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a calorimeter contained 75.0 g of water at 16.95 C. A 93.3-g sample of iron at 65.58 C was placed in it, giving a final temperat
Nostrana [21]

Answer:- \frac{382.69J}{0C} .

Solution:- Mass of Iron added to water is 93.3 g. Initial temperature of iron metal is 65.58 degree C and final temperature of the system is 19.68 degree C.

temperature change, \Delta T for iron metal = 65.58 - 19.68 = 45.9 degree C

specific heat for the metal is given as 0.444 J per g per degree C.

let's calculate the heat lost by iron metal using the equation:

q=mc\Delta T

where, q is the heat energy, m is mass, c is specific heat and delta T is change in temperature. let's plug in the values and calculate q for iron metal:

q=93.3g(45.9^0C)(\frac{0.444J}{g.^0C})

q = 1901.42 J

Using same equation we will calculate the heat gained by water.

mass of water is 75.0 g.

\Delta T for water = 19.68 - 16.95 = 2.73 degree C

specific heat for water is 4.184 J pr g per degree C. Let's plug in the values:

q=75.0g(\frac{4.184J}{g.^0C})(2.73^0C)

q = 856.674 J

Total heat lost by iron metal is the sum of heat gained by water and calorimeter.

So, heat gained by calorimeter = heat lost by iron metal - geat gained by water

heat gained by calorimeter = 1901.42 J - 856.674 J = 1044.746 J

Change in temperature for calrimeter is same as for water that is 2.73 degree C

For calorimeter, q=C.\Delta T

C=\frac{q}{\Delta T}

C=\frac{1044.746J}{2.73^0C}

C=\frac{382.69J}{0C}

So, the heat capacity of calorimeter is \frac{382.69J}{0C} .


4 0
3 years ago
Water is 11% hydrogen and 89% oxygen by mass. How many grams of oxygen are in a 250 g glass of water?
mylen [45]

Answer:

mass O2 = 222.5 g

Explanation:

  • %wt = ((mass compound)/(mass sln))×100

balance reaction:

  • 2H2 + O2 ↔ 2H2O

∴ %wt H2 = 11 % = ((mass H2)/(mass H2O))×100

∴ %wt O2 = 89 % = ((mass O2)/(mass H2O))×100

∴ mass H2O 250 g

⇒ mass O2 = (0.89)(250 g)

⇒ mass O2 = 222.5 g

3 0
4 years ago
If 0.756L of gas exerts a pressure of 94.6kPa, what would the volume be at standard pressure (101.33kPa)?
yaroslaw [1]

Answer:

The answer to your question is Volume = 0.7 l

Explanation:

Data

Volume 1 = V1 = 0.746 l

Pressure 1 = P1 = 94.6 kPa

Volume 2 = V2 = ?

Pressure 2 = 101.33 kPa

Process

To solve this process use Boyle's law.

           P1V1 = P2V2

-Solve for V2

           V2 = P1V1/P2

-Substitution

          V2 = (94.6 x 0.746) / 101.33

-Simplification

           V2 = 70.57 / 101.33

-Result

            V2 = 0.696 l ≈ 0.7 l

4 0
4 years ago
Which equation shows how to calculate how many grams (g) of KCI would be
Semenov [28]

Answer:

C or D

Explanation:

Hope it helped a little bit

3 0
3 years ago
I want someone to kelp me with the biochemistry exam today please can any one help me sent me a message"(​
Dvinal [7]

Answer:

where your question i will help you if im know the answer

6 0
3 years ago
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