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3 years ago

How much heat is required to raise the temperature of a 6.21 g sample of iron from 25.0 oC to 79.8 oC?

Chemistry

1 answer:

telo118 [61]3 years ago

5 0

Answer:

151.1J

Explanation:

Given parameters:

Mass of iron = 6.21g

Initial temperature of iron = 25°C

Final temperature of iron = 79.8°C

Unknown:

Amount of heat = ?

Solution:

The amount of heat require to cause this temperature can be determined using the expression below;

H = m c (T₂ - T₁)

H is the amount of heat

m is the mass

c is the specific heat capacity

T is the temperature

Specific heat capacity of iron 0.444J/g°C

Insert the parameters and solve;

H = 6.21 x 0.444 x (79.8 - 25)

H = 151.1J

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1. An object was carefully weighed on three different balances. Each balance was zeroed

Hoochie [10]

Answer:

10.335

Explanation:

An object was carefully weighed on three different balances

Each of these balances were zeroed before weighing

The masses that were weighed are as follows

10.35 g , 10.355 g, 10.30 g

Therefore the average value of these measurements can be calculated as follows

The total number of mass is 3

= 10.30 + 10.355 + 10.30/3

= 31,005/3

= 10.335

Hence the average value of these measurements is 10.335

7 0

3 years ago

An aqueous KNO3 solution is made using 72.5 g of KNO3 diluted to a total solution volume of 2.00 L. Calculate the molarity, mola

Sonja [21]

Answer:

The molarity is 0.359 $\frac{moles}{L}$

The molality is 0.354 $\frac{moles}{kg}$

The mass percent of the solution es 3.45%

Explanation:

Molarity is a unit of concentration that indicates the amount of moles of solute that appear dissolved in each liter of the mixture. It is determined by:

$Molarity (M)=\frac{number of moles of solute}{volume}$

Being:

K: 39 g/mole
N: 14 g/mole
O: 16 g/mole

The molar mass of KNO₃ is:

KNO₃= 39 g/mole + 14 g/mole + 3*16 g/mole= 101 g/mole

You can apply the following rule of three: if 101 grams of KNO₃ are present in 1 mole, 72.5 grams in how many moles are present?

$moles of KNO_{3}=\frac{72.5 grams*1 mole}{101 grams}$

moles of KNO₃= 0.718

So you have:

moles of KNO₃= 0.718
volume= 2 L

Applying this quantity in the definition of molarity:

$molarity=\frac{0.718 moles}{2 L}$

Molarity= 0.359 $\frac{moles}{L}$

The molarity is 0.359 $\frac{moles}{L}$

Molality is a way of measuring the concentration of solute in solvent and indicates the amount of moles of solute in each kilogram of solvent.

Then the molality is calculated by:

$Molality=\frac{moles of solute}{mass of solvent in kilograms}$

Density is defined as the property that matter, whether solid, liquid or gas, has to compress in a given space. That is, it is the amount of mass per unit volume. So, if the density of 1.05 g / mL for the solution indicates that in 1 mL of solution there are 1.05 grams of solution, in 2000 mL (where 2L = 2000 mL, because 1 L = 1000mL) how much mass is there?

$mass=\frac{2000 mL*1.05 grams}{1 mL}$

mass= 2100 grams

Since mass solution = mass water + mass KNO₃

then mass water = mass solution - mass KNO₃

Being mass solution 2100 grams and mass KNO₃ 72.5 grams, and replacing you get: mass water= 2100 grams - 72.5 grams

mass water= 2,027.5 grams

Then, being:

moles of KNO₃= 0.718
mass of solvent in kilograms= 2.0275 kg (being 2,027.5 grams= 2.0275 kilograms because 1,000 grams= 1 kilogram)

Replacing in the definition of molality:

$molality=\frac{0.718 moles}{2.0275 kg}$

molality= 0.354 $\frac{moles}{kg}$

The molality is 0.354 $\frac{moles}{kg}$

The mass percent of a solution is the number of grams of solute per 100 grams of solution. Then the mass percent is the mass of the element or solute divided by the mass of the compound or solute and the result of which is multiplied by 100 to give a percentage.

$mass percent=\frac{mass of solute}{mass of solution} *100$

So, in this case:

$mass percent=\frac{72.5 grams}{2100 grams} *100$

mass percent= 3.45 % KNO₃ by mass

The mass percent of the solution es 3.45%

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2 years ago

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Art [367]

You can determine the number of electrons by valence electrons

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3 years ago

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Chlorine gas was bubbled in a solution of sodium bromide. Explain what was observed.

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Answer:

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