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3 years ago

4.45 kcal of heat was added to increase the temperature of a sample of water from 23.0 °C to 57.8 °C. Calculate

Chemistry

1 answer:

Alona [7]3 years ago

4 0

Answer:

m = 4450 g

Explanation:

Given data:

Amount of heat added = 4.45 Kcal ( 4.45 kcal ×1000 cal/ 1kcal = 4450 cal)

Initial temperature = 23.0°C

Final temperature = 57.8°C

Specific heat capacity of water = 1 cal/g.°C

Mass of water in gram = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 57.8°C - 23.0°C

ΔT = 34.8°C

4450 cal = m × 1 cal/g.°C × 34.8°C

m = 4450 cal / 1 cal/g

m = 4450 g

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In the lab you weigh out 76.02 g of Iron (Fe). How many moles of Iron do you have in the sample. (Your answer must have a unit..

son4ous [18]

Answer:

We have 1.361 moles in the sample

Explanation:

Mass of iron = 76.02g

Molar mass of iron = 55.845 g/ mole ( This we can find in the periodic table, and menas that 1 mole of iron has a mass of 55.845 g).

To calculate the number of moles we will use following formula:

moles (n) = mass / molar mass

moles iron = 76.02g / 55.845 g/ mole

moles iron = 1.36127 moles

To use the correct number of significant digits we use the following rule for multiplication and division :

⇒ the number with the least number of significant figures decides the number of significant digits.

⇒76.02 has 4 digits ( 2 after the comma) and 55.845 has 5 digits (3 after the comma).

⇒ this means 1.361 moles

We have 1.361 moles in the sample

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3 years ago

How many molecules are there in 10.5 grams of iron (111) sulfate trihydrate? With work shown

MrRa [10]

Fe2(SO4)3.3H2O = 56*2 + 3*(32+16*4)+3*(1*2+16) = 454 g/mol
Mol =m/M = 10.5/454 = 0.023 mol
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3 years ago

Two elements are studied. One has atomic number X and one has atomic number X+2. It is known that element X is a halogen. To wha

vladimir2022 [97]

Answer:

Group 1 (or IA)

Explanation:

If element X is a halogen, then it belongs to the group 17 (or VIIA, under a different notation).

For each extra unit of atomic number, the group number increases by 1. That means that the X+1 element would belong to the group 18 (or VIIIA). <em>The X+2 element would thus belong in the group 1 </em>(or IA) one period higher (higher as in numeric value, not as in position in the periodic table).

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2 years ago

A 2.20 mol sample of NO 2 ( g ) is added to a 3.50 L vessel and heated to 500 K. N 2 O 4 ( g ) − ⇀ ↽ − 2 NO 2 ( g ) K c = 0.513

igor_vitrenko [27]

Answer:

[NO₂] = 0.434 M

[N₂O₄] = 0.0971 M

Explanation:

The equilibrum is: N₂O₄(g) ⇆ 2NO₂ (g)

1 moles of nitrogen (IV) oxide is in equilibrium with 2 moles of nitrogen dioxide.

Initally we only have 2.20 moles of NO₂. So let's write the equilibrium again:

2NO₂ (g) ⇆ N₂O₄(g)

Initially 2.20 mol -

React x x/2

X amount has reacted, and the half has been formed, according to stoichiometry.

Eq (2.20-x) / 3.50L (x/2)/ 3.50L

We divide by the volume because we need molar concentrations. Let's make the Kc's expression:

Kc = [N₂O₄] / [NO₂]²

0.513 = ((x/2)/ 3.50L) / [(2.20-x) / 3.50L]

0.513 = ((x/2)/ 3.50L) / [(2.20-x)² / 3.50L²]

0.513 = ((x/2)/ 3.50L) / [2.20-x)² / 3.50L²]

0.513 = ((x/2)/ 3.50L) / (4.84 - 4.40x + x²) / 12.25)

0.513 / 12.25 (4.84 - 4.40x + x²) = x/2 / 3.50

0.203 - 0.184x + 0.0419x² = x/2 / 3.50

3.50(0.203 - 0.184x + 0.0419x²) = x/2

7 (0.203 - 0.184x + 0.0419x²) - x = 0

1.421 - 2.288x + 0.2933x² = 0 → Quadratic formula

a = 0.2933 ; b = -2.288 ; c = 1.421

(-b +- √(b²-4ac)) / (2a)

x₁ = 7.12

x₂ = 0.68 → We consider this value, so we can have a (+) concentration.

Concentrations in the equilibrium are:

[NO₂] = (2.20-0.68) / 3.50 = 0.434 M

[N₂O₄] = (0.68/2) / 3.50 = 0.0971 M

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What are atoms made of

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Protons, neutrons, and electrons.

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