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3 years ago

In two or more sentences describe how a constant volume Calorimetry is made, and how it is used to measure energy transfer

Chemistry

1 answer:

Lady_Fox [76]3 years ago

6 0

a constant volume calorimeter= THE BOMB CALORIMETER!

bomb calorimeter: A bomb calorimeter is a type of constant-volume calorimeter used in measuring the heat of combustion of a particular reaction. calorie: The amount of energy needed to raise the temperature of 1 gram of water by 1 °C. It is a non-SI unit of energy equivalent to approximately 4.18 Joules.

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What is the percent composition by mass of oxygen in Ca(NO3)2 (gram-formula= 164 g/mol)?

Neko [114]

To find this, we will use this formula:

Molar mass of element
------------------------------------ x 100
Molar mass of compound

So, first lets calculate the mass of the compound as a whole. We use the atomic masses on the periodic table to determine this.

Ca: 40.078 g/mol
N2 (there is two nitrogens): 28.014 g/mol
O6 (there are six nitrogens: 3 times 2): 95.994 g/mol

When we add all of those numbers up together, we get 164.086. That is the molar mass for the whole compound. However, we are trying to figure out what percent of the compound oxygen makes up. From the molar mass, we know that 95.994 of the 164.086 is oxygen. Lets plug those numbers into our equation!

95.994
-----------
164.086

When we divide those two numbers, we get .585. When we multiply that by 100, we get 58.5.

So, the percent compostition of oxygen in Ca(NO3)2, or, calcium nitrate, is 58.5%.

5 0

3 years ago

Omg pls help i dunno what the frick frack this is

snow_lady [41]

Answer:

1. Mass of KCl produced = 774.8 g of KCl

2. Mass of KNO₃ produced = 13.837g

3. Moles of NaOH made = 0.846 moles

4. Moles of LiCl produced = 0.846 moles

5. Moles of CO₂ produced = 207.6 moles

Explanation:

1. From the equation of reaction, 1 mole of ZnCl₂ produces, 2 moles of KCl.

5.02 moles of ZnCl₂ will produce, 2 × 5.02 moles of KCl = 10.4 moles of KCl

Molar mass of KCl = (39 + 35.5) g/mol = 74.5 g/mol

10.4 moles of KCl = 10.4 × 74.5 g

Mass of KCl produced = 774.8 g of KCl

2. Mole ratio of KNO₃ and KOH = 1:1

O.137 moles of KOH will produce 0.137 moles of KNO₃

Molar mass of KNO₃ = 101 g/mol

Mass of KNO₃ produced = 0.137 × 101 g = 13.837g

3. Molar mas of Ca(OH)₂ = 74.0 g

Moles of Ca(OH)₂ in 31.3 g = 31.3/74.0 = 0.423 moles of Ca(OH)₂

Mole ratio of NaOH and Ca(OH)₂ in the reaction = 2 : 1

Moles of NaOH made = 2 × 0.423 = 0.846 moles

4. Molar mass of MgCl₂ = 95.0 g

Moles of MgCl₂ in 40.2 g = 40.2/95.0 = 0.423 moles

From the reaction equation, mole ratio of MgCl₂ and LiCl = 1:2

Moles of LiCl produced = 2 × 0.423 = 0.846 moles

5. From the equation of reaction, 1 mole of C₆H₁₀O₅ produces 6 moles of cO₂

34.6 moles of C₆H₁₀O₅ will produce 34.6 × 6 moles of CO₂

Moles of CO₂ produced = 207.6 moles

4 0

3 years ago

Balance the nuclear reaction equation for the beta minus decay of nickel-63 by completing the missing numbers.

Marta_Voda [28]

Answer: 63 & 29

Explanation:

7 0

3 years ago

How many molecules are in 7.62 L of CH4, at 87.5°C and 722 torr

pickupchik [31]

Answer: There are $1.469 \times 10^{23}$ molecules present in 7.62 L of $CH_4$ at $87.5^{o}C$ and 722 torr.

Explanation:

Given : Volume = 7.62 L

Temperature = $87.5^{o}C = (87.5 + 273) K = 360.5 K$

Pressure = 722 torr

1 torr = 0.00131579

Converting torr into atm as follows.

$722 torr = 722 torr \times \frac{0.00131579 atm}{1 torr}\\= 0.95 atm$

Therefore, using the ideal gas equation the number of moles are calculated as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

$PV = nRT\\0.95 atm \times 7.62 L = n \times 0.0821 L atm/mol K \times 360.5 K\\n = \frac{0.95 atm \times 7.62 L}{0.0821 L atm/mol K \times 360.5 K}\\= \frac{7.239}{29.59705}\\= 0.244 mol$

According to the mole concept, 1 mole of every substance contains $6.022 \times 10^{23}$ atoms. Hence, number of atoms or molecules present in 0.244 mol are calculated as follows.

$0.244 mol \times 6.022 \times 10^{23}\\= 1.469 \times 10^{23}$

Thus, we can conclude that there are $1.469 \times 10^{23}$ molecules present in 7.62 L of $CH_4$ at $87.5^{o}C$ and 722 torr.

5 0

3 years ago

John roller skates 48 km in 3 hours. What is his average speed in kilometers

Nadya [2.5K]

Answer:16km/hr

Explanation:

48km/3hr= 16km/hr

7 0

3 years ago

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