Answers
Answer:
a)C2HBrClF3 = 197.35 g/mol
b)C12H14N2CL2 = 229.06g/mol
c)C8H10N4O2 = 194.22g/mol
d) CO(NH2)2=60.07 g/mol
e)C17H35CO2Na = 306.52 g/mol
Explanation:
Molar mass of a compound is equal to the sum of the atomic masses of the constituent elements.
a) C2HBrClF3
b) C12H14N2CL2
c) C8H10N4O2
d) CO(NH2)2
e) C17H35CO2Na
Related Questions
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If you want to melt ice into liquid water, does the ice absorbs heat or release heat? Is this endothermic or exothermic?
Use the weather map to answer the question.Weather map with blue line with triangles pointing down and to the right. Red line with semi-circles pointing up and to the right. Red line is to the right of the blue line.© Dorling Kindersley / Universal Images Group / Image Quest 2016What does the red line indicate?A. A cold front is moving to the north and east.B. A cold front is moving to the south and west.C. A warm front is moving to the north and east.D. A warm front is moving to the south and west.
How many moles of h2 are need to produce 11.21 g of nh3
Answers
Answer:
fe+s is a mixture of iron and sulphur and it can be separated by magnetic separation. But FeS is a compound formed by heating iron and sulphur together.it can't be separated by any method.
Explanation:
Now write an equation below that shows how to calculate Kp from Kc for this reaction at an absolute temperature T. You can assume T is comfortably above room temperature. If you include any common physical constants in your equation be sure you use their standard symbols, found in the ALEKS Calculator.
Answers
Answer:
Kp = [CH₄(g)]/[H₂(g)]²(RT)⁻¹
Explanation:
C(s) + 2H₂(g) => CH₄(g)
Kp = Kc(RT)ⁿ
n= change in molar volumes of gas = 1 - 2 = -1
R = 0.08206 L·Atm/mol·K (gas constant)
T = Kelvin Temperature (arbitrary in problem) K =°C + 273
Kc = [CH₄(g)]/[H₂(g)]² (measured values need to be in moles/L)
Kp = [CH₄(g)]/[H₂(g)]²(RT)⁻¹
Answers
The heat that is required to change the temperature of two cups of water (500 g) from room temperature (25◦C) to boiling
C) 157 kJ
Heat
Heat required= Mass of water x specific heat capacity of water x change in temperature of water required
Q=m* c* delta T
M = 500g
C = 4.184 g°C
Delta T = 100 - 25(room temp) = 75°C
Heat = 500 x 4.184 x 75
Heat = 156900 J
Heat = 156.9 KJ
Heat ~ 157.0 KJ (3.D.P)
Thus, the correct answer is C.
Learn more about "Heat":
Answer:
C
Explanation:
Heat required= Mass of water x specific heat capacity of water x change in temperature of water required
M = 500g
C = 4.184 g°C
Delta T = 100 - 25(room temp) = 75°C
Heat needed= 500 x 4.184 x 75
= 156900 J
= 156.9 KJ
~ 157.0 KJ (3.D.P)
Answers
Answer:
124 g (3 sig figs)
or
124.011 g (6 sig figs
Explanation:
Step 1: Calculate g/mol for AgNO₃
Ag - 107.868 g/mol
N - 14.01 g/mol
O - 16.00 g/mol
107.868 + 14.01 + 16.00(3) = 169.878 g/mol
Step 2: Multiply 0.73 moles by molar mass
0.73 mol (169.979 g/mol)
124 grams of AgNO₃
Answers
Final answer:
Using Boyle's Law of gases which states that the pressure and volume of a gas have an inverse relationship when temperature is kept constant, we find that when the pressure of the gas increases from 5.0 to 7.0 atmospheres, the volume of the gas decreases to approximately 3.57 liters.
Explanation:
The question pertains to the application of Boyle's Law, a fundamental concept in the field of physics dealing with gases. Boyle's Law states that the pressure and volume of a gas have an inverse relationship when the temperature is held constant. This means if the pressure of a gas increases, the volume decreases, and vice versa.
In this case, you have 5.0 liters of a gas under an initial pressure of 5.0 atmospheres. The pressure is then increased to 7.0 atmospheres, and you are asked to determine the new volume of the gas. To solve this problem, we use the formula for Boyle's Law, which is P1V1 = P2V2. We know P1 (initial pressure) is 5.0 atmospheres and V1 (initial volume) is 5.0 liters. P2 (final pressure) is increased to 7.0 atmospheres and V2 (final volume) is what we are trying to find.
So, we plug the numbers into the equation and get: 5.0 atmospheres * 5.0 liters = 7.0 atmospheres * V2. Solving for V2, we find V2 to be approximately 3.57 liters. Therefore, when the pressure of the gas is increased from 5.0 atmospheres to 7.0 atmospheres, the volume decreases to around 3.57 liters, while the temperature remains constant.
Learn more about Boyle's Law here:
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Answers
Answer:
333.6 atm
Explanation:
The following data were obtained from the question:
Mole of O2 (nO2) = 3.96 moles
Mole of N2 (nN2) = 7.49 moles
Mole of CO2 (nCO2) = 1.19 moles
Total pressure = 563 mmHg
Partial pressure of N2 =..?
Next, we shall determine the total number of mole in the container.
This can be obtained as follow:
Mole of O2 (nO2) = 3.96 moles
Mole of N2 (nN2) = 7.49 moles
Mole of CO2 (nCO2) = 1.19 moles
Total mole =?
Total mole = nO2 + nN2 + nCO2
Total mole = 3.96 + 7.49 + 1.19
Total mole = 12.64 moles
Next, we shall determine the mole fraction of N2.
This can be obtained as follow:
Mole fraction = mole of substance/total mole
Mole of N2 (nN2) = 7.49 moles
Total mole = 12.64 moles
Mole fraction of N2 =?
Mole fraction of N2 = 7.49/12.64
Finally, we shall determine the partial pressure of N2.
This can be obtained as follow:
Mole fraction of N2 = 7.49/12.64
Total pressure = 563 mmHg
Partial pressure of N2 =..?
Partial pressure = mole fraction x total pressure
Partial pressure of N2 = 7.49/12.64 x 563
Partial pressure of N2 = 333.6 atm
The, the partial pressure of nitrogen, N2 is 333.6 atm