An unknown substance used in the Agricultural sector has the formula CN3H5 .Calculate the percent by mass of nitrogen in the unknown substance to the nearest three significant digits? (N-14.007 g/mol; C-12.011 g/mol; H-1.008g/mol).
To calculate the percent by mass of nitrogen in the unknown substance, we need to first find the molar mass of CN3H5:
M(CN3H5) = (1 x 12.011 g/mol) + (3 x 14.007 g/mol) + (5 x 1.008 g/mol) = 70.078 g/mol
Next, we need to find the mass of nitrogen in CN3H5:
Mass of N = 3 x 14.007 g/mol = 42.021 g/mol
Finally, we can calculate the percent by mass of nitrogen:
%N = (mass of N/total mass of CN3H5) x 100
%N = (42.021 g/mol/70.078 g/mol) x 100
%N = 59.99%
Therefore, the percent by mass of nitrogen in the unknown substance is approximately 60%, to the nearest three significant digits.
Calculate the root mean square velocity of the oxygen gas (O2) molecules at 77 degrees Celsius. (R = 8.314 Kg.m²/S².mol.K).
To calculate the root mean square velocity of oxygen gas (O2) molecules at 77 degrees Celsius, we can use the following formula:
urms = √(3RT/M)
where urms is the root mean square velocity, R is the gas constant, T is the temperature in Kelvin, and M is the molar mass of the gas.
First, we need to convert the temperature from Celsius to Kelvin:
T = (77 + 273) K
T = 350 K
The molar mass of oxygen (O2) is 32 g/mol, or 0.032 kg/mol.
Now we can plug in the values and solve for urms:
urms = √(3 x 8.314 Kg.m²/S².mol.K x 350 K / 0.032 kg/mol)
urms = √(964.7 m²/S²)
urms = 31.04 m/s (rounded to two decimal places)
Therefore, the root mean square velocity of oxygen gas (O2) molecules at 77 degrees Celsius is approximately 31.04 m/s.
How much of heat is needed to warm 500 g of water from 25 degrees Celsius near to it's boiling point of 99 degrees Celsius? The specific heat of capacity is 4.18j/g-k.
To calculate the heat needed to warm 500 g of water from 25 degrees Celsius to its boiling point of 99 degrees Celsius, we can use the following formula:
Q = mcΔT
where Q is the heat needed (in joules), m is the mass of the water (in grams), c is the specific heat capacity of water (in joules per gram per degree Celsius), and ΔT is the change in temperature (in degrees Celsius).
First, we need to calculate ΔT:
ΔT = 99°C - 25°C
ΔT = 74°C
Next, we can plug in the values and solve for Q:
Q = (500 g) x (4.18 J/g°C) x (74°C)
Q = 155,660 J
Therefore, it would take 155,660 joules of heat to warm 500 g of water from 25 degrees Celsius to its boiling point of 99 degrees Celsius.
The temperature of a chemistry laboratory during summer was recorded to be 29.32. A student used a thermometer to measure the temperature four times and recorded the temperatures shown: 28.62, 27.09, 29.01, 29.00, calculate the percentage error
To calculate the percentage error, we need to first find the absolute error and then divide it by the true value and multiply by 100 to convert it to a percentage.
The true value of the temperature is 29.32, and the average of the four measurements is:
(28.62 + 27.09 + 29.01 + 29.00) / 4 = 28.93
The absolute error is the absolute value of the difference between the true value and the measured value:
|29.32 - 28.93| = 0.39
So the absolute error is 0.39 degrees Celsius.
To calculate the percentage error, we divide the absolute error by the true value and multiply by 100:
percentage error = (absolute error / true value) x 100
percentage error = (0.39 / 29.32) x 100
percentage error = 1.33%
Therefore, the percentage error in this measurement is approximately 1.33%.
The internal energy delta E of a system is -230 kj. If the system released 300 kj, how much work was done in the system?
According to the first law of thermodynamics:
delta E = Q - W
where delta E is the change in internal energy of the system, Q is the heat transferred to the system, and W is the work done by the system.
In this case, we are given delta E as -230 kJ and Q as -300 kJ (since the system released 300 kJ of heat). We can plug these values into the formula and solve for W:
-230 kJ = -300 kJ - W
W = -300 kJ - (-230 kJ)
W = -300 kJ + 230 kJ
W = -70 kJ
Therefore, the work done in the system is 70 kJ (note that the negative sign indicates that work was done by the system, which means that the system lost energy to the surroundings).