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).
An unknown compound has percent composition of 79.95% C, 17.72% N, and 6.33% H by mass with a molar mass of about 240 g/mol. Determine the molecular formula of the dye. (Atomic mass of C-12.011 g/mol; N=14.007g/mol; H-1.008g/mol).
To determine the molecular formula of the dye, we first need to find the empirical formula by converting the percent composition to a formula.
To do this, we assume we have a 100 g sample of the compound, which means we have:
- 79.95 g C
- 17.72 g N
- 6.33 g H
We can convert these masses to moles by dividing by the molar masses of the elements:
- 79.95 g C / 12.011 g/mol = 6.656 mol C
- 17.72 g N / 14.007 g/mol = 1.265 mol N
- 6.33 g H / 1.008 g/mol = 6.286 mol H
Next, we need to find the simplest whole-number mole ratio of the elements by dividing by the smallest number of moles (in this case, 1.265 mol N):
- C: 6.656 mol / 1.265 mol = 5.27 ≈ 5
- N: 1.265 mol / 1.265 mol = 1
- H: 6.286 mol / 1.265 mol = 4.97 ≈ 5
Therefore, the empirical formula is C5H5N.
To find the molecular formula, we need to divide the molar mass of the compound (240 g/mol) by the empirical formula mass, which we can calculate as follows:
- C5H5N: 5(12.011 g/mol) + 5(1.008 g/mol) + 1(14.007 g/mol) = 79.098 g/mol
Dividing 240 g/mol by 79.098 g/mol gives us approximately 3, so the molecular formula is:
( C5H5N ) × 3 = C15H15N3
Therefore, the molecular formula of the dye is C15H15N3.
Use -394, -286, -890 to calculate the standard enthalpy change of the reaction C(s) +2H2(g) = CH4(g)
To calculate the standard enthalpy change of the reaction C(s) + 2H2(g) → CH4(g), we need to use Hess's law, which states that if a reaction can be expressed as the sum of two or more reactions, then the enthalpy change of the overall reaction is the sum of the enthalpy changes of the individual reactions.
In this case, we can use the following reactions:
1. C(s) + O2(g) → CO2(g) ΔH° = -394 kJ/mol
2. H2(g) + 1/2 O2(g) → H2O(l) ΔH° = -286 kJ/mol
3. CO2(g) + 2H2(g) → CH4(g) + H2O(l) ΔH° = ?
We need to reverse reaction 1 and multiply reaction 2 by 2 so that we can cancel out CO2 and H2O.
Revised reactions:
1. CO2(g) → C(s) + O2(g) ΔH° = +394 kJ/mol
2. 2H2O(l) → 2H2(g) + O2(g) ΔH° = +572 kJ/mol (2 x -286 kJ/mol)
3. CO2(g) + 2H2(g) → CH4(g) + H2O(l) ΔH° = ?
We can add revised reactions 1 and 2 to get:
CO2(g) + 2H2O(l) → C(s) + 2H2(g) + O2(g) ΔH° = -178 kJ/mol
We can then subtract revised reaction 3 from this to get:
ΔH° = (-178) kJ/mol - (-890) kJ/mol
ΔH° = +712 kJ/mol
Therefore, the standard enthalpy change of the reaction C(s) + 2H2(g) → CH4(g) is +712 kJ/mol.
What happens to the volume of a fixed amount of gas if the pressure is divided by a factor of 3 at constant temperature
For a fixed amount of gas at constant temperature, the relationship between pressure and volume is described by Boyle's law, which states that the product of the pressure and volume is constant:
P1V1 = P2V2
where P1 and V1 are the initial pressure and volume, respectively, and P2 and V2 are the final pressure and volume, respectively.
If the pressure is divided by a factor of 3, then the final pressure is 1/3 of the initial pressure (P2 = P1/3). We can substitute this into the equation above and solve for V2:
P1V1 = P2V2
V2 = (P1V1) / P2
V2 = (P1V1) / (P1/3)
V2 = (3P1V1) / P1
V2 = 3V1
Therefore, the volume of the fixed amount of gas would triple if the pressure is divided by a factor of 3 at constant temperature.
When 15.0 g of sodium nitrate (NaNO3) was dissolved in water in a bomb calorimeter, the temperature reduced from 25.00 to 21.56. If the heat capacity of the solution and the calorimeter is 1071kj/°C. What is the enthalpy change for the process per mol of sodium nitrate that dissolves in water
The enthalpy change for the dissolution of NaNO3 in water can be calculated using the heat absorbed by the solution and the calorimeter, which is given by:
q = CΔT
where q is the heat absorbed, C is the combined heat capacity of the solution and calorimeter, and ΔT is the change in temperature.
In this case, we can calculate the heat absorbed as:
q = CΔT = 1071 kJ/°C × (25.00°C - 21.56°C) = 3701.16 kJ
We also need to determine the number of moles of NaNO3 that were dissolved, which we can calculate using the molar mass of NaNO3:
n = 15.0 g / 85.00 g/mol = 0.1765 mol
Finally, we can calculate the enthalpy change per mole of NaNO3 that was dissolved in water using the following formula:
ΔH/mol = q / n
where ΔH/mol is the enthalpy change per mole of NaNO3, and n is the number of moles of NaNO3.
Substituting the values we have calculated, we get:
ΔH/mol = 3701.16 kJ / 0.1765 mol = -20961 kJ/mol
Therefore, the enthalpy change per mole of NaNO3 that dissolves in water is -20,961 kJ/mol (note that the negative sign indicates that the process is exothermic).
When a 7.50 g sample of solid NaOH dissolves in 100.00 g of water in a coffee cup calorimeter, the temperature rises from 21.6 to 37.8. Calculate the change in H for solution
To calculate the change in enthalpy for the solution of NaOH in water, we can use the following formula:
ΔH = -q / n
where ΔH is the change in enthalpy, q is the heat absorbed by the solution and the calorimeter, and n is the number of moles of NaOH that dissolved.
First, we need to calculate the heat absorbed by the solution and the calorimeter, which can be calculated using the following formula:
q = Csolution x msolution x ΔT
where Csolution is the specific heat capacity of the solution (we can assume it to be the same as water, which is 4.18 J/g°C), msolution is the mass of the solution (which is the mass of the water plus the mass of the NaOH), and ΔT is the change in temperature.
msolution = 100.00 g + 7.50 g = 107.50 g
ΔT = 37.8°C - 21.6°C = 16.2°C
q = 4.18 J/g°C x 107.50 g x 16.2°C
q = 7454.77 J
Next, we need to calculate the number of moles of NaOH that dissolved. This can be calculated using the molar mass of NaOH:
molar mass of NaOH = 23.00 g/mol + 16.00 g/mol + 1.008 g/mol = 40.01 g/mol
n = 7.50 g / 40.01 g/mol = 0.1875 mol
Finally, we can substitute these values into the formula for ΔH:
ΔH = -q / n = -7454.77 J / 0.1875 mol = -3.97 x 10^4 J/mol
Therefore, the change in enthalpy for the solution of NaOH in water is -3.97 x 10^4 J/mol, or -39.7 kJ/mol (rounded to two significant figures). Since the reaction is exothermic (heat is released), the value is negative.