Answer:
The pressure in the container is 1,7 atm.
Explanation:
We convert the unit of temperature in Celsius into Kelvin:
0°C= 273K --> 45°C= 45 + 273= 318 K
We use the formula:
PV=nRT ---> P =nRT/V
P= 1,20 mol x 0,082 l atm/K mol x 318 K/18,0 L
<em>P= 1, 7384 atm</em>
Answer:
Change in velocity is -4m\s
acceleration is -2m\s^2 cause Change in velocity = final v - initial v
Explanation:
Calculating acceleration involves dividing velocity by time — or in terms of SI units, dividing the meter per second [m/s] by the second [s]. Thus the SI unit of acceleration is the meter per second squared .
Answer:
<h2>Density = 3.9 g/ mL</h2>
Explanation:
The density of a substance can be found by using the formula
![Density = \frac{mass}{volume}](https://tex.z-dn.net/?f=Density%20%3D%20%20%5Cfrac%7Bmass%7D%7Bvolume%7D%20)
From the question
mass = 43 g
volume = 11.1 mL
Substitute the values into the above formula and solve for the density
That's
![Density = \frac{43}{11.1} \\ = 3.873873...](https://tex.z-dn.net/?f=Density%20%3D%20%20%5Cfrac%7B43%7D%7B11.1%7D%20%20%5C%5C%20%20%3D%203.873873...)
We have the final answer as
Density = 3.9 g/ mL to one decimal place
Hope this helps you
Answer:
a. 300 Kg of fertilizer
b. 225 Kg of fertilizer
c. 400 Kg of fertilizer
d. 600 Kg of fertilizer
Explanation:
The percentage composition ratio of Nitrogen, Phosphorus and Potassium in a 1 Kg bag of the given fertilizer is 40:15:10.
Therefore a 1 Kg bag contains;
40/100 * 1 Kg = 0.4 Kg of Nitrogen;
15/100 * 1 Kg = 0.15 Kg of phosphorus;
10/100 * 1 Kg = 0.1 Kg of potassium
Quantity of fertilizer required to add to a hectare to supply;
a. Nitrogen at 120 kg/ha = 120/0.4 = 300 Kg of fertilizer
b.. Nitrogen at 90 Kg/ha = 90/0.4 = 225 Kg of fertilizer
c. Phosphorus at a rate of 60 kg/ha = 60/0.15 = 400 Kg of fertilizer
d. Potassium at a rate of 60 kg/ha = 60/0.1 = 600 Kg of fertilizer
Hello!
We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100g solution)
C: 70.79% = 70,79 g
H: 8.91% = 8.91 g
N: 4.59% = 4.59 g
O: 15.72% = 15.72 g
Let us use the above mentioned data (in g) and values will be converted to amount of substance (number of moles) by dividing by molecular mass (g / mol) each of the values, lets see:
![C: \dfrac{70.79\:\diagup\!\!\!\!\!g}{12\:\diagup\!\!\!\!\!g/mol} \approx 5.89\:mol](https://tex.z-dn.net/?f=C%3A%20%5Cdfrac%7B70.79%5C%3A%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21g%7D%7B12%5C%3A%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21g%2Fmol%7D%20%5Capprox%205.89%5C%3Amol)
![H: \dfrac{8.91\:\diagup\!\!\!\!\!g}{1\:\diagup\!\!\!\!\!g/mol} = 8.91\:mol](https://tex.z-dn.net/?f=H%3A%20%5Cdfrac%7B8.91%5C%3A%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21g%7D%7B1%5C%3A%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21g%2Fmol%7D%20%3D%208.91%5C%3Amol)
![N: \dfrac{4.59\:\diagup\!\!\!\!\!g}{14\:\diagup\!\!\!\!\!g/mol} \approx 0.328\:mol](https://tex.z-dn.net/?f=N%3A%20%5Cdfrac%7B4.59%5C%3A%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21g%7D%7B14%5C%3A%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21g%2Fmol%7D%20%5Capprox%200.328%5C%3Amol)
![O: \dfrac{15.72\:\diagup\!\!\!\!\!g}{16\:\diagup\!\!\!\!\!g/mol} = 0.9825\:mol](https://tex.z-dn.net/?f=O%3A%20%5Cdfrac%7B15.72%5C%3A%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21g%7D%7B16%5C%3A%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21g%2Fmol%7D%20%3D%200.9825%5C%3Amol)
We note that the values found above are not integers, so let's divide these values by the smallest of them, so that the proportion is not changed, let's see:
![C: \dfrac{5.89}{0.328}\to\:\:\boxed{C\approx 18}](https://tex.z-dn.net/?f=C%3A%20%5Cdfrac%7B5.89%7D%7B0.328%7D%5Cto%5C%3A%5C%3A%5Cboxed%7BC%5Capprox%2018%7D)
![H: \dfrac{8.91}{0.328}\to\:\:\boxed{H\approx 27}](https://tex.z-dn.net/?f=H%3A%20%5Cdfrac%7B8.91%7D%7B0.328%7D%5Cto%5C%3A%5C%3A%5Cboxed%7BH%5Capprox%2027%7D)
![N: \dfrac{0.328}{0.328}\to\:\:\boxed{N\approx 1}](https://tex.z-dn.net/?f=N%3A%20%5Cdfrac%7B0.328%7D%7B0.328%7D%5Cto%5C%3A%5C%3A%5Cboxed%7BN%5Capprox%201%7D)
![O: \dfrac{0.9825}{0.328}\to\:\:\boxed{O\approx 3}](https://tex.z-dn.net/?f=O%3A%20%5Cdfrac%7B0.9825%7D%7B0.328%7D%5Cto%5C%3A%5C%3A%5Cboxed%7BO%5Capprox%203%7D)
T<span>hus, the minimum or empirical formula found for the compound will be:
</span>
I hope this helps. =)