Answer:
Mass = 51 g
Explanation:
Given data:
Mass of nitrogen = 41.93 g
Mass of ammonia formed = ?
Solution:
Chemical equation:
N₂ + 3H₂ → 2NH₃
Number of moles of nitrogen:
Number of moles = mass/molar mass
Number of moles = 41.93 g/ 28 g/mol
Number of moles = 1.5 mol
now we will compare the moles of nitrogen and ammonia.
N₂ : NH₃
1 : 2
1.5 : 2/1×1.5 = 3 mol
Mass of ammonia formed:
Mass = number of moles × molar mass
Mass = 3 mol × 17 g/mol
Mass = 51 g
Explanation:
there is 2 nitrogen but if you mean nitrate is 6
Answer:
The method is accurate in the calculation of the ![Cu^+2](https://tex.z-dn.net/?f=Cu%5E%2B2)
Explanation:
As a first step we have to calculate the <u>average concentration </u>of
find it by the method.
![\frac{0.782+0.762+0.825+0.838+0.761 }{5} =0.79 ppm](https://tex.z-dn.net/?f=%5Cfrac%7B0.782%2B0.762%2B0.825%2B0.838%2B0.761%20%7D%7B5%7D%20%3D0.79%20ppm)
Then we have to find the<u> standard deviation:</u>
![s=\sqrt{\frac{1}{N-1}\sum_{i=1}^N(x_i-\bar{x})^2}=0.0359](https://tex.z-dn.net/?f=s%3D%5Csqrt%7B%5Cfrac%7B1%7D%7BN-1%7D%5Csum_%7Bi%3D1%7D%5EN%28x_i-%5Cbar%7Bx%7D%29%5E2%7D%3D0.0359)
For the confidence interval we have to use the formula:
μ=Average±![\frac{t*s}{\sqrt{n} }](https://tex.z-dn.net/?f=%5Cfrac%7Bt%2As%7D%7B%5Csqrt%7Bn%7D%20%7D)
Where:
t=t student constant with 95 % of confidence and 5 data=2.78
μ=
± ![\frac{2.78*0.0359}{\sqrt{5} }](https://tex.z-dn.net/?f=%5Cfrac%7B2.78%2A0.0359%7D%7B%5Csqrt%7B5%7D%20%7D)
upper limit: 0.84
lower limit: 0.75
If we compare the limits of the value obtanied by the method (Figure 1 Red line) with the reference material (Figure 1 blue line) we can see that the values obtained by the method are within the values suggested by the reference material. So, it's method is accurate.
Answer:
double replacement
synthesis
double replacement
i think this one is decomposition
synthesis
double replacement
single replacement
single replacement
double replacement
single replacement .....
hopefully i help
Grass is biotic- living.
please vote my answer brainliest. thanks!