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Sav [38]
3 years ago
13

Give the result of the following expression with the correct number of significant figures

Chemistry
1 answer:
Marianna [84]3 years ago
4 0
84 m/s x 31.221 s = 2600 m
-rounded from 2622.564 m because 84 m/s has only two significant figures
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PLS HELP ME, use the periodic table to write the electron configuration of selenium
EastWind [94]

Answer:

[Ar] 3d10 4s2 4p4

Explanation:

1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p4

6 0
2 years ago
You are provided with a compound fertilizer, 40-15-10.Calculate the quantity of fertilizer to add to a one hectare field supply
victus00 [196]

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 bag of the given fertilizer is 40:15:10.

The percentages can be expressed as fractions as follows:

For nitrogen; 40/100 = 0.4

For phosphorus; 15/100 = 0.15

For potassium; 10/100 = 0.1

To find the quantity of fertilizer required to add to a hectare to supply the given amount of nutrients, the amount to be provided is divided by the percentage or fractional compostion of each nutrient.

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  60 kg/ha = 60/0.15 = 400 Kg of fertilizer

d. Potassium at 60 kg/ha = 60/0.1 = 600 Kg of fertilizer

5 0
2 years ago
2. What volume will 3.00 moles of nitrogen gas occupy at STP?
Amiraneli [1.4K]

Answer:

Answer: V=67.2 L

Explanation:

Ideal Gas Law: PV=nRT

P=1.00 atm (STP)

V=?

n=3.00 mol

R=0.08206Latm/Kmol

T=273.15 K (STP)

To find V, we would manipulate the equation to V=nRT/P

With significan figures, our answer is V=67.2 L.

4 0
2 years ago
The table shows the percentages of hydrocarbons that are found in a sample of crude oil. Hydrocarbons Percentage Paraffins 30 Na
ioda
This will be classified as light on the API scale due to the large percentage of lighter fractions such as paraffins and naphthenes. 
4 0
3 years ago
Read 2 more answers
Consider the following reaction: A(g)⇌2B(g). Find the equilibrium partial pressures of A and B for each of the following differe
Illusion [34]

Answer:

a. Kp=1.4

P_{A}=0.2215 atm

P_{B}=0.556 atm

b.Kp=2.0 * 10^-4

P_{A}=0.495atm

P_{B}=0.00995 atm

c.Kp=2.0 * 10^5

P_{A}=5*10^{-6}atm

P_{B}=0.9999 atm

Explanation:

For the reaction  

A(g)⇌2B(g)

Kp is defined as:

Kp=\frac{(P_{B})^{2}}{P_{A}}

The conditions in the system are:

          A                    B

initial   0                1 atm

equilibrium x       1atm-2x

At the beginning, we don’t have any A in the system, so B starts to react to produce A until the system reaches the equilibrium producing x amount of A. From the stoichiometric relationship in the reaction we get that to produce x amount of A we need to 2x amount of B so in the equilibrium we will have 1 atm – 2x of B, as it is showed in the table.    

Replacing these values in the expression for Kp we get:

Kp=\frac{(1-2x)^{2}}{x}

Working with this equation:

x*Kp=(1-2x)^{2} - -> x*Kp=4x^{2}-4x+1- - >4x^{2}-(4+Kp)*x+1=0

This last expression is quadratic expression with a=4, b=-(4+Kp) and c=1

The general expression to solve these kinds of equations is:

x=\frac{-b(+-)*\sqrt{(b^{2}-4ac)}}{2a} (equation 1)

We just take the positive values from the solution since negative partial pressures don´t make physical sense.

Kp = 1.4

x_{1}=\frac{(1.4+4)+\sqrt{(-(1.4+4)^{2}-4*4*1)}}{2*4}=1.128

x_{1}=\frac{(1.4+4)-\sqrt{(-(1.4+4)^{2}-4*4*1)}}{2*4}=0.2215

With x1 we get a partial pressure of:

P_{A}=1.128 atm

P_{B}=1-2*1.128 = -1.256 atm

Since negative partial pressure don´t make physical sense x1 is not the solution for the system.

With x2 we get:

P_{A}=0.2215 atm

P_{B}=1-2*0.2215 = 0.556 atm

These partial pressures make sense so x2 is the solution for the equation.

We follow the same analysis for the other values of Kp.

Kp=2*10^-4

X1=0.505

X2=0.495

With x1

P_{A}=0.505atm

P_{B}=1-2*0.505 = -0.01005 atm

Not sense.

With x2

P_{A}=0.495atm

P_{B}=1-2*0.495 = 0.00995 atm

X2 is the solution for this equation.  

Kp=2*10^5

X1=50001

X2=5*10^{-6}

With x1

P_{A}=50001atm

P_{B}=1-2*50001=-100001atm

Not sense.

With x2

P_{A}= 5*10^{-6}atm

P_{B}=1-2*5*10^{-6}= 0.9999 atm

X2 is the solution for this equation.  

8 0
3 years ago
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