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3 years ago

How many significant digits in 0.23100

Chemistry

1 answer:

Reil [10]3 years ago

4 0

Answer:

There are 5 significant digits in 0.23100.

Explanation:

This is because all non-zero digits are considered significant and zeros after decimal points are considered significant.

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A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce

spin [16.1K]

Explanation:

The given data is as follows.

Volume of lake = $15 \times 10^{6} m^{3}$ = $15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}$

Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = $5.6 \times 15 \times 10^{9} mg$

= $84 \times 10^{9}$ mg

= $84 \times 10^{3}$ kg

Flow rate of river is 50 $m^{3} sec^{-1}$

Volume of water in 1 day = $50 \times 10^{3} \times 86400 liter$

= $432 \times 10^{7}$ liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are $2.9792 \times 10^{10} mg$ or $2.9792 \times 10^{4} kg$

Flow rate of sewage = $0.7 m^{3} sec^{-1}$

Volume of sewage water in 1 day = $6048 \times 10^{4}$ liter

Concentration of sewage = 300 mg/L

Total amount of pollutants = $1.8144 \times 10^{10} mg$ or $1.8144 \times 10^{4}kg$

Therefore, total concentration of lake after 1 day = $\frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l$

= 6.8078 mg/l

$k_{D}$ = 0.2 per day

$L_{o}$ = 6.8078

Hence, $L_{liquid}$ = $L_{o}(1 - e^{-k_{D}t}$

$L_{liquid}$ = $6.8078 (1 - e^{-0.2 \times 1})$

= 1.234 mg/l

Hence, the remaining concentration = (6.8078 - 1.234) mg/l

= 5.6 mg/l

Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

5 0

3 years ago

CH3 + HCl <=> CH3Cl + H2O

dmitriy555 [2]

Answer:

The pressure of CH3OH and HCl will decrease.

The final partial pressure of HCl is 0.350038 atm

Explanation:

Step 1: Data given

Kp = 4.7 x 10^3 at 400K

Pressure of CH3OH = 0.250 atm

Pressure of HCl = 0.600 atm

Volume = 10.00 L

Step 2: The balanced equation

CH3OH(g) + HCl(g) <=> CH3Cl(g) + H2O(g)

Step 3: The initial pressure

p(CH3OH) = 0.250atm

p(HCl) = 0.600 atm

p(CH3Cl)= 0 atm

p(H2O) = 0 atm

Step 3: Calculate the pressure at the equilibrium

p(CH3OH) = 0.250 - X atm

p(HCl) = 0.600 - X atm

p(CH3Cl)= X atm

p(H2O) = X atm

Step 4: Calculate Kp

Kp = (pHO * pCH3Cl) / (pCH3* pHCl)

4.7 * 10³ = X² /(0.250-X)(0.600-X)

X = 0.249962

p(CH3OH) = 0.250 - 0.249962 = 0.000038 atm

p(HCl) = 0.600 - 0.249962 = 0.350038 atm

p(CH3Cl)= 0.249962 atm

p(H2O) = 0.249962 atm

Kp = (0.249962 * 0.249962) / (0.000038 * 0.350038)

Kp = 4.7 *10³

The pressure of CH3OH and HCl will decrease.

The final partial pressure of HCl is 0.350038 atm

4 0

3 years ago

When mixed, solutions of silver nitrate, AgNO3, and sodium sulfate, Na2SO4, form a precipitate of silver sulfate, Ag2SO4. The ba

Maslowich

Answer:

2 AgNO3 + Na2SO4 → Ag2SO4 + 2 NaNO3

Explanation:

The general schemefor a reaction is given as;

Reactants --> Products

In this question, the reactants are AgNO3 and Na2SO4. The product is Ag2SO4.

The equation is given as;

AgNO3 + Na2SO4 --> Ag2SO4

The other poduct formed in this reaction is NaNO3.

The full reaction is given as;

AgNO3 + Na2SO4 --> Ag2SO4 + NaNO3

The above reaction is not balanced because there are unequal number of atoms of the elements on both sides of the reaction.

The balanced equation is given as;

2 AgNO3 + Na2SO4 → Ag2SO4 + 2 NaNO3

In this equation, there are equal number of moles of the atoms on both sides.

3 0

3 years ago

What are some possible explanations you think a researcher might find to explain why the fish are washing up on the shore of Lak

8_murik_8 [283]

Answer:

the temperature or weather ,

5 0

3 years ago

PLEASE SOMEONE HELP ME WITH THIS!

Bad White [126]

Answer:

1) 1.202 L , 2) 1.291 dg , 3) 204.877 and 4) 1.04x $10^{3\\}$

Explanation:

You need to review about conversion factors and how to use them in the correct order. You can cancel the units and get the ones that you need if you use the appropriate conversion factors, remember is a number that you can use to multiply or divide.

For your exercise:

1) The conversion factor is: 1 L = 1000 mL

You will need to divide by 1000 mL to obtain liters L

1202.57120 mL x $\frac{1 L}{1000 mL}$ = 1.202 L

2) The conversion factor is: 1 g = 10 dg

0.1290743 g x $\frac{10 dg}{ 1 g}$ = 1.291 dg

For the next exercises, you need to follow some rules:

1. All numbers that are different from Zero (non-zero digits) are significant figures.

2.The Zeros between non-zeros digits (Imbedded zeros) always are significant, 2007.

3. If you want to be specific and want some zeros to be significant you need to add a decimal point. For example 500. or 500.0

4. Leading zeros (to the left) are not significant.

5. Trailing zeros (zeros to the right) in a whole number without decimal point are not significant.

3) 843.062 - 638.1848 = 204.8772

Now if we round to 6 significant figures we get 204.877

4)123.0 x 8.43 = 1036.89

Now we round to 3 significant figures because 8.43 has the least significant figures.

1.04x $10^{3}$

4 0

3 years ago