Titan is bigger than mercury. true or false.

Groundwater in Pherric, New Mexico, initially contains 1.800 mg/L of iron as Fe3+. What must the pH be raised to in order to pre

xenn [34]

Answer : The pH will be, 3.2

Explanation :

As we known that the value of solubility constant of ferric hydroxide at $25^oC$ is, $2.79\times 10^{-39}$

Amount or solubility of iron consumed = (1.800 - 0.3) mg/L = 1.5 mg/L

The given solubility of iron convert from mg/L to mol/L.

$1.5mg/L=\frac{1.5\times 10^{-3}g/L}{56g/mol}=2.7\times 10^{-7}mol/L$

The chemical reaction will be:

$Fe(OH)_3\rightarrow Fe^{3+}+3OH^-$

The expression of solubility constant will be:

$K_{sp}=[Fe^{3+}]\times [3OH^-]^3$

Now put all the given values in this expression, we get the concentration of hydroxide ion.

$2.79\times 10^{-39}=(2.7\times 10^{-7})\times [3OH^-]^3$

$[OH^-]=1.5\times 10^{-11}M$

Now we have to calculate the pOH.

$pOH=-\log [OH^-]$

$pOH=-\log (1.5\times 10^{-11})$

$pOH=10.8$

Now we have to calculate the pH.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-10.8\\\\pH=3.2$

Therefore, the pH will be, 3.2

4 0

3 years ago

A sample of nitrogen gas expands in volume from 16 L to 5.4 L at constant

xxTIMURxx [149]

a. W = 0 J

b. W = - 308.028 J

<h3>Further explanation</h3>

Given

Nitrogen gas expands in volume from 1.6 L to 5.4 L

Required

The work done

Solution

Isothermal :

W = -P . ΔV

Input the value :

a. At a vacuum, P = 0

So W = 0

b. At pressure = 0.8 atm

W = - 0.8 x ( 5.4 - 1.6)

W = -3.04 L.atm ( 1 L.atm = 101.325 J)

W = - 3.04 x 101.325

W = - 308.028 J

6 0

3 years ago

Each square of the periodic table shows information about one element. The square usually shows the element's

omeli [17]

D, all of the above.

7 0

3 years ago

How much heat does a 100. g sample of copper absorb when its temperature increases by 30.0°C? The specific heat of copper is 0.3

erica [24]

Answer:

$\boxed {\boxed {\sf B. \ 1170 \ Joules }}$

Explanation:

We are asked to find how much heat a sample of copper absorbs when the temperature is increased.

Since we know the mass, temperature increase, and specific heat capacity, we can use the following formula to calculate heat.

$q= mc \Delta T$

The mass of the copper sample is 100 grams, the temperature is changed or increased by 30.0 degrees Celsius, and the specific heat of copper is 0.39 Joules per gram degrees Celsius.

m= 100 g
c= 0.39 J/g °C
ΔT= 30.0 °C

Substitute the values into the formula.

$q= (100 \ g )(0.39 \ J/g \textdegree C ) (30.0 \textdegree C )$

Multiply the first two values. Note that the units of grams cancel.

$q= 39 \ J/ \textdegree C (30.0 \textdegree C )$

Multiply again, this time the units of degrees Celsius cancel.

$q= 1170 \ J$

The copper sample absorbs <u>1170 Joules</u> of heat and <u>Choice B </u>is correct.

7 0

3 years ago

Vishwanath had some money. he spent 3 upon 4 part of money to buy goods for his birthday,1 upon 5 part of money give to his sist

lord [1]

Answer:

The correct answer is - 800.

Explanation:

Given:

Total amount = ? or assume x

spend in buying birthday item = 3/4 of x

given to sister = 1/5 of x

remaining to mother = 40

solution:

the remaning amount = x- (3x/4+x/5) = 4=

=> x- 19x/20 = 40

=> x = 20*40

=> x = 800

thus, the correct answer is = 800

8 0

2 years ago