All categories

3 years ago

Determine the pH of a 5x10^-4 M solution of Ca(OH)2

1 answer:

5 0

Ca(OH)₂ ==> Ca²⁺ + 2 OH-

Ca(OH)₂ is strong Bases

Therefore, the [OH-] equals 5 x 10⁻⁴ M. For every Ca(OH)₂ you produce 2 OH⁻.

pOH = - log[ OH⁻]

pOH = - log [ 5 x 10⁻⁴ ]

pOH = 3.30

pH + pOH = 14

pH + 3.30 = 14

pH = 14 - 3.30

pH = 10.7

hope this helps!

You might be interested in

Which of the following elements is most likely to form a covalent bond?

Alecsey [184]

Bromine, it's the only element that isn't a metal

7 0

3 years ago

The Kelvin temperature required for 0.0470 mol of helium gas to fill a balloon to 1.20 L under 0.878 atm is

4vir4ik [10]

From the ideal gas law, PV = nRT, we can rearrange the equation to solve for T given the other parameters.

T = PV/nR

where P = 0.878 atm, V = 1.20 L, n = 0.0470 moles, and R = 0.082057 L•atm/mol•K. Plugging in our values, we obtain the temperature in Kelvin:

T = (0.878 atm)(1.20 L)/(0.0470 mol)(0.082057 L•atm/mol•K)
T = 273 K

So, the second answer choice would be correct.

7 0

3 years ago

Using the van der Waals equation, calculate the pressure for a 1.25 mol sample of xenon contained in a volume of 1.000L at 75Â°C

alina1380 [7]

Answer:

ABC

Explanation:

hrdjyt

5 0

2 years ago

CH4 with pressure 1 atm and volume 10 liter at 27°C is passed into a reactor with 20% excess oxygen, how many moles of oxygen is

BaLLatris [955]

Answer : The moles of $O_2$ left in the products are 0.16 moles.

Explanation :

First we have to calculate the moles of $CH_4$ .

Using ideal gas equation:

$PV=nRT$

where,

P = pressure of gas = 1 atm

V = volume of gas = 10 L

T = temperature of gas = $27^oC=273+27=300K$

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

$(1atm)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)$

$n=0.406mole$

Now we have to calculate the moles of $O_2$ .

The balanced chemical reaction will be:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that,

As, 1 mole of $CH_4$ react with 2 moles of $O_2$

So, 0.406 mole of $CH_4$ react with $2\times 0.406=0.812$ moles of $O_2$

Now we have to calculate the excess moles of $O_2$ .

$O_2$ is 20 % excess. That means,

Excess moles of $O_2$ = $\frac{(100 + 20)}{100}$ × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × 0.812 = 0.97 mole

Now we have to calculate the moles of $O_2$ left in the products.

Moles of $O_2$ left in the products = Excess moles of $O_2$ - Required moles of $O_2$

Moles of $O_2$ left in the products = 0.97 - 0.812 = 0.16 mole

Therefore, the moles of $O_2$ left in the products are 0.16 moles.

7 0

3 years ago

How does distance affect work pls quick its 15 points pls keep it short pls help!!!!

ra1l [238]

The simple equation used to calculate work is force multiplied by distance, thus as this is the case increasing the distance by a certain amount, assuming the force applied to the object is constant, the amount of work you are doing on the box for instance pushing it, is going to be greater

Since you are pushing the box with the same force covering a greater distance with the force.

4 0

3 years ago