CHEMISTRY HIGH SCHOOL

Would the concentration of hydroxide ion in a 0.30 m solution of potassium hydroxide (koh) be greater than, less than, or equal to 0.30 m?

Answers

Answer 1
Answer:

Potassium hydroxide KOH is a strong base which can completely ionize to form potassium and hydroxide ions.

The equation representing ionization of KOH,

K^{+}(aq)+OH^{-}(aq)" alt="KOH(aq)-->K^{+}(aq)+OH^{-}(aq)" align="absmiddle" class="latex-formula">

The mole ratio of hydroxide to KOH from the above equation is

The given concentration of KOH solution=0.30M

Calculating the concentration of hydroxide ions present in 0.30M KOH solution:

Therefore the concentration of hydroxide ion will be equal to 0.30 M in a 0.30 M KOH solution.


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HIGH SCHOOL

What volume of 0.123 m naoh, in milliliters, contains 25.0 g of naoh?

Answers

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MIDDLE SCHOOL

What are the difference between a red giant and a red super giant

Answers

Red giants are very large stars which are highly luminous and have an inflated outer surface with an approximate temperature of about 5000 degrees K or lower.

Whereas, Red supergiants are the largest stars in the whole universe in terms of volume but they are simply bright, red stars unlike the red giants.

Having said that, there are no agreed theoretical distinctions between the red giants and red supergiants so there is still a room for confusion to discriminate between the two.

So, unlike red giants, red supergiants are simply bright, red stars. It so happens that they may be in the same evolutionary state, but it is also possible that they have moved on. For example, most massive stars will appear as red supergiants while helium is fused into carbon in the core.



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HIGH SCHOOL

why can an increase in temperature lead to more effective collisions between reactant particles and an increase in the rate of a chemical reaction?

Answers

The reason why is because particles are moving faster and possess greater kinetic energy due to the increase in temperature, this causes greater frequency or number of collisions to take place and also allows for a higher likelihood that upon reactant molecules colliding they will undergo the reaction process as they will have achieved the required energy needed for the reaction to start. Thus overall having a greater change in quantity conversion of reactants to products within a shorter period of time, thus increasing the reaction rate.
MIDDLE SCHOOL

How is the scientific definition of "heat" different from how we use it every day? PLEASE NEED THE ANSWER ASAP

Answers

Heat is energy, that can be transferred to another item, which is “less hot”.
HIGH SCHOOL

Carbon disulfide burns in oxygen to yield car- bon dioxide and sulfur dioxide according to the chemical equation cs2(l) 3 o2(g) −→ co2(g) 2 so2(g). if 0.91 mol of cs2 is combined with 1.52 mol of o2, identify the limiting reactant.

Answers

Answer: Oxygen is the limiting reagent.

Explanation:

As can be seen from the given balanced equation:

3 moles of reacts with 1 mole of

1.52 moles of reacts with= of

Thus is the limiting reagent as it limits the formation of products. (0.91-0.51)= 0.40 moles of will remain as such and thus   is an excess reagent.

HIGH SCHOOL

ASAP PLEASE HELP ILL GIVE BRAINLEST!!!!!!!!!! There are 25 elements found in living things. How many of these elements are found in some organisms but not all?
A.1
B. 6
C.19
D.25

Answers

The awnser is (C, 19.

Answer:

B. 6

Explanation:

COLLEGE

Chloroflurocarbons (CFCs) are no longer used as refrigerants because they help destroy the ozone layer. Trichlorofluoromethane (CCl3F) boils are 23.8C and its molar heat of vaporization is 24.8 kj/mol. Calculate the molar entropy of evaporation of CC3F (l)

Answers

Answer:

83.60 J/k mol

Explanation:

The molar entropy of vaporization,ΔS vap   is given by the expression:

ΔS vap = ΔHvap / Tb  where

ΔHvap = molar enthalpy change of vaporization, J/mol

Tb = boiling temperature, K

We have both  ΔHvap, and Tb so lets solve our question.

T= (23.8 + 273) = 296.8 K

ΔHvap = 24.8 kJ = 24.8 x 10³ J

ΔS vap = ΔHvap / Tb = 24.8 x 10³ J/mol / 296.8 K = 83.6 J/k mol

HIGH SCHOOL

A pure gold bar is made of 19.55 mol of gold. What is the mass of the bar in grams

Answers

The mass of one mole of gold is 196.9 g/mol. Therefore, the mass of 19.95 mol of gold is 3849.3 g.

What is gold?

Gold is 79th element in periodic table. Gold is a transition metal and exhibit all the metallic properties such as malleability, ductility, luster, conductivity by thermally and electrically.

The atomic mass of an element is the its weight for one mole of the element thus, containing 6.02 × 10²³ atoms.

The atomic mass of gold = 196.9 g/mol.

This is the mass of one mole of gold.

The mass of 19.55 moles of gold is thus its atomic weight multiplied by 19.55.

mass of 19.5 moles of gold = 19.55 × 196.9 g/mol = 3849.3 g.

Therefore, the mass of the gold bar containing 19.55 mole of gold is 3849.9 g.

To find more on gold, refer here:

brainly.com/question/4838993

#SPJ3

(19.55 mol Au) / ( 1 ) x (196.97 g Au) / ( 1 mol Au) =
19.55 x 196.97 = 
3850.76 g Au

I hope this helps you and have a great day!! :)
COLLEGE

A mixture of ch4 and h2o is passed over a nickel catalyst at 1000. k. the emerging gas is collected in a 5.00-l flask and is found to contain 8.32 g of co, 2.63 g of h2, 42.3 g of ch4, and 49.2 g of h2o. assuming that equilibrium has been reached, calculate kc and kp for the reaction.

Answers

                    
According to that Kc is an equilibrium constant in terms of molar concentrations.
and Kc = [C]^c *[D]^d / [A]^a * [B]^b >>>> (1)
in the general reaction:
aA + bB ↔ cC + dD 
and, from our balanced equation:
CH4 + H2O ⇔ Co + 3H2 >>> (2)
So, we need to calculate the concentrations (molarity) of the products and reactants:
the Molarity of CH4 = no. of moles/volume (L)
 and no. of moles = weigh / Molecular weight = 42.3 / 16 = 2.643 moles
so the molarity of CH4 = 2.643 / 5 = 0.528 molar
the molarity of H20 = (49.2 / 18) / 5 =  0.546 molar
the molarity of CO = (8.32/28) / 5 = 0.059 molar
the molarity of H2 = (2.63 / 2) / 5 = 0.263 molar 
By substitution in (1) according to (2);
∴ Kc = [0.059]*[0.263]^3 / ( [0.528]*[0.546]) = 3.7 * 10 ^-3  >>>> (3)
Kp = Kc (RT)^(Δn) >>> (4)
where R is the gas constant = 0.0821,
and Δn is the change in moles in gas= (3(H2) + 1 (CO) - (1 H2O + 1 CH4) = 2
by substition in (4):
∴ Kp = 3.7*10^-3 (0.0821* 1000)^2= 24.939



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