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avanturin [10]
3 years ago
5

at 298 k the H+ ion concentration of a aqueous solution is 1.00 x 10-5 M what is the pH of the solution

Chemistry
1 answer:
Basile [38]3 years ago
7 0

Answer:

pH = 5.00

Explanation:

The pH is an unit of concentration widely used in chemistry and quality control of aqueous solutions. It serves to find the security of a product to human or pet consumption. Is defined as:

pH = -log [H+]

That means, the pH of a solution that is [H+] = 1.00x10⁻⁵M:

<h3>pH = 5.00</h3>
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Consider the reaction: CH4 + 2O2 = CO2 + 2H2O
crimeas [40]

Answer:

The answer to your question is:

a)  80 g of O2

b) O2, 15.13 g of CO2

c) It's not posible to know which is the limiting reactant.

Explanation:

Reaction                             CH4   +   2O2   ⇒   CO2   +   2H2O

a. Calculate the grams of O2 needed to react with 20.00 grams of CH4. _____________

MW CH4 = 16 g

MW O2 = 32 g

                               16 g of CH4 ----------------  2(32) g of O2

                               20 g              --------------    x

                               x = (20 x 64) / 16 = 80 g of O2

b. Given 15.00 g. of CH4 and 22.00 g. of O2, identify the limiting reactant and calculate the grams of CO2 that can be produced. LR _________ grams CO2 _________ .  

                                CH4   +   2O2   ⇒   CO2   +   2H2O

                                15 g         22 g

                                16 g of CH4 ----------------  64 g of O2

                                15 g of CH4  ---------------   x

                               x = (15 x 64) / 16 = 60 g of O2

The Limiting reactant is O2 because it is necessary 60g of O2 for 16 g of CH4 and there are only 22.

                                 CH4   +   2O2   ⇒   CO2   +   2H2O

                        64 g of O2 ------------------  44 g of CO2

                        22 g of O2 ------------------   x

                        x = (22 x 44)/ 64 = 15. 13 g of CO2

c. For the reaction CH4 + 2O2 = CO2 + 2H2O, if you have 10.31 g. of CH4 and an unknown amount of oxygen, and form 20.00 g. of CO2, i. Identify if there is a limiting reactant ______________ ii. Calculate the number of grams of the limiting reactant present if there is one. ______________  

                           CH4   +   2O2   ⇒   CO2   +   2H2O                              

                           10.31 g                     20 g

We can identify the limiting reactant if we know the quantity of the reactants, if we only know the quantity of one it is not posible to which is the limiting reactant.

4 0
3 years ago
Suppose of potassium sulfate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of potassium
dimulka [17.4K]

Answer:

This question is incomplete, here's the complete question:

<em><u>"Suppose 0.0842g of potassium sulfate is dissolved in 50.mL of a 52.0mM aqueous solution of sodium chromate. Calculate the final molarity of potassium cation in the solution. You can assume the volume of the solution doesn't change when the potassium sulfate is dissolved in it. Round your answer to 2 significant digits."</u></em>

Explanation:

Reaction :-

K2SO4 + Na2CrO4 ------> K2CrO4 + Na2SO4

Mass of K2SO4 = 0.0842 g, Molar mass of K2SO4 = 174.26 g/mol

Number of moles of K2SO4 = 0.0842 g / 174.26 g/mol = 0.000483 mol

Concentration of Na2CrO4 = 52.0 mM = 52.0 * 10^-3 M = 0.052 mol/L

Volume of Na2CrO4 solution = 50.0 ml = 50 L / 1000 = 0.05 L

Number of moles of Na2CrO4 = 0.05 L * 0.052 mol/L = 0.0026 mol

Since number of moles of K2SO4 is smaller than number of moles Na2CrO4, so 0.000483 mol of K2SO4 will react with 0.000483 mol of Na2CrO4 will produce 0.000483 mol of K2CrO4.

0.000483 mol of K2CrO4 will dissociate into 2* 0.000483 mol of K^+

Final concentration of potassium cation

= (2*0.000483 mol) / 0.05 L = 0.02 mol/L = 0.02 M

8 0
3 years ago
When energy is transferred to ocean water from moving air above it what happens to most of the energy
Free_Kalibri [48]

c trust me i did the test

3 0
3 years ago
(((NEED ANSWER QUICK!!!)))<br><br> Which is the stronger conjugate base, CN- or OCN-? Explain
Shkiper50 [21]

Answer:

The stronger conjugate base will be the weaker acid; i.e., the acid with the smaller Ka-value.

Explanation:

Given conjugate base CN⁻ => weak acid => HCN =>  Ka =4.9 x 10⁻¹⁰

Given conjugate base OCN⁻ => weak acid=> HOCN => Ka = 3.5 x 10⁻⁴

Ka(HCN) << Ka(HOCN) => CN⁻ is a much stronger conjugate base than OCN⁻

6 0
3 years ago
A powdered drink mix is stirred into a glass of water to form a solution. If any more is added it will not dissolve
Brut [27]
SaturTED, hope this helps, m8
6 0
3 years ago
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