Explanation:
firstly find for the molar mass of kcl and molar mass of k
and then
molar mass of k = x
molar mass of kcl= 40
cross mutiply and then simplify you will get your answer
Answer:
C₄H₁₀O + 6O₂ ⇒ 4CO₂ + 5H₂O
Explanation:
Match the amount of reactants and products on both sides of the equation.
Answer:
b. potassium.
Explanation:
Potassium-sparing diuretics and salt substitutes are diuretics that eliminate salt and water but save potassium. They act by inhibiting the conducting sodium channels in the collecting tubule, such as amiloride and triamterene, or by blocking aldosterone, such as spironolactone.
Concomitant use of potassium-sparing diuretics together with salt substitutes may result in dangerously high blood levels of serum potassium. For this reason, it is important to consult a physician before taking these substances at the same time to avoid potential problems with potassium accumulation.
<u>Answer:</u> The net ionic equation for the given reaction is 
<u>Explanation:</u>
Net ionic equation of any reaction does not include any spectator ions.
Spectator ions are the ions which do not get involved in a chemical equation. It is also defined as the ions that are found on both the sides of the chemical reaction when it is present in ionic form.
The chemical equation for the reaction of hydrochloric acid and potassium sulfite is given as:

Ionic form of the above equation follows:

As, potassium and chloride ions are present on both the sides of the reaction, thus, it will not be present in the net ionic equation.
The net ionic equation for the above reaction follows:

Hence, the net ionic equation for the given reaction is written above.
<h3>
Answer:</h3>
200 mL
<h3>
Explanation:</h3>
Concept tested: Dilution formula
We are given;
- Concentration of stock solution as 1.00 M
- Volume of the stock solution as 50 mL
- Molarity of the dilute solution as 0.25 M
We are required to calculate the volume of diluted solution;
- The stock solution is the original solution before dilution while diluted solution is the solution after dilution.
- Using the dilution formula we can determine the volume of diluted solution;
M1V1 = M2V2
Rearranging the formula;
V2 = M1V1 ÷ M2
= (1.00 M × 50 mL) ÷ 0.25 M
= 200 mL
Therefore, a volume of 200mL of 0.25 M solution could be made from the stock solution.