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How may milliliters of 0.200 moles Sodium Hydroxide are required to completely neutralize 5.00 milliliters of 0.100 moles Phosphoric acid?
Molarity, not moles. Simple equality will do here.(0.200 M NaOH)(X milliliters) = (0.100 M H3PO4)(5.00 milliliters)0.200X = 0.5X = 2.5 milliliters sodium hydroxide solution needed====================================
What volume of NaOH solution (in mL) is needed to neutralize a given acid solution?
To determine the volume of NaOH solution needed to neutralize an acid solution, you would need to know the concentration of the acid solution and the volume of the acid solution. Using the equation n1V1 n2V2, where n represents the number of moles and V represents the volume, you can calculate the volume of NaOH solution needed.
What volume of a 0.160 M solution would have to be added to 550.0 mL of the acidic solution to neutralize completely all of the acid?
To neutralize the acidic solution completely, you need to add a solution of basic nature. The volume of the basic solution required can be calculated using the formula: ( V_1 \times C_1 = V_2 \times C_2 ), where ( V_1 ) and ( C_1 ) are the volume and concentration of the acidic solution, and ( V_2 ) and ( C_2 ) are the volume and concentration of the basic solution, respectively. Substituting the known values, you can find the volume of the basic solution needed.
What volume of a 0.152 M potassium hydroxide solution is required to neutralize 10.2 ml of a 0.198 M hydrochloric acid solution?
To determine the volume of potassium hydroxide solution needed to neutralize the hydrochloric acid solution, you can use the formula M1V1 = M2V2. By plugging in the given values, you can calculate the volume of the potassium hydroxide solution required. In this case, the volume of the 0.152 M potassium hydroxide solution needed to neutralize 10.2 ml of the 0.198 M hydrochloric acid solution would be 7.43 ml.
How many milliliters of a 1.00 M solution of sodium chloride are needed to completely react with 100 grams of water according to the chemical reaction shown below?
There is no chemical reaction between sod chloride solution and water, it would just dilute the sod chloride solution.
How many ml of 110 M of HCl are needed to completely neutralize 54ml of 106 M of BaOH2 solution?
A concentration of 110 M or 106 M doesn't exist.
How may milliliters of 0.200 moles Sodium Hydroxide are required to completely neutralize 5.00 milliliters of 0.100 moles Phosphoric acid?
Molarity, not moles. Simple equality will do here.(0.200 M NaOH)(X milliliters) = (0.100 M H3PO4)(5.00 milliliters)0.200X = 0.5X = 2.5 milliliters sodium hydroxide solution needed====================================
How many liters of 5.00 M hydrochloric acid solution are needed to completely neutralize the magnesium hydroxide?
0.0532
How many liters of 5.00 M hydrochloric acid solution are needed to completely neutralize the strontium hydroxide?
0.0932 L
How many milliliters of a 5.00M solution of hydrochloric acid are needed to completely react with 31.5g of zinc sulfide?
129
What volume of NaOH solution (in mL) is needed to neutralize a given acid solution?
To determine the volume of NaOH solution needed to neutralize an acid solution, you would need to know the concentration of the acid solution and the volume of the acid solution. Using the equation n1V1 n2V2, where n represents the number of moles and V represents the volume, you can calculate the volume of NaOH solution needed.
How many milliliters of a 7.00 M solution of hydrochloric acid are needed to completely react with 56.5 g of zinc sulfide according to the chemical reaction shown below?
166
What volume of a 0.160 M solution would have to be added to 550.0 mL of the acidic solution to neutralize completely all of the acid?
To neutralize the acidic solution completely, you need to add a solution of basic nature. The volume of the basic solution required can be calculated using the formula: ( V_1 \times C_1 = V_2 \times C_2 ), where ( V_1 ) and ( C_1 ) are the volume and concentration of the acidic solution, and ( V_2 ) and ( C_2 ) are the volume and concentration of the basic solution, respectively. Substituting the known values, you can find the volume of the basic solution needed.
What volume of a 0.152 M potassium hydroxide solution is required to neutralize 10.2 ml of a 0.198 M hydrochloric acid solution?
To determine the volume of potassium hydroxide solution needed to neutralize the hydrochloric acid solution, you can use the formula M1V1 = M2V2. By plugging in the given values, you can calculate the volume of the potassium hydroxide solution required. In this case, the volume of the 0.152 M potassium hydroxide solution needed to neutralize 10.2 ml of the 0.198 M hydrochloric acid solution would be 7.43 ml.
How many milliliters of a 1.00 M solution of sodium chloride are needed to completely react with 100 grams of water according to the chemical reaction shown below?
There is no chemical reaction between sod chloride solution and water, it would just dilute the sod chloride solution.
What is the molarity of HNO3 if 20.0 ml of the solution is needed to exactly neutralize 10.0 ml of a 1.67 M NaOH solution?
The reaction between HNO3 and NaOH is a 1:1 molar ratio. This means that the moles of HNO3 required to neutralize the NaOH is the same as the moles of NaOH. Given that 20.0 ml of HNO3 is needed to neutralize 10.0 ml of a 1.67 M NaOH solution, the molarity of the HNO3 solution is twice the molarity of the NaOH solution, which is 3.34 M.
An acid solution is 0.120 m in hcl and 210 m in h2so4 what volume of a 150 m koh solution would have to be added to 500.0 ml of the acidic solution to neutralize completely all of the acid?
To find the volume of KOH solution needed to neutralize the acidic solution, you'll need to determine the moles of acid present and use the stoichiometry of the neutralization reaction. Calculate the moles of HCl and H2SO4 separately, and then find the limiting reactant. Finally, use the balanced chemical equation to determine the moles of KOH needed, which can then be converted to volume using the concentration of the KOH solution.
