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Why you can use 4.9035 g K2Cr2O7 to prepare 0.1 n solution?
To prepare a 0.1 N solution of K2Cr2O7, you need to calculate the molar mass of K2Cr2O7 and use the formula for normality. By dividing the given weight by the molar mass, you can determine the number of moles present, and then calculate the normality using the volume of the solution.
How many grams of potassium are present in 21.6 g of K2Cr2O7?
To find the amount of potassium in K2Cr2O7, calculate the molar mass of K2Cr2O7 first. The molar mass of K2Cr2O7 is 294.2 g/mol. Potassium accounts for 239.1 g/mol = 78.2 g/mol in K2Cr2O7, so in 21.6 g of K2Cr2O7, there are 21.6 g * (239.1 g/mol / 294.2 g/mol) = 5.77 g of potassium.
What is the compound for K2Cr207?
The compound K2Cr2O7 is potassium dichromate. It is a bright orange solid commonly used in laboratories as an oxidizing agent and for various chemical reactions.
What is the name of K2Cr2O7?
Potassium dichromate
What is difference between K2Cr2O7 and K2Cr2O4?
K2Cr2O7 is potassium dichromate, while K2Cr2O4 is potassium chromate. K2Cr2O7 is an orange-red compound used as an oxidizing agent, while K2Cr2O4 is a yellow compound. K2Cr2O7 tends to be more powerful as an oxidizing agent compared to K2Cr2O4.
Why you can use 4.9035 g K2Cr2O7 to prepare 0.1 n solution?
To prepare a 0.1 N solution of K2Cr2O7, you need to calculate the molar mass of K2Cr2O7 and use the formula for normality. By dividing the given weight by the molar mass, you can determine the number of moles present, and then calculate the normality using the volume of the solution.
How do you make 0.25N K2Cr2O7 solution?
To prepare a 0.25N K2Cr2O7 solution, you would dissolve 12.43 grams of K2Cr2O7 in enough water to make 1 liter of solution. This will give you a molarity of 0.25N for potassium dichromate (K2Cr2O7) in the solution.
How do you prepare the molar solution of K2Cr2O7?
Dissolve 294,185 g of potassium dichromate in 1L demineralized water.
What is the formula for patassium dichromate?
The chemical formula of potassium dichromate is K2Cr2O7
How many grams of potassium are present in 21.6 g of K2Cr2O7?
To find the amount of potassium in K2Cr2O7, calculate the molar mass of K2Cr2O7 first. The molar mass of K2Cr2O7 is 294.2 g/mol. Potassium accounts for 239.1 g/mol = 78.2 g/mol in K2Cr2O7, so in 21.6 g of K2Cr2O7, there are 21.6 g * (239.1 g/mol / 294.2 g/mol) = 5.77 g of potassium.
Name the product of ovi dation of etanol with k2Cr2O7?
Aldehydes is the product name of ovi dation of ethanol with k2Cr2O7.
What is the compound for K2Cr207?
The compound K2Cr2O7 is potassium dichromate. It is a bright orange solid commonly used in laboratories as an oxidizing agent and for various chemical reactions.
When was N. Ravi born?
N. Ravi was born on 1948-01-01.
What is the name of K2Cr2O7?
Potassium dichromate
What is difference between K2Cr2O7 and K2Cr2O4?
K2Cr2O7 is potassium dichromate, while K2Cr2O4 is potassium chromate. K2Cr2O7 is an orange-red compound used as an oxidizing agent, while K2Cr2O4 is a yellow compound. K2Cr2O7 tends to be more powerful as an oxidizing agent compared to K2Cr2O4.
What is the dissociation equation for K2Cr2O7?
K2Cr2O7(aq) ------> 2K+(aq)+Cr2O72-(aq)
How many milliliters of 0.325 N HCl are nescessary to reduce 37.6 milliliters of a 0.526 N solution of K2Cr2O7?
First:Comparing the electropotential values of both redox couples (being equal, see below 1. and 2.)AND considering that Cl2 gas might escape the mixture giving toxic fumes,it does NOT look a good titration method.2Cl- --> Cl2,g + 2e- (delta)Eo= -1.36VCr2O72- + 14H+ + 6e- --> 2Cr3+ + 7H2O (delta)Eo= +1.36VSecond:There is no exact answer possible because the titration end point can never be established, you'll never know when the last Cl2 is escaped, and there is no sharp (sudden) change in colour, and the reaction at the endpoint is very slow.Third: The answer is 60.9 mLThe equivalent point (theorethical end point) can be calculated as follows:mEq reductor (HCl) = mEq oxidator (K2Cr2O7)0.325 (mEq/mL HCl) * VHCl (mL) = 0.526 (mEq/mL K2Cr2O7) * 37.6 (mL K2Cr2O7) So VHCl (mL) = 0.526 (mEq/mL K2Cr2O7) * 37.6 (mL K2Cr2O7) / 0.325 (mEq/mL HCl) = 60.854 = 60.9 mL
