The compound RX3 decomposes according to the equation

1 answer:

8 0

If you are given the compound RX3 and it decomposes according to the equation, 3RX3 → R + R2X3 + 3X2. The average rate at which RX3 is disappearing over the entire experiment is -0.0484 mol per liter per second. I used excel to solve for the equation given the data above.

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Using the periodic table, find the neutral atom that has the same electron configuration as 1s²2s22p.

aleksley [76]

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2 years ago

Hydrogen reacts with 0.771 g of carbon to form 0.90 g of a compound. What is the mass percent of hydrogen in the compound

morpeh [17]

Answer: 14.3%

Explanation: In order to find the mass percent of hydrogen in this compound, you must determine how many grams of hydrogen you'd get in 100 g of compound.

In your case, you know that an unknown mass of hydrogen reacts with 0.771 g of carbon to form 0.90 g of hydrocarbon, which is a compound that contains only carbon and hydrogen.

Use the total mass of the hydrocarbon to determine how many grams of hydrogen reacted with the carbon.

Now, if 0.90 g of this compound contain 0.129 g of hydrogen, it follows that 100 g of this compound will contain.

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3 years ago

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Suppose a student titrates a 10.00-ml aliquot of saturated ca(oh)2 solution to the equivalence point with 16.08 ml of 0.0199 m h

Alborosie

Following chemical reaction is involved upon titration of Ca(OH)2 with HCl,
Ca(OH)2 + 2HCl ↔ CaCL2 + 2H2O

Above is an example of acid-base titration to generate salt and water. Here, H+ ions of acid (HCl) combines with OH- (ions) of base [Ca(OH)2] to generated H2O

Given,
concentration of HCl = 0.0199 M
Total volume of HCl consumed during titration = 16.08 mL = 16.08 X 10^(-3) L

∴, number of moles of H+ consumed = Molarity X Vol. of HCl (in L)
= 0.0199 X 16.08 X 10^(-3)
= 3.1999 X 10^-4 mol
Thus, total number of moles of [OH-] ions present initial = 3.1999 X 10-4 mol
So, initial conc. [OH-] ion = $\frac{number of moles of [OH-]}{volume of solution (L)}$ = $\frac{3.1999 X 10^(^-^4^)}{10 X 10^(^-^3^)}$ = 0.03199 M