Register
If a solution containing 86.04 g of mercury(II) nitrate is allowed to react completely with a solution containing 15.488 g of sodium sulfide, how many grams of solid precipitate will be formed
121k views
8 votes
If a solution containing 86.04 g of mercury(II) nitrate is allowed to react completely with a solution containing 15.488 g of sodium sulfide, how many grams of solid precipitate will be formed

User Vlad Ankudinov
by
3.9k points

1 Answer

6 votes

Answer:

Top answer

Given: Weight of mercury perchlorate is 100.82 g Weight of sodium sulphide is 17.796 g Solution: The chemical equation for the reaction of mercury(II)perchl ... More

User Nkem
by
3.3k points
Ask a Question
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

3.6m questions

4.5m answers

Other Questions

What is the variable for this number 22.4L?
1. If temperature is increased, the number of collisions per second do what
A helium balloon containing 0.100 mol of gas occupies a volume of 2.4 L at 25 C and 1.0 atm. how many moles have we added if we inflate it to 5.6 L?
Which of the following is NOT a benefit of wildfires? * O Insects moving in after a fire provide more food for birds. O Wildfires help undergrowth in a forest grow denser. O Fires can clean out diseased
Calculate the number of moles of sodium hydroxide in 8g of sodium hydroxide (NaOH). Na=23 O=16 H=1
Link Copied!