The answer is A. Water
Bronsted-Lowry base compounds are those that can accept protons
Bronsted-Lowry Acid Compounds are those that can recieve one
Water / H2O is an Amphoteric compund which mean that its molecul can act as a Base and Acid compound, so the answer is A.
Answer:
The law of conservation of mass states that mass cannot be created or destroyed only transfered. So when a size change takes place the mass is being transfered into something else. During a state change the mass is being transfered into another state of matter. When a substance dissolves into another liquid the mass is still their just into another state.
Explanation:
The number of formula units in 2.50 mol of the compound is 15.1 * 10^23.
The question is unclear whether NaNO2 or NaNO3 is implied. However,in either case, the solution applies equally.
6.02 * 10^23 formula units of the compound are contained in 1 mole
x formula units are contained in 2.5 moles of the compound
x = 6.02 * 10^23 formula units * 2.5 moles/ 1 mole
x = 15.1 * 10^23 formula units of the compound.
Learn more; brainly.com/question/9743981
There are 3 significant figures. Significant numbers are the numbers that build up your total number. 1-9 always count, 0 only counts if it’s after another number. For example: 0,901 has 3 significant numbers as does 0,910. 9,10 also has 3. 0,09 has just 1.
Answer:
Hydrogen H₂ will be the limiting reagent.
The excess reactant that will be left after the reaction is 3.45 moles.
4.3 moles of water can be produced.
Explanation:
The balanced reation is:
2 H₂ + O₂ → 2 H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:
- H₂: 2 moles
- O₂: 1 mole
- H₂O: 2 moles
To determine the limiting reagent, you can use a simple rule of three as follows: if by stoichiometry 1 mole of O₂ reacts with 2 moles of H₂, how much moles of H₂ will be needed if 5.6 moles of O₂ react?

moles of H₂= 11.2 moles
But 11.2 moles of H₂ are not available, 4.3 moles are available. Since you have less moles than you need to react with 5.6 moles of O₂, <u><em>hydrogen H₂ will be the limiting reagent</em></u> and oxygen O₂ will be the excess reagent.
Then you can apply the following rules of three:
- If by reaction stoichiometry 2 moles of H₂ react with 1 mole of O₂, 4.3 moles of H₂ will react with how many moles of O₂?

moles of O₂= 2.15 moles
The excess reactant that will be left after the reaction can be calculated as:
5.6 moles - 2.15 moles= 3.45 moles
<u><em>The excess reactant that will be left after the reaction is 3.45 moles.</em></u>
- If by reaction stoichiometry 2 moles of H₂ produce 2 moles of H₂O, 4.3 moles of H₂ produce how many moles of H₂O?

moles of H₂O= 4.3 moles
<u><em>4.3 moles of water can be produced.</em></u>