Answer:
B. Warm water rises within the pot.?
Explanation
<em>There wasn't enough information given for me to safely determine the correct answer.</em>
0.003 moles of NaOH was used in the titration.
<h3>What is titration?</h3>
The concentration of an identified analyte can be found using a simple laboratory technique called titration. As a standard solution with a given concentration and volume, a reagent known as the titrant or titrator is created.
By using a solution with a known concentration to measure the concentration of an unknown solution, this process is known as titration. To a known volume of the analyte (the unknown solution), the titrant (the known solution) is typically added from a buret until the reaction is finished. To ascertain the unknown concentration of an identifiable analyte, titration, commonly referred to as titrimetry, is a widely used quantitative laboratory analytical technique (Medwick and Kirschner, 2010). Volume measurements are a crucial component of titration
Concentration in mol/dm3 =
Amount of solution mol
= concentration in mol/dm3 × volume in dm3
Amount of sodium hydroxide
= 0.100 × 0.0250
= 0.00250 mol
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Answer:
10 molecules of NH₃.
Explanation:
N₂ + 3H₂ --> 2NH₃
As the N₂ supply is unlimited, what we need to do to solve this problem is <u>convert molecules of H₂ into molecules of NH₃</u>. To do so we use the <em>stoichiometric coefficients</em> of the balanced reaction:
- 15 molecules H₂ *
= 10 molecules NH₃
10 NH₃ molecules could be prepared from 15 molecules of H₂ and unlimited N₂.
2NH4ClO4 --------> N2 + Cl2 + 2O2 + 4H2O
from reaction 2 mol 1 mol
given x mol 0.10 mol
Proportion:
<u>2 mol NH4ClO4 </u>= <u>1 mol Cl2</u>
x mol NH4ClO4 0.10 mol Cl2
x= (2*0.10)/1 = 0.20 mol NH4ClO4
Explanation:tr
a) Molar mass of HF = 20 g/mol
Atomic mass of hydrogen = 1 g/mol
Atomic mass of fluorine = 19 g/mol
Percentage of an element in a compound:
Percentage of fluorine:
Percentage of hydrogen:
b) Mass of hydrogen in 50 grams of HF sample.
Moles of HF = 
1 mole of HF has 1 mole of hydrogen atom.
Then 2.5 moles of HF will have:
of hydrogen atom.
Mass of 2.5 moles of hydrogen atom:
1 g/mol × 2.5 mol = 2.5 g
2.5 grams of hydrogen would be present in a 50 g sample of this compound.
c) As we solved in part (a) that HF molecules has 5% of hydrogen by mass.
Then mass of hydrogen in 50 grams of HF compound we will have :
5% of 50 grams of HF = 
