<h3>What is pH?</h3>
This is defined as the degree of acidity or alkalinity of a substance and pH from 1.0 - 6.9 is acidic, 7 is neutral while 7.1 - 14.0 is basic.
The aqueous solutions and their appropriate pH can be seen above which is the right choice.
Read more about pH here brainly.com/question/22390063
Answer:
The answer to your question is 0.67 moles of Glucose
Explanation:
Data
120 g of Glucose
convert to moles
Process
1.- Calculate the molar mass of Glucose
C₆H₁₂O₆ = (12 x 6) + (1 x 12) + (16 x 6)
= 72 + 12 + 96
= 180 g
2.- Remember that the molar mass of each molecule equals 1 mol. Use the rule of three to find the answer.
180 g ------------------- 1 mol
120 g ------------------ x
x = (120 x 1)/180
x = 120/180
x = 0.67 moles
3.- Conclusion
120 g of Glucose are equal to 0.67 moles.
Answer:
C
Explanation:
It can already be seperated because the mixture hasn't disolved into the water yet.
Answer:
7.37 mL of KOH
Explanation:
So here we have the following chemical formula ( already balanced ), as HNO3 reacts with KOH to form the products KNO3 and H2O. As you can tell, this is a double replacement reaction,
HNO3 + KOH → KNO3 + H2O
Step 1 : The moles of HNO3 here can be calculated through the given molar mass ( 0.140 M HNO3 ) and the mL of this nitric acid. Of course the molar mass is given by mol / L, so we would have to convert mL to L.
Mol of NHO3 = 0.140 M 
30 / 1000 L = 0.140 M 0.03 L = .0042 mol
Step 2 : We can now convert the moles of HNO3 to moles of KOH through dimensional analysis,
0.0042 mol HNO2 ( 1 mol KOH / 1 mol HNO2 ) = 0.0042 mol KOH
From the formula we can see that there is 1 mole of KOH present per 1 moles of HNO2, in a 1 : 1 ratio. As expected the number of moles of each should be the same,
Step 3 : Now we can calculate the volume of KOH knowing it's moles, and molar mass ( 0.570 M ).
Volume of KOH = 0.0042 mol ( 1 L / 0.570 mol ) ( 1000 mL / 1 L ) = 7.37 mL of KOH
Answer:
The pressure of the gas increased by a factor of 4
