The concentration of the hydroxide ions after 50 ml of 0.250M NaOH is added to 120ml of 0.200M Na2SO4 is 7.35 x 10^-2 M.
What is meant by concentration?
Concentration is the total amount of solute present in the given volume of solution. this is expressed in terms of molarity, molality, mole fraction, normality etc. The term concentration mostly refers to the solvents and solutes present in the solution.
Concentration of hydroxide ions can be calculated by,
M (OH^-) = V (NaOH) x M (NaOH) / V (total) = 50ml x 0.250M / 50ml + 120ml = 0.0735M = 7.35 x 10^-2 M.
where M (OH^-) = concentration of hydroxide ions, V(NaOH) = volume of NaOH, M(NaOH) = concentration of NaOH.
Therefore, the concentration of the hydroxide ions after 50 ml of 0.250M NaOH is added to 120ml of 0.200M Na2SO4 is 7.35 x 10^-2 M.
To learn more about concentration click on the given link brainly.com/question/17206790
#SPJ4
Oxygen because it is on the left of the periodic table so it has a strong pull.
<span>
<span>
</span></span>Volume = 4/3 * PI * r^3
1.59 x 10^24 copper atoms = 2.64 moles of copper
Atomic Mass of copper = 63.55
2.64 * 63.55 = 167.77 grams of copper
Volume of Copper = Mass / Density
Volume of Copper = 167.77 grams / 8.96
Volume of Copper = <span>
<span>
18.72</span></span> cubic centimeters
r^3 = Volume / (4/3 * PI)
r^3 = 18.72 / 4.188
r^3 =
<span>
<span>
<span>
4.47
radius = </span></span></span><span><span><span>1.647</span> centimeters
</span></span>
Answer:
her results has good precision
Explanation:
accuracy and precision
CₓHₐ + (x+a/4)O₂ = xCO₂ + a/2H₂O
n(CO₂)=4.69 mg/44.01 mg/mmol = 0.1066 mmol
n(H₂O)=1.10 mg/18.02 mg/mmol = 0.06104 mmol
C : H = 0.1066 : 0.06104*2 = 0.1066 : 0.12208 = 1 : 1.1453 = 7 : 8
C₇H₈ + 9O₂ = 7CO₂ + 4H₂O
C₇H₈
