Suspensions
Explanation:
Suspensions are heterogeneous mixtures that contains large particles that can settle out or be filtered.
- Suspensions are mixtures of small insoluble particles of a solid in a liquid or gas.
- Examples are:
- powdered chalk in water
- muddy water
- harmattan
The particles in suspension can settle on standing
Learn more:
Suspension brainly.com/question/1557970
heterogeneous mixture brainly.com/question/1446244
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The methyl orange would be the best indicator for titrating the weak base sodium bicarbonate using HCl titrant
When weak base is titrated with strong acid then , then solution is slightly acidic at end point . If weak acid is titrated with strong base then , the solution is slightly basic because salt formed will hydrolyzed to a certain extent .
In acid base titration at the end point the amount of the acid becomes chemically equivalent to the amount of the base present .The methyl orange would be the best indicator for titrating the weak base with strong acid .
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Many elements show very strong similarities to each other.<span>For example, lithium (Li), sodium (Na), and potassium (K) are all soft, very reactive metals.
</span>
Answer:
8.934 g
Step-by-step explanation:
We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 192.12 44.01
H₃C₆H₅O₇ + 3NaHCO₃ ⟶ Na₃C₆H₅O₇ + 3H₂O + 3CO₂
m/g: 13.00
For ease of writing, let's write H₃C₆H₅O₇ as H₃Cit.
(a) Calculate the <em>moles of H₃Cit
</em>
n = 13.00 g × (1 mol H₃Cit /192.12 g H₃Cit)
n = 0.067 67 mol H₃Cit
(b) Calculate the <em>moles of CO₂
</em>
The molar ratio is (3 mol CO₂/1 mol H₃Cit)
n = 0.067 67 mol H₃Cit × (3 mol CO₂/1 mol H₃Cit)
n = 0.2030 mol CO₂
(c) Calculate the <em>mass of CO₂
</em>
m = 0.2030 mol CO₂ × (44.01 g CO₂/1 mol CO₂)
m = 8.934 g CO₂
Answer: 20 gallons of 16% and 8 gallons of 23% solution is usedfor 28 gallons of the 18 % solution.
Explanation:
According to the dilution law,
where,
= concentration of ist disinfectant solution = 16 %
= volume of pure acid solution = x gallons
= concentration of another disinfectant solution= 23%
= volume of another acid solution= (28-x) gallons
= concentration of resulting disinfectant solution = 18 %
= volume of resulting acid solution = 28 gallons
![x=20galllons](https://tex.z-dn.net/?f=x%3D20galllons)
(28-x) = (28-20) = 8 gallons
Thus 20 gallons of 16% and 8 gallons of 23% solution is usedfor 28 gallons of the 18 % solution.