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alexdok [17]
3 years ago
6

why mixture of stone pebbles And water is concerned considered suspension you are provided with a mixture of sand salt and camph

or how would you separate these particles explain in detail​
Chemistry
1 answer:
baherus [9]3 years ago
6 0

Answer:

camphor sublimates salt is soluble in water while sand does not sublime and does not dissolve in water you first heat the mixture in a beaker covered with a watch glass camphor will then accumulate on the watch glass then you dissolve the remaining mixture of sand and salt salt will dissolve forming a salt solution then you filter using a filter paper and a beaker the residue on the filter paper is sand while the filtrate is salt solution you then heat the salt solution so that it can evaporate leaving salt particles thus you will have obtained salt sand and camphor

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If the molar heat of combustion of liquid benzene at constant volume and 300k is -3272KJ. Calculate the heat of combustion at co
vladimir2022 [97]

Answer:

The heat at constant pressure is -3,275.7413 kJ

Explanation:

The combustion equation is 2C₆H₆ (l) + 15O₂ (g)  → 12CO₂ (g) + 6H₂O (l)

\Delta n_g = (12 - 15)/2 = -3/2

We have;

\Delta H = \Delta U + \Delta n_g\cdot R\cdot T

Where R and T are constant, and ΔU is given we can write the relationship as follows;

H = U + \Delta n_g\cdot R\cdot T

Where;

H = The heat at constant pressure

U = The heat at constant volume = -3,272 kJ

\Delta n_g = The change in the number of gas molecules per mole

R = The universal gas constant = 8.314 J/(mol·K)

T = The temperature = 300 K

Therefore, we get;

H = -3,272 kJ + (-3/2) mol ×8.314 J/(mol·K) ×300 K) × 1 kJ/(1000 J) = -3,275.7413 kJ

The heat at constant pressure, H = -3,275.7413 kJ.

4 0
2 years ago
a prescriber has ordered childrens motrin 400 mg po q6h for a child who weighs 60 kg. how many mg/kg is the child receiving
DedPeter [7]

Answer:

6.67 mg/kg     per dose   ( 26.67 mg/kg   per day)

Explanation:

400 mg / 60 kg = 6 2/3  mg/kg  per dose

   per <em>DAY</em>  is  four times this number

5 0
2 years ago
Identify the Lewis acid and Lewis base from among the reactants in each of the following equations. Match the words in the left
goldfiish [28.3K]

Answer:

1. Lewis acid: F. Fe₃⁺, Lewis base: B. CN⁻

2. Lewis acid: A. AlCl₃, Lewis base: D. Cl⁻

3. Lewis acid: C. AlBr₃, Lewis base: E. NH₃

Hope this helps.

8 0
3 years ago
12. Photosynthesis builds sugars out of small molecules, making it an
JulsSmile [24]

Answer:

Condensation reaction/ direct synthesis reaction

Explanation:

Combines simple molecules to form complex molecules producing water

8 0
3 years ago
Calculate the ph of a 0.20 m solution of kcn at 25.0 ∘c. express the ph numerically using two decimal places
storchak [24]
The pH of the solution is basically the negative logarithm of the concentration of hydrogen ions or H+. In equation form, pH = -log[H+]. It could also be in terms of the concentration of hydroxide ions or OH- as pOH, where pOH = -log[OH-]. The sum of pH and pOH is 14. These are the important equations to know when it comes to equilibrium pH problems.

KCN is a basic salt coming from the reaction of a weak acid, HCN, and a strong base, KOH. In the hydrolysis of KCN, only the strong conjugate base (SCB) is involved. Since HCN is the weak acid, the SCB is CN-. The reaction would be

CN- + H2O ⇔ HCN + OH-

The important data is the equilibrium constant of acidity of the weak acid. Ka for HCN is 6.2×10^-10. Then, let's do the ICE(Initial-Change-Equilibrium) analysis.

          CN-    +    H2O    ⇔    HCN +    OH-

I      0.2 m             ∞                   0             0
C      -x                 ∞                   +x        +x
-----------------------------------------------------------
E      0.2-x                               +x         +x

The value x denotes the number of moles CN- reacted. There is no value for H2O because the solution is dilute such that H2O>>>CN-. Then, we apply the ratio:

K_{H} = \frac{ K_{W} }{ K_{A} } = \frac{[HCN][OH-]}{[CN-]}

where K,H is the equilibrium constant of hydrolysis and Kw is equilibrium constant for water solvation which is equal to 1×10^-14. Therefore,

K_{H} = \frac{ 1x10^-14}{ 6.2x10^-10 } = \frac{[X][X]}{[0.2-X]}

x = 0.001788 m
Since the value of OH- is also x, then OH-=0.001788 m. Consequently,

pOH = -log(0.001788) = 2.75
pH = 14 - pOH = 14 - 2.75
pH = 11.25

4 0
3 years ago
Read 2 more answers
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