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lbvjy [14]
3 years ago
6

State property of the liquids that caused them to separate into layers

Chemistry
1 answer:
nydimaria [60]3 years ago
6 0

Explanation:

Decantation is a process for the separation of mixtures of immiscible liquids or of a liquid and a solid mixture such as a suspension.

Source: https://en.m.wikipedia.org/wiki/Decantation

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What are the differences between sand and potting soil? Are they both mixtures? How do you know?
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What class is this for because it depends
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3 years ago
What is the molecular formula of a compound with the empirical formula
ipn [44]

Answer:  C_2H_2O_2

Explanation:

Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.  

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.

The empirical formula is CHO

The empirical weight of CHO = 1(12)+1(1)+1(16)= 29 g.

The molecular weight = 60 g/mole

Now we have to calculate the molecular formula:

n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{60}{29}=2

The molecular formula will be=2\times CHO=C_2H_2O_2

Thus molecular formula will be C_2H_2O_2

5 0
3 years ago
How many grams of CO2 could be formed with 2.09 x 1023 atoms of O?
choli [55]

Answer:

r u in high school this is hard

Explanation:

6 0
3 years ago
A 200.0 mL solution of 0.40 M ammonium chloride was titrated with 0.80 M sodium hydroxide. What was the pH of the solution after
choli [55]

Answer:

9.25

Explanation:

Let first find the moles of NH_4Cl and NaOH

number of moles of NH_4Cl = 0.40  mol/L × 200 ×  10⁻³L

= 0.08 mole

number of moles of NaOH = 0.80  mol/L × 50 ×  10⁻³L

= 0.04 mole

The equation for the reaction is expressed as:

NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}

The ICE Table is shown below as follows:

                            NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}

Initial (M)              0.08            0.04                            0

Change (M)         - 0.04          -0.04                          + 0.04

Equilibrium (M)      0.04             0                                0.04

K_a*K_b = 10^{-14} \ at \ 25^0C

K_a = \frac{10^{-14}}{K_b}

K_a = \frac{10^{-14}}{1.76*10^{-5}}

K_a= 5.68*10^{10}

pK_a = - log \ (K_a)

pK_a = - log \ (5.68*10^{-10})

pK_a = 9.25

pH = pKa + \ log (\frac{HB}{HA} )   for buffer solutions

pH = pKa + \ log (\frac{moles \ of \ base }{ moles\ of \ acid} ) since they are in the same solution

pH = 9.25 + \ log (\frac{0.04 }{ 0.04} )

pH = 9.25

8 0
3 years ago
Students in a chemistry class are drawing molecular models based on molecular formulas they are given by the teacher.
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Sorry I need this marks
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