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kolbaska11 [484]
3 years ago
14

Mixtures that appear to be homogeneous but actually contain particles that are intermediate in size between solutions and suspen

sions arescience
Chemistry
1 answer:
vovikov84 [41]3 years ago
8 0

Colloids

Explanation:

Mixtures appearing to be homogeneous but appearing to be particles that are intermediate in size between solutions and suspensions are called colloids.

They are homogeneous mixtures of two phases; the disperse phase and the dispersion medium.

  • The disperse phase is like the solute of a solution.
  • The dispersion medium is similar to the solvent.

Examples of colloid include, starch, foams, gels e.t.c

Learn more:

Mixtures brainly.com/question/1446244

#learnwithBrainly

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Since the number of atoms in a substance is so large , a unit to count them was created. This unit is the number of atoms in 12
larisa86 [58]

The number of atoms in one mole of any substance is measured by Avogadro's number. The value of Avogadro's number is 6.023 x 10 ^23. It is named after scientist Avogadro who proposed this number. 12 grams of carbon-12 represents 1 mole of carbon-12. For this reason, the number of atoms present in 1 mole of any substance is 6.023 x 10 ^23. Therefore, the number of atoms present in 1 mole carbon-12 is 6.023 x 10^23.


(Answer) This unit is the number of atoms in 12 grams of carbon-12 and known as Avogadro's number.

7 0
3 years ago
What would indicate that a physical change takes place when cooper is drawn into wire
amid [387]

Answer:

When a copper is drawn into wire the only change that occurs is change in its shape and size no change will take place into its composition that is the wires are still possessing the properties of copper metal. Thus, a physical change takes place when copper is drawn into wire.

4 0
2 years ago
Which of these substances acts as a buffer? <br> a. blood <br> b. lemon juice <br> c. curd
il63 [147K]
<span>The correct answer is( A) blood.

when the buffer solution its PH value changes very little when a small amount 

of strong acid or base is added to it, and here the bicarbonate buffering system is used to regular the PH of the blood that keeping the PH at nearly constant value by maintaining the original acidity or basicity of the solution.</span>
7 0
3 years ago
3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 15
shutvik [7]

Answer:

C₅H₁₀O₅

Explanation:

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

\text{Mass of C} = \text{5.13 g CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{1.400 g C}

(b) Mass of H

\text{Mass of H} = \text{2.10 g H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.2349 g H}

(c) Mass of O

Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

\text{Moles of C = 1400  mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

\text{C: } \dfrac{116.6}{116.6}= 1\\\\\text{H: } \dfrac{233.1}{116.6} = 1.999\\\\\text{O: } \dfrac{116}{116.6} = 1.00

4. Round the ratios to the nearest integer

C:H:O = 1:2:1

5. Write the empirical formula

The empirical formula is CH₂O.

6. Calculate the molecular formula.

EF Mass = (12.01 + 2.016  + 16.00) u  = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{150 u}}{\text{30.03 u}} = 5.00  \approx 5

MF = (CH₂O)₅ = C₅H₁₀O₅

The molecular formula of X is C₅H₁₀O₅.

8 0
3 years ago
The Michaelis constant for pancreatic lipase is 5 mM. At 60 C, lipase is subject to deactivation with a half-life of 8 min. Fat
Dmitriy789 [7]

Explanation:

According to the given data, we will calculate the following.

 Half life of lipase t_{1/2} = 8 min x 60 s/min

                                       = 480 s

Rate constant for first order reaction is as follows.

         k_{d} = \frac{0.6932}{480}

                        = 1.44 \times 10^{-3}s^{-1}&#10;

Initial fat concentration S_{o} = 45 mol/m^{3}

                                                = 45 mmol/L

Rate of hydrolysis V_m_{o} = 0.07 mmol/L/s

Conversion X = 0.80

Final concentration (S) = S_{o} (1 - X)

                                      = 45 (1 - 0.80)

                                      = 9 mol/m^{3}

or,                                  = 9 mmol/L

It is given that K_{m} = 5mmol/L

Therefore, time taken will be calculated as follows.

                    t = -\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]

Now, putting the given values into the above formula as follows.

            t = -\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]  

             = -\frac{1}{1.44 \times 10^{-3}s^{-1}}ln[1 - \frac{1.44 \times 10^{-3}s^{-1}}{0.07 mmol/L/s&#10;}{K_{M} ln (\frac{45 mmol/L&#10;}{9 mmol/L&#10;}) + (45 mmol/L - 9 mmol/L&#10;)]

              = 1642.83 s \times \frac{1 min}{60 sec}

              = 27.38 min

Therefore, we can conclude that time taken by the enzyme to hydrolyse 80% of the fat present is 27.38 min.

6 0
3 years ago
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