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Paraphin [41]
3 years ago
10

Which of the following are macro molecular colloids ?

Chemistry
1 answer:
fredd [130]3 years ago
4 0
The following of the macro molecular collide is all of the above
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he decomposition of acetaldehyde, CH3CHO, was determined to be a second order reaction with a rate constant of 0.0771 M-1 s-1. I
nataly862011 [7]

Answer:

The concentration is [-1 + sqrt(1+0.11t)]/0.1542 M

Explanation:

Let the concentration of CH3CHO after selected reaction times be y

Rate = Ky^2 = change in concentration of CH3CHO/time

K = 0.0771 M^-1 s^-1

Change in concentration of CH3CHO = 0.358 - y

0.0771y^2 = 0.358-y/t

0.0771ty^2 = 0.358 - y

0.0771ty^2 + y - 0.358 = 0

The value of y must be positive and is obtained in terms of t using the quadratic formula

y = [-1 + sqrt(1^2 -4(0.0771t)(-0.358)]/2(0.0771) = [-1 + sqrt(1+0.11t)]/0.1542 M

6 0
3 years ago
Which element will have the greatest attraction for the calcium electrons?
ratelena [41]
Its would be fluorine as its the most electronegative 

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6 0
3 years ago
PLZ I NEED THIS NOW A chemical change can also be called a chemical _________.
posledela
A chemical change can also be called a chemical reaction
6 0
3 years ago
Read 2 more answers
Rank these elements according to FIRST ionization energy, from highest to lowest:
Rzqust [24]
The trend of ionization energy in the periodic table is decreasing from right to left and from top to bottom. In this case, we are given with elements <span>K, Ca, Ge, Se, Br, Kr and see the periodic table to check the trend. The answer from highest to lowest Kr, Br, Se, Ge, Ca, and K</span>
3 0
3 years ago
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What volume of water vapor would be produced from the combustion of 815.74 grams of propane (C3H8) with 1,006.29 grams of oxygen
d1i1m1o1n [39]

3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane (C_3H_8) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Stoichiometric calculations:

C_3H_8(g) + 5 O_2(g)→ 3 CO_2(g) + 4 H_2O(g)

From the equation of the reaction, the mole ratio of propane to oxygen is 1:5.

Mole of 815.74 grams of propane = \frac{ 815.74}{44.1 }

Mole of 815.74 grams of propane = 18.49750567 moles

Mole of  1,006.29 grams of oxygen =\frac{ 1,006.29}{32 }

Mole of  1,006.29 grams of oxygen = 31.4465625 moles

Going by the mole ratio, it appears propane is limiting while oxygen is in excess.

From the equation, 1 mole of propane produces 4 moles of water vapour. Thus, the equivalent mole of water vapour will be:

18.49750567 moles x 4 = 73.99 moles.

Using the ideal gas equation:

PV = nRT

v = (73.99  x 0.08206 x 623) ÷ 0.96

v =  3940.2

Hence, 3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane (C_3H_8) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.

Learn more about the ideal gas here:

brainly.com/question/27691721

#SPJ1

7 0
2 years ago
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