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sergeinik [125]
3 years ago
8

Describe 3 methods to increase the rate of dissolution for sugar in a cup of tea

Chemistry
1 answer:
Xelga [282]3 years ago
8 0
1. Stirring. This will agitate the sugar and encourage it to dissolve. Stirring creates friction and motion, which in turn will create heat that pulls the sugar particles apart as they are given energy. 

2. Applying heat. As mentioned in the previous method. The application of heat gives extra energy to the sugar molecules which encourages it to pull apart and dissolve. 

3. Increase the surface area. Where more sugar particles are fully submerged in the tea or being surrounded by the tea, the rate of dissolution will increase. 
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Chlorine has two naturally stable isotopes: 35Cl (34.968853 amu) and 37Cl (36.965903 amu). The natural abundance of each isotope
allochka39001 [22]

Answer:

35Cl ⇒ 34.968853 amu  has an abundance of 30.05%

Explanation:

The molar mass of chlorine, (which is the average of all its naturally stable isotope masses), is 36.36575 amu.

There are 2 naturally stable isotopes, this means together they have an abundance of 100%

The isotopes are:

35Cl ⇒ 34.968853 amu  has an abundance of X %

37Cl ⇒ 36.965903 amu  has an abundance of Y %

X + Y = 100%   OR X = 100% - Y

36.36575 = 34.968853X + 36.965903Y  

36.36575 = 34.968853(1-Y) + 36.936.96590365903Y

36.36575 = 34.968853 -34.968853Y + 36.965903Y

1.396897 = 1.99705Y

Y = 0.699 = 69.95%

X = 100-69.9 = 30.05%

To control, we can plug in the following equation:

34.968853 * 0.3005 + 36.965903 * 0.6995 = 36.3658

This means

37Cl ⇒ 36.965903 amu  has an abundance of 69.95 %

35Cl ⇒ 34.968853 amu  has an abundance of 30.05%

7 0
3 years ago
Consider the reaction 2NO(g) 1 O2(g) ¡ 2NO2(g) Suppose that at a particular moment during the reaction nitric oxide (NO) is reac
GalinKa [24]
<h2>a) The rate at which NO_2 is formed is 0.066 M/s</h2><h2>b) The rate at which molecular oxygen O_2 is reacting is 0.033 M/s</h2>

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

2NO(g)+O_2(g)\rightarrow 2NO_2(g)

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate in terms of disappearance of NO = -\frac{1d[NO]}{2dt} = 0.066 M/s

Rate in terms of disappearance of O_2 = -\frac{1d[O_2]}{dt}

Rate in terms of appearance of NO_2= \frac{1d[NO_2]}{2dt}

1. The rate of formation of NO_2

-\frac{d[NO_2]}{2dt}=\frac{1d[NO]}{2dt}

\frac{1d[NO_2]}{dt}=\frac{2}{2}\times 0.066M/s=0.066M/s

2. The rate of disappearance of O_2

-\frac{1d[O_2]}{dt}=\frac{d[NO]}{2dt}

-\frac{1d[O_2]}{dt}=\frac{1}{2}\times 0.066M/s=0.033M/s

Learn more about rate law

brainly.com/question/13019661

https://brainly.in/question/1297322

7 0
3 years ago
What is the mass of 100 mL of corn oil?
prisoha [69]

Answer:

m=(0.92g/ml)(100ml)

=92g

hope this helps

5 0
3 years ago
Help needed ASAP, I will mark your answer as brainliest.
beks73 [17]

Answer:

D

Explanation:

^

6 0
2 years ago
Read 2 more answers
What is the pressure of a sample of gas at a volume of .335 L if it occupies 1700 mL at 850 mm Hg?
dangina [55]

Answer:

Pressure = 4313.43mmHg

Explanation:

P1 = ?

V1 = 0.335L

V2 = 1700mL =1700*10^-3L = 1.7L

P2 = 850mmhg

From Boyle's law, the volume of a fixed mass of gas is inversely proportional to its pressure provided that temperature remains constant.

P = k / v

K = pv. P1V1 = P2V2 = P3V3 =........=PnVn

P1V1 = P2V2

Solve for P1,

P1 = (P2*V2) / V1

P1 = (850 * 1.7) / 0.335

P1 = 4313.43mmHg

The pressure of the gas was 4313.43mmHg

7 0
3 years ago
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