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Molodets [167]
3 years ago
10

Rain showers have continued over a particular area for several days which of these weather factors is most likely responsible A.

a high-pressure system B.a stationary front C.low humidity D.a warm front
Chemistry
2 answers:
Ede4ka [16]3 years ago
7 0
D. a warm front causes rain showers to occur over a particular area for several days
yanalaym [24]3 years ago
7 0

The answer is B. a stationary Front.

Hope this helps!

*Explanation*

A red line is responsible for keeping the weather good. A blue line is not responsible.

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What is the vapor pressure (in kPa) of ethanol, CH3CH2OH, over a solution which is composed of 18.00 mL of ethanol and 12.55 g o
ladessa [460]

Answer:

The vapor pressure of ethanol in the solution is 10,27 kPa

Explanation:

To obtain the vapor pressure of a solution it is necessary to use Raoult's law:

P_{solution} = X{solvent}P_{0solvent} <em>(1)</em>

The moles of ethanol are:

18,00mL×\frac{0,789g}{1mL}×\frac{1 mol}{46,07g} = 0,3083 mol Ethanol.

Moles of benzoic acid:

12,55 g×\frac{1mol}{122,12g} = 0,1028 mol benzoic acid.

Thus, mole fraction of solvent, X, is:

\frac{0,3083 mol}{0,3083mol+0,1028mol} =<em> 0,7499</em>

Replacing this value in (1):

P_{solution} = 0,7499*13,693kPa = <em>10,27 kPa</em>

<em></em>

I hope it helps!

7 0
3 years ago
g Suppose you are titrating vinegar, which is an acetic acid solution of unknown concentration, with a sodium hydroxide solution
Elza [17]

Answer: The molar concentration of acetic acid in the vinegar is 0.539 M.

Explanation:

The formula used is:

M_1V_1=M_2V_2

where,

M_1 and V_1 are the concentration and volume of base.

M_2 and V_2 are the concentration and volume of an acid.

Given:

Molar concentration of NaOH = 0.1798 M

Volume of NaOH = 30.01 mL

Volume of acetic acid = 10.0 mL

Now putting all the given values in the above formula, we get:

M_1V_1=M_2V_2\\\\0.1798M\times 30.01mL=M_2\times 10.0mL\\\\M_2=0.539M

Thu, the molar concentration of acetic acid in the vinegar is 0.539 M.

4 0
3 years ago
How did Rutherford's experimental evidence lead to the development of a new atomic model?
Basile [38]
In his Gold Foil experiment, few particles were deflected strongly, where as he thought all particles will go straight through the foil (some did though, which are called electrons.) J. J. Thompson, Rutherford's former teacher, proposed that if there are negative charge particles (he named them electrons), there must also be a positive charge particles; Rutherford proved his theory right, and he called the positive charge protons. He also found that inside the atom, there must be a positive charge that is clustered in a tiny region in its center, which is called the nucleus.
3 0
4 years ago
A binary compound created by reaction of bismuth and an unknown element E contains 52.07% Bi and 47.93% E by mass. If the formul
yulyashka [42]

Answer:

Atomic mass of E is 128.24

Explanation:

  • The percentage composition by mass of an element in a compound is given by dividing the mass of the element by the total mass of the compound and expressing it as a percentage.
  • In this case; the compound Bi₂E₃

Percentage composition of bismuth = 52.07%

Percentage composition of E = 47.93%

Mass Bismuth in the compound is (2×208.9804) = 417.96 g

Therefore,

To calculate the atomic mass of E

52.07% = 417.96 g

47.93% = ?

            = (47.93 × 417.96 ) ÷ 52.07 %

            = 384.729

         E₃ = 384.729

Therefore; E = 384.729 ÷ 3

                     = 128.24  

The atomic mass of E is 128.24

6 0
3 years ago
Which of the following is true?
Umnica [9.8K]

Answer:

The answer should be D

Explanation:

because turning 1 mole of propane to grams means its still one mole of propane just in a different unit.

8 0
2 years ago
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