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max2010maxim [7]
3 years ago
15

Which choice can be classified as a pure substance?

Chemistry
2 answers:
Lyrx [107]3 years ago
8 0

Answer:

Salt

Explanation:

Pure substances are those that contain atoms or molecules of the same type. For eg: elements are pure (gold, hydrogen etcl.) and compounds (such as water, sodium chloride, etc.), etc.

Rest of choice are impure because they contain a mixture of different things, like pie has eggs, or all sorts of other molecules, salt water has water and salt molecules, and iced tea has water and other molecules associated with tea.

pentagon [3]3 years ago
3 0

Answer:

Salt

Explanation:

It is because it has a defined composition.

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Question 231.15 pts A 0.200 M solution of a weak monoprotic acid HA is found to have a pH of 3.00 at room temperature. What is t
MariettaO [177]

Answer:

The correct answer is: Ka= 5.0 x 10⁻⁶

Explanation:

The ionization of a weak monoprotic acid HA is given by the following equilibrium: HA ⇄ H⁺ + A⁻. At the beginning (t= 0) we have 0.200 M of HA. Then, a certain amount (x) is dissociated into H⁺ and A⁻, as is detailed in the following table:

               HA               ⇄        H⁺        +          A⁻

t= 0      0.200 M                     0                     0

t              -x                             x                       x

t= eq      0.200M -x               x                       x

At equilibrium, we have the following ionization constant expression (Ka):

Ka= \frac{ [H^{+}]  [A^{-} ]}{ [HA]}

Ka= \frac{x x}{0.200 M -x}

Ka= \frac{x^{2} }{0.200 M - x}

From the definition of pH, we know that:

pH= - log  [H⁺]

In this case, [H⁺]= x, so:

pH= -log x

3.0= -log x

⇒x = 10⁻³

We introduce the value of x (10⁻³) in the previous expression and then we can calculate the ionization constant Ka as follows:

Ka= \frac{(10^{-3})^{2}  }{0.200 - (10^{-3}) }= \frac{10^{-6} }{0.199}= 5.025 x 10⁻⁶= 5.0 x 10⁻⁶

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What is the acceleration of an object that goes from 3 m/s to 5 m/s in 8 s?
nikdorinn [45]

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Explanation:

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Explanation:

The  scientists used the term electromagnetic spectrum to describe the entire range of light that exists in the universe. From gamma rays to radio waves, most of the light  present in the universe invisible to us.

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What is the electric field (in N/C) at a point 5.0 cm from the negative charge and along the line between the two charges?
Natali [406]

Answer: E = 2.455 x 10^5 N/C

Explanation:

q1 = 1.2x10^-7C

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since the location is (27 - 5)cm from q1

hence electric field, E1 = k*q1/r²

E1= (9x10^9 x 1.2x10^-7)/(0.22)² = 22314.05 N/C

for q2:

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E2 at 5cm

E2 = (9x10^9 x 6.2x10^-8)/(0.05)² = 223200 N/C

Hence, the total electric field at 5cm position is

E = E1 + E2

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E = 2.455 x 10^5 N/C

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