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Marianna [84]
2 years ago
7

What are some examples of HIGH resistance?

Chemistry
1 answer:
zloy xaker [14]2 years ago
6 0

Answer:

Its long wire narrow wire more resistance

Explanation:

I took the same k12 test

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State, in terms of electrons, why the radius of Na+ ion in the table salt is smaller than the radius of a Na atom.
Anastaziya [24]

Answer:

• The Na atom is formed when the Na+ ion gains an electron.

• A Na+ ion has 10 electrons and a Na atom has 11 electrons.

• A sodium atom has one more electron shell than a sodium ion.

• A Na+ ion is formed when a Na atom loses an electron

3 0
2 years ago
Which of the following represents an increase in kinetic energy?
Tom [10]

Answer:

C

Explanation:

5 0
3 years ago
1. The picture above shows salt (NaCl).
Gwar [14]

Answer: A Chloride is attracted to the hydrogen atoms in a water molecule.

Explanation: Salt dissociates in water because Chlorine which is an anion gets attracted to the positively charged atoms of Hydrogen in water. This attraction causes Cl to bind with Hydrogen allowing Na to be left behind.

7 0
3 years ago
A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3. Determine the pH of the solution before the addition of any HNO3.
alex41 [277]

Answer:

pH=11.12

Explanation:

Hello,

In this case, ammonia dissociation is:

NH_3(aq)+H_2O(l)\rightleftharpoons NH_4^+(aq)+OH^-(aq)

So the equilibrium expression:

Kb=\frac{[NH_4^+][OH^-]}{[NH_3]}

That in terms of the reaction extent and the initial concentration of ammonia is written as:

1.8x10^{-5}=\frac{x*x}{0.10M-x}

Thus, solving by using solver or quadratic equation we find:

x=0.00133M

Which actually equals the concentration of hydroxyl ion, therefore the pOH is computed:

pOH=-log([OH^-])=-log(0.00133)=2.88

And the pH from the pOH is:

pH=14-pOH=14-2.88\\\\pH=11.12

Best regards.

6 0
3 years ago
Freon-12 CCl2F2, is prepared from CCl4 by reaction with HF. The other product of this reaction is HCl. Outline the steps needed
LekaFEV [45]

Answer:

percent yield = 40.6 %

Explanation:

The question asks to determine the percent yield, which can be defined as:

percent yield = \frac{actual yield}{theoretical yield} *100

where the actual yield is how much product was obtained, in this case 12.5 g of CCl₂F₂, and the theoretical yield is how much product could be obtained with the given reactants theoretically.

So we know already the actual yield, we need to <em>calculate the theoretical yield.</em>

First we need to <em>write the reaction chemical equation</em>:

CCl₄ + HF → CCl₂F₂ + HCl

and <em>balance the equation</em>:

CCl₄ + 2 HF → CCl₂F₂ + 2 HCl

In the equation we can see that <em>for every mol of CCl₄ we should get 1 mol of CCl₂F₂</em> (molar ratio 1:1). So if we <u>calculate the moles of CCl₄</u> in the given 39.2 g of CCl₄ we could know how many moles of CCl₂F₂ (assuming HF is in excess).

  • Moles of CCl₄ = mass CCl₄ / molar mass CCl₄
  • Molar Mass CCl₄ = 12.011 + 4 * 35.45 = 153.811 g/mol
  • Moles of CCl₄ = 39.2 g / 153.811 g/mol = 0.2549 moles

From the molar ratio we know:

Moles of CCl₂F₂ = moles of CCl₄ = 0.2549 moles

Now we need to <u>convert these moles into grams</u> to get the theoretical yield of CCl₂F₂ in grams:

  • mass CCl₂F₂ = moles CCl₂F₂ * molar mass CCl₂F₂
  • Molar Mass CCl₂F₂ = 12.011 + 2 * 35.45 + 2 * 18.998 = 120.907 g/mol
  • Mass CCl₂F₂ = 0.2549 moles * 120.907 g/mol = 30.81 g
  • Theoretical yield CCl₂F₂ = 30.81 g

Percent yield = (12.5 g / 30.81 g) * 100 = 40.6 %

8 0
3 years ago
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