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andrezito [222]
3 years ago
12

I need a tutor. Anyone who could be my tutor please message me. I am in 6th grade and rarely have trouble but when I do I would

need help. Whenever I post the tutor would have to help asap. Thanks guys!✌❤️
Chemistry
2 answers:
Aleksandr-060686 [28]3 years ago
4 0
I'd love to be. I have 4 younger siblings and am always helping them! Lol
dlinn [17]3 years ago
3 0
If you are looking for a math tutor google sofatutor.com. They only charge 2 dollars a month and they have helped my by so much! I went from getting Ds to A+.
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IF 500 mL of air contains 2.50x10^22 particles (atoms and molecules). how much particles do you inhale in one day if you breath
vaieri [72.5K]

Answer:

I think it is 0.00192

Explanation:

28.795705

15000

=0.00192

8 0
3 years ago
Explain how impurities are selected for doping in group 14 semiconductors
aleksklad [387]

Impurities selection for doping in group 14 semiconductors is: based on their ability to add more holes and fewer electrons or to add more electrons and reduce the holes.

<h3>Meaning of Semiconductors</h3>

Semiconductors can be defined as any material that has the ability to exhibit some properties of a conductor and some properties of an insulator.

A semiconductor can be used as either a conductor or an insulator when worked upon.

In conclusion, Impurities selection for doping in group 14 semiconductors is: based on their ability to add more holes and fewer electrons or to add more electrons and reduce the holes.

Learn more about semiconductors: brainly.com/question/1918629

#SPJ1

6 0
2 years ago
What chemical reaction involve acid and bases​
svetoff [14.1K]

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Neutralization reaction

5 0
3 years ago
Read 2 more answers
Which would most easily form negative one ions.
irinina [24]

Answer:

Cl

Explanation:

chlorine (2,8,7) is a non metal with highest electronegativity. Hence, it is most likely to form a negative ion with charge −1.

I hope it helps you

8 0
3 years ago
What volume of 0.307 m naoh must be added to 200.0ml of 0.425m acetic acid (ka = 1.75 x 10-5 ) to produce a buffer of ph = 4.250
Blababa [14]

The buffer solution target has a pH value smaller than that of pKw (i.e., pH < 7.) The solution is therefore acidic. It contains significantly more protons \text{H}^{+} than hydroxide ions \text{OH}^{-}. The equilibrium equation shall thus contain protons rather than a combination of water and hydroxide ions as the reacting species.

Assuming that x \; \text{L} of the 0.307 \text{mol} \cdot \text{dm}^{-3} sodium hydroxide solution was added to the acetic acid. Based on previous reasoning, x is sufficiently small that acetic acid was in excess, and no hydroxide ion has yet been produced in the solution. The solution would thus contain 0.2000 \times 0.425 - 0.307 \; x = 0.085 - 0.307 \; x moles of acetic acid and 0.307 \; x moles of acetate ions.

Let \text{HAc} denotes an acetic acid molecule and \text{Ac}^{-} denotes an acetate ion. The RICE table below resembles the hydrolysis equilibrium going on within the buffer solution.

\begin{array}{lccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\end{array}

The buffer shall have a pH of 4.250, meaning that it shall have an equilibrium proton concentration of 10^{4.250}\; \text{mol}\cdot \text{dm}^{-3}. There were no proton in the buffer solution before the hydrolysis of acetic acid. Therefore the table shall have an increase of 10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3} in proton concentration in the third row. Atoms conserve. Thus the concentration increase of protons by 10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3} would correspond to a decrease in acetic acid concentration and an increase in acetate ion concentration by the same amount. That is:

\begin{array}{lcccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\text{C} & - 10^{-4.250} & & +10^{-4.250} & & +10^{-4.250} \\\text{E} & 0.085 - 10^{-4.250} - 0.307 \; x& & 10^{-4.250} & & 10^{-4.250} + 0.307 \; x\end{array}

By definition:

\text{K}_{a} = [\text{H}^{+}] \cdot [\text{Ac}^{-}] / [\text{HAc}]\\\phantom{\text{K}_{a}} = 10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x)

The question states that

\text{K}_{a} = 1.75 \times 10^{-5}

such that

10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x) = 1.75 \times 10^{-5}\\6.16 \times 10^{-5} \; x = 1.48 \times 10^{-6}\\x = 0.0241

Thus it takes 0.0241 \; \text{L} of sodium hydroxide to produce this buffer solution.

6 0
3 years ago
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