All categories

3 years ago

Help Asap!! I will try to mark Brainliest!✨

Chemistry

2 answers:

lbvjy [14]3 years ago

7 0

If I remember correctly from my science class, it is Protons and Neutrons.

Hope this helps, if not, comment below please!!!

mezya [45]3 years ago

4 0

Protons and electrons,
I apologize if I’m wrong

You might be interested in

What are the observed effects of sleep deprivation on leptin and ghrelin?

Afina-wow [57]

Answer:

Lack of sleep increases hunger hormone ghrelin and decreases the satiety hormone leptin which may help explain the association between inadequate sleep and overweight.

Explanation:

6 0

4 years ago

PH CHEM, PLEASE HELP QUICK! NO LINKS/VIRUSES PLEASE!

Contact [7]

Answer:

The pH of the solution is 11.48.

Explanation:

The reaction between NaOH and HCl is:

NaOH + HCl → H₂O + NaCl

From the reaction of 3.60x10⁻³ moles of NaOH and 5.95x10⁻⁴ moles of HCl we have that all the HCl will react and some of NaOH will be leftover:

$n_{NaOH}} = n_{i_{NaOH}} - n_{HCl} = 3.60 \cdot 10^{-3} moles - 5.95 \cdot 10^{-4} moles = 3.01 \cdot 10^{-3} moles$

Now, we need to find the concentration of the OH⁻ ions.

$[OH^{-}] = \frac{n_{NaOH}}{V}$

Where V is the volume of the solution = 1.00 L

$[OH^{-}] = \frac{n_{NaOH}}{V} = \frac{3.01 \cdot 10^{-3} moles}{1.00 L} = 3.01 \cdot 10^{-3} mol/L$

Finally, we can calculate the pH of the solution as follows:

$pOH = -log([OH^{-}]) = -log(3.01 \cdot 10^{-3}) = 2.52$

$pH + pOH = 14$

$pH = 14 - pOH = 14 - 2.52 = 11.48$

Therefore, the pH of the solution is 11.48.

I hope it helps you!

3 0

3 years ago

What is a negatively charged ion

dem82 [27]

an electrically charged atom or group of atoms formed by the loss or gain of one or more electrons, as a cation (positive ion) , which is created by electron loss and is attracted to the cathode in electrolysis, or as an anion (negative ion) , which is created by an electron gain and is attracted to the anode.

7 0

3 years ago

A 25.00-mL sample of propionic acid, HC3H5O2, of unknown concentration was titrated with 0.141 M KOH. The equivalence point was

Nastasia [14]

Answer:

Concentration of hydroxide-ion at equivalence point = $8.3\times 10^{-6}M$

Explanation:

$HC_{3}H_{5}O_{2}+KOH\rightarrow C_{3}H_{5}O_{2}^{-}K^{+}+H_{2}O$

1 mol of $HC_{3}H_{5}O_{2}$ reacts with 1 mol of KOH to produce 1 mol of $C_{3}H_{5}O_{2}^{-}$

At equivalence point, all $HC_{3}H_{5}O_{2}$ gets converted to $C_{3}H_{5}O_{2}^{-}$ .

Moles of $C_{3}H_{5}O_{2}^{-}$ produced at equivalence point is equal to moles of KOH added to reach equivalence point.

So, moles of $C_{3}H_{5}O_{2}^{-}$ produced = $\frac{43.76\times 0.141}{1000}moles=0.00617moles$

Total volume of solution at equivalence point = (25.00+43.76) mL = 68.76 mL

Concentration of $C_{3}H_{5}O_{2}^{-}$ at equivalence point = $\frac{0.00617\times 1000}{68.76}M=0.0897M$

$OH^{-}$ produced at equivalence point is due to hydrolysis of $C_{3}H_{5}O_{2}^{-}$ . We have to construct an ICE table to calculate concentration of $OH^{-}$ at equivalence point.