Can anyone help me please this is due and i need help someone check over my answers

The diagram below shows changes to the nucleus of an atom.

anzhelika [568]

Answer:

maybe 1 or 3 im not sure

Explanation:

i didn't study it yet sorry for not helping but try asking someone else

8 0

3 years ago

Read 2 more answers

Calculate either [H3O+] or [OH−] for each of the solutions at 25 °C. Solution A: [OH−]=1.55×10−7 M Solution A: [H3O+]= M Solutio

Ahat [919]

Answer: A. $[H_3O^+]=0.64\times 10^{-7}M$

B. $[OH^-]=0.11\times 10^{-5}M$

C. $[OH^-]=\frac{10^{-14}}{0.000775}=1.3\times 10^{-11}M$

Thus solution B is basic in nature.

Explanation:

pH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

pH for acidic solutions is less than 7, for basic solutions it is more than 7 and for neutral solutions it is equal to 7.

$pH=-\log [H^+]$

$pOH=-log[OH^-]$

$pH+pOH=14$

or $[H^+][OH^-]=10^{-14}$

A. $[OH^-]=1.55\times 10^{-7}M$

$[H_3O^+]=\frac{10^{-14}}{1.55\times 10^{-7}}=0.64\times 10^{-7}M$

$pH=-log[H_3O^+]=-log[0.64\times 10^{-7}]=7$

B. $[H_3O^+]=9.43\times 10^{-9}M$

$[OH^-]=\frac{10^{-14}}{9.43\times 10^{-9}}=0.11\times 10^{-5}M$

$pH=-log[H_3O^+]=-log[9.43\times 10^{-9}]=8$

C. $[H_3O^+]=0.000775M$

$[OH^-]=\frac{10^{-14}}{0.000775}=1.3\times 10^{-11}M$

$pH=-log[H_3O^+]=-log[0.000775]=3$

Thus solution B is basic.

4 0

4 years ago

Need help ASAP!!!! what is the value (angle) for the C=C=O bond in Ketene i.e. CH2=C=O

stepan [7]

Answer:

$180^\circ$ by the VSEPR theory.

Explanation:

This question is asking for the bond angle of the $\rm C=C=O$ bond in $\rm H_2C=C=O$ . The VSEPR (valence shell electron pair repulsion) theory could help. Start by considering: how many electron domains are there on the carbon atom between these two bond?

Note that "electron domains" refer to covalent bonds and lone pairs collectively.

Each nonbonding pair (lone pair) of valence electrons counts as one electron domain.
Each covalent bond (single bond, double bond, or triple bond) counts as exactly one electron domain.

For example, in $\rm H_2C=C=O$ , the carbon atom at the center of that $\rm C=C=O$ bond has two electron domains:

This carbon atom has two double bonds: one $\rm C=C$ bond and one $\rm C=O$ bond. Even though these are both double bonds, in VSEPR theory, each of them count only as one electron domain.
Keep in mind that there are only four valence electrons in each carbon atom. It can be shown that all four valence electrons of this carbon atom are involved in bonding (two in each of the two double bonds.) Hence, there would be no nonbonding pair around this atom.

In VSEPR theory, electron domains around an atom repel each other. As a result, they would spread out (in three dimensions) as far away from each other as possible. When there are only two electron domains around an atom, the two electron domains would form a straight line- with one domain on each side of the central atom. (To visualize, consider the three atoms in this $\rm C=C=O$ bond as three spheres on a stick. The central $\rm C$ atom would be between the other $\rm C$ atom and the $\rm O$ atom.)

This linear geometry corresponds to a bond angle of $180^\circ$ .

3 0

3 years ago

Enough of a monoprotic acid is dissolved in water to produce a 1.35 M solution. The pH of the resulting solution is 2.93. Calcul

Ede4ka [16]

The acid dissociation constant is defined as Ka = [H+][A-]/[HA] where [H+], [A-] and [HA] are the concentrations of protons, conjugate base, and acid in solution respectively. Assuming this is a weak acid as the pH is quite high for a 1.35 M solution, we can assume that the change in [HA] is negligible and therefore [HA] = 1.35 M.
To calculate [H+] we can use the relationship pH = -log[H+], rearranging to give: [H+] = 10^(-pH) = 10^(-2.93) = 1.17 x 10^(-3).
Since the acid is relatively concentrated we can assume therefore that [H+] = [A-] as for each proton dissociated, a conjugate base is formed.
Therefore, we can calculate Ka as:
Ka = [H+]^2/[HA] = (1.17 x 10^-3 M)^2/1.35 = 1.01 x 10^-6 M

6 0

3 years ago

A liquid medicine is labeled as having 5% active ingredient, by volume. If

enyata [817]

Answer:

2.5 mL

Explanation:

Step 1: Given data

Concentration of the active ingredient (C): 5% v/v

Volume of the medicine (V): 50 mL

Step 2: Calculate the amount of the active ingredient in the medicine

The concentration of the active ingredient is 5% v/v, that is, there are 5 mL of the active ingredient per 100 mL of the medicine. The volume of the active ingredient is 50 mL of the medicine is:

50 mL Med × 5 mL AI/100 mL Med = 2.5 mL AI

4 0

3 years ago