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2 years ago

9

When a radioactive magnesium (Mg) atom decays by emitting a positron, what is the identity of the resulting atom? A. Aluminum (A

l) B. Magnesium (Mg) C. Neon (Ne) D. Sodium (Na)

1 answer:

iragen [17]2 years ago

5 0

The answer is: D. Sodium (Na).

Beta decay is radioactive decay in which a beta ray and a neutrino are emitted from an atomic nucleus.

There are two types of beta decay: beta minus and beta plus.

In beta plus decay (atomic number Z is decreased by one), a proton is converted to a neutron and positron and an electron neutrino, so mass number does not change.

In this example, magnesium (atomic number 12) is converted to dosium (atomic number 11).

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14) The central iodine atom in the ICl4- ion has __________ nonbonded electron pairs and __________ bonded electron pairs in its

masha68 [24]

Answer:

Two non bonded electron pairs and four bonded electron pairs

Explanation:

An image of the compound as obtained from chemlibretext is attached to this answer.

The ion ICl4- ion, is an AX4E2 ion. This implies that there are four bond pairs and two lone pairs of electrons. As expected, the shape of the ion is square planar since the lone pairs are found above and below the plane of the square. This is clear from the image attached.

7 0

2 years ago

A 600.0 mL sample of 0.20 MHF is titrated with 0.10 MNaOH. Determine the pH of the solution after the addition of 600.0 mL of Na

Leona [35]

Answer: pH=12.69

Explanation:

${\text{Moles of HF}=Molarity\times {\text{Volume of solution in liters}}$

${\text{Moles of HF}=0.20M\times 0.6L=0.12 moles$

$HF\rightarrow H^++F^-$

Initial 0.12 0 0

Eqm 0.12-x x x

$K_a=\frac{[H^+][F^-]}{[HF]}$

$3.5\times 10^{-4}=\frac{x^2}{0.12-x}$

(neglecting small value of x in comparison to 0.12)

$x=4.2\times 10^{-5}$

Moles of $H^+=4.2\times 10^{-5}$

$NaOH\rightarrow Na^++OH^-$

${\text{Molesof NaOH}}=Molarity\times {\text{Volume of solution in liters}}$

${\text{Moles of NaOH}}=0.10M\times 0.6L=0.06 moles$

0.06 moles of NaOH will give 0.06 moles of $[OH^-]$

Now $4.2\times 10^{-5}$ moles of $OH^-$ will be neutralized by $4.2\times 10^{-5}$ moles of $H^+$ and $(0.06-4.2\times 10^{-5})=0.059$ moles of $OH^-$ will be left.

Molarity of $OH^-=\frac{0.059moles}{1.2L}=0.049M$

$pOH=-\log[OH^-]=-\log[0.049]=1.31$

pH = 14 - pOH= 14 - 1.31 = 12.69

5 0

2 years ago

What is the ph of a solution made by combining 157 ml of 0.35 m nac2h3o2 with 139 ml of 0.46 m hc2h3o2? the ka of acetic acid is

ExtremeBDS [4]

The first step is to calculate the molarity of each compound:
final volume of solution = 157 + 139 = 296 mL
molarity of <span>nac2h3o2 = (157 x 0.35) / 296 = 0.1856 molar
molarity of </span><span>hc2h3o2 = (139 x 0.46) / 296 = 0.216 molar

Then, we calculate the pH as follows:
pKa of acetic acid = -log(</span><span>1.75 × 10^-5) = 4.7569
pH = pKa + </span><span> log ([salt] / [acid])
= </span>4.7569 + log(0.1856 / 0.216)
= 4.691

6 0

2 years ago

What is the final pressure (expressed in atm) of a 3.05 l system initially at 724 mm hg and 298 k, that is compressed to a final

krok68 [10]

<u>Answer:</u>

P2 = 778.05 mm Hg = 1.02 atm

<u>Explanation:</u>

We are to find the final pressure (expressed in atm) of a 3.05 liter system initially at 724 mm hg and 298 K which is compressed to a final volume of 2.60 liter at 273 K.

For this, we would use the equation:

$\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}$

where P1 = 724 mm hg

V1 = 3.05 L

T1 = 298 K

P2 = ?

V2 = 2.6 L

T2 = 173 K

Substituting the given values in the equation to get:

$\frac{(724)(3.05)}{298} =\frac{P_2(2.6)}{173}$

P2 = 778.05 mm Hg = 1.02 atm

7 0

2 years ago

You have a 16.0-oz. (473-mL) glass of lemonade with a concentration of 2.66 M. The lemonade sits out on your counter for a coupl

salantis [7]

<u>Answer:</u> The new concentration of lemonade is 3.90 M

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

Molarity of lemonade solution = 2.66 M

Volume of solution = 473 mL

Putting values in equation 1, we get:

$2.66M=\frac{\text{Moles of lemonade}\times 1000}{473}\\\\\text{Moles of lemonade}=\frac{2.66\times 473}{1000}=1.26mol$

Now, calculating the new concentration of lemonade by using equation 1:

Moles of lemonade = 1.26 moles

Volume of solution = (473 - 150) mL = 323 mL

Putting values in equation 1, we get:

$\text{New concentration of lemonade}=\frac{1.26\times 1000}{323}\\\\\text{New concentration of lemonade}=3.90M$

Hence, the new concentration of lemonade is 3.90 M

7 0

2 years ago