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2 years ago

The volume of a metal rod was determined to be 38.6 mL. Its

1 answer:

3 0

On the basis of significant figures, the accuracy of measurement made using an analytical balance is more reliable.

<h3>What is an analytical balance?</h3>

An analytical balance is a special type of weighing balance that has a reputation for obtaining the mass of very small objects even as low as milligrams. It consists of a transparent glass casing and a scale.

On the basis of significant figures, the use of an analytical balance to determine the weight will improve the accuracy of the determination since it determines the mass to more significant figures.

Learn more about analytical balance: brainly.com/question/795053

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3,7-dibromo-4,6-dichloro oct-5-ene-1,2-diol. --- show its structure by photo.

VikaD [51]

Refer to the attachment for answer.

SOME EXTRA INFORMATION:

Bromo is used for Bromine (Br)

Chloro is used for Chlorine(Cl)

'ene' represents double bond between carbon atoms.

'ol' is used for alcohol

HOPE IT IS USEFUL

4 0

3 years ago

Read 2 more answers

What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb

Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

$(\text{CH}_3)_3\text{N}$ in this question acts as a weak base. As seen in the equation in the question, $(\text{CH}_3)_3\text{N}$ produces $\text{OH}^{-}$ rather than $\text{H}^{+}$ when it dissolves in water. The concentration of $\text{OH}^{-}$ will likely be more useful than that of $\text{H}^{+}$ for the calculations here.

Finding the value of $[\text{OH}^{-}]$ from pH:

Assume that $\text{pK}_w = 14$ ,

$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

$[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}$ .

Solve for $[(\text{CH}_3)_3\text{N}]_\text{initial}$ :

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}$

Note that water isn't part of this expression.

The value of Kb is quite small. The change in $(\text{CH}_3)_3\text{N}$ is nearly negligible once it dissolves. In other words,

$[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}$ .

Also, for each mole of $\text{OH}^{-}$ produced, one mole of $(\text{CH}_3)_3\text{NH}^{+}$ was also produced. The solution started with a small amount of either species. As a result,

$[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}$ .

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}$ ,

$[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}$ ,

$[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}$ .

8 0

3 years ago

State two procedures which should be used to properly handle a light microscope

babunello [35]

hold at base and arm

5 0

3 years ago

A sample of sodium-24 with an activity of 14 mCi is used to study the rate of blood flow in the circulatory system. If sodium-24

SOVA2 [1]

Answer:

See explanation

Explanation:

From;

0.693/t1/2 = 2.303/t log (Ao/A)

Where;

t1/2 = half life of the sodium-24

t = time taken

Ao = initial activity of sodium-24

A= activity of sodium-24 at time t

0.693/15 = 2.303/15 log (14/A)

0.0462 = 0.1535 log (14/A)

0.0462/0.1535 = log (14/A)

log (14/A) = 0.0462/0.1535

14/A = Antilog(0.3)

14/A= 1.995

A = 14/1.995

A = 7.0 mCi

0.693/15 = 2.303/30 log (14/A)

0.0462 =0.0768 log(14/A)

0.0462/0.0768 =log (14/A)

(14/A) =Antilog (0.6)

A = 14/Antilog (0.6)

A = 3.5 mCi

0.693/15 = 2.303/45 log (14/A)

0.0462= 0.0512 log (14/A)

log (14/A) = 0.0462/0.0512

log (14/A) = 0.9

(14/A) = Antilog (0.9)

A= 14/Antilog (0.9)

A = 14/7.9

A = 1.77 mCi

2.5 days = 2.5 * 24 hours = 60 hours

0.693/15 = 2.303/60 log (14/A)

0.0462 = 0.03838 log (14/A)

log (14/A) = 0.0462/0.03838

(14/A) = Antilog(1.2)

A= 14/Antilog(1.2)

A = 14/15.8

A = 0.886 mCi

Note that activity (A) decreases as time increases.

5 0

3 years ago

Luda [366]

Bit.my/3a8Nt8n here’s the link for the answer

3 0

3 years ago