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3 years ago

Actual value= -273 Experimental value= -274

Chemistry

1 answer:

Maurinko [17]3 years ago

4 0

Answer:

<h2><u><em>When measuring data, the result often varies from the true value. Error can arise due to many different reasons that are often related to human error, but can also be due to estimations and limitations of devices used in measurement. Regardless, in cases such as these, it can be valuable to calculate the percentage error. The computation of percentage error involves the use of the absolute error, which is simply the difference between the observed and the true value.</em></u></h2>

Explanation:

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Which best describes the motion of molecules in the gaseous state?

artcher [175]

Answer:

They spread apart to fill the enclosed object. In other words they move freely...

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3 years ago

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

Rank the species CH2NH, CH3I, HF, C2H6 from most to least miscible in ammonia (NH3) 1. CH3I > C2H6 > HF > CH2NH 2. CH2N

777dan777 [17]

Answer:

$HF>CH_{2}NH>CH_{3}I>C_{2}H_{6}$

Explanation:

$NH_{3}$ is an weak base. HF is an weak acid. Hence an acid-base reaction will ocuur between $NH_{3}$ and HF resulting a homogeneous mixture.

$NH_{3}$ is a polar protic molecule. Hence it can form hydrogen bonding with another polar protic molecule $CH_{2}NH$ . Also dipole-dipole attraction force along with london dispersion force exists between $CH_{2}NH$ and $NH_{3}$ .