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3 years ago

12

Expressed to the correct number of significant figures, what is the correct sum of (3.04g + 4.134g + 6.1g)?

1 answer:

Lera25 [3.4K]3 years ago

4 0

Answer:

13.3g

Explanation:

3.04g + 4.134g + 6.1g = 13.274g

=13.3g

When expressing the correct number of significant figures in addition, you go to the last common decimal place. In this case that is 0.1 because all the numbers have this decimal place.

This is because your measurement can only be as precise as the least precise measurement. The more decimal places in the measurement, the more precise the measurement.

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Calculate the [ OH − ] and the pH of a solution with an [ H + ] = 0.090 M at 25 °C . [ OH − ] = M pH = Calculate the [ H + ] and

Valentin [98]

Answer: a) $[OH^-]=1.09\times 10^{-13}$ and pH = 1.04

b) $[H^+]=1.02\times 10^{-11}$ and $pH=10.99$

c) $[H^+]=7.08\times 10^{-11}$ and $[OH^-]=1.41\times 10^{-4}$

Explanation:

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

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

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

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

$pH+pOH=14$

a) $[H^+]=0.090M$

$pH=-\log [0.090]=1.04$

$pOH=14-1.04=12.96$

$12.96=-log[OH^-]$

$[OH^-]=1.09\times 10^{-13}$

b) $[OH^-]=0.00098M$

$pOH=-\log [0.00098]=3.01$

$pH=14-3.01=10.99$

$10.99=-log[H^+]$

$[H^+]=1.02\times 10^{-11}$

c) $pH=10.15$

$10.15=-\log [H^+]$

$[H^+]=7.08\times 10^{-11}$

$pOH=14-10.15=3.85$

$3.85=-log[OH^-]$

$[OH^-]=1.41\times 10^{-4}$

3 0

3 years ago

How much energy is needed to vaporize 75.0 g of diethyl ether (c4h10o) at its boiling point (34.6°c), given that δhvap of diethy

malfutka [58]

Answer: 26.8 kJ of energy is needed to vaporize 75.0 g of diethyl ether

Explanation:

First we have to calculate the moles of diethyl ether

$\text{Moles of diethyl ether}=\frac{\text{Mass of diethyl ether}}{\text{Molar mass of diethyl ether}}=\frac{75.0g}{74g/mole}=1.01moles$

As, 1 mole of diethyl ether require heat = 26.5 kJ

So, 1.01 moles of diethyl ether require heat = $\frac{26.5}{1}\times 1.01=26.8kJ$

Thus 26.8 kJ of energy is needed to vaporize 75.0 g of diethyl ether

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