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3 years ago

An aqueous solution contains 0.100 m naoh at 25.0 °c. the ph of the solution is ________.

1 answer:

7 0

Recall; pH + pOH = 14
In this case [OH-] =0.100 m
therefore;
pOH = -LOG[OH-]
= - Log (0.100)
= 1.00
Therefore; the pOH is 1.00
And since, pH +pOH = 14
Then pH = 14-pOH
= 14 -1
= 13
Thus the pH is 13.00

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An electron can be added to halogen atom to force a halide ion with

Lilit [14]

An electron can be added to halogen atom to force a halide ion with 8 valence electrons

An atom can be defined as the smallest part of an element which can take part in a chemical reaction.

However whenever, an electron is added to halogen atom to force a halide ion with 8 different valence electrons

So therefore; an electron can be added to halogen atom to force a halide ion with 8 valence electrons

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2 years ago

Which of the following metals will react with water to produce a metal hydroxide and hydrogen gas? a. Mg b. Li c. Al d. Pb

Stells [14]

I believe the answer is A) Mg.

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3 years ago

Read 2 more answers

What is the total number of liters of SO3 produced when 32.0 of SO2 reacts completely

ahrayia [7]

Write the chemical equation for reaction
that is
2SO2+O2 --->2SO2

find the moles of SO2 used = moles=mass/molar mass of so2

= 32g/80g/mol=0.4 moles

by use of reacting ratio between SO2 and SO3 which is 2:2 therefore the moles of so3 is also = 0.4 moles

STP 1 mole = 22.4L.
what about 0.4moles

= 0.4 /1 x22.4=8.96 liters

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3 years ago

Which of the following would be a quantitative observation? a.The solution has a volume of 25 mL b.The solution turned blue c.Th

alexandr402 [8]

The correct answer for this question is this one: " a.The solution has a volume of 25 mL "

The observation that shows a quantitative observation is when you are talking about numeric data. Just like this one, <em>The solution has a volume of 25 mL </em>
Hope this helps answer your question and have a nice day ahead.

5 0

3 years ago

A sample of Xe gas is observed to effuse through a pourous barrier in 4.83 minutes. Under the same conditions, the same number o

Solnce55 [7]

Answer:

28.93 g/mol

Explanation:

This is an extension of Graham's Law of Effusion where $\frac{R1}{R2} = \sqrt{\frac{M2}{M1} } = \frac{t2}{t1}$

We're only talking about molar mass and time (t) here so we'll just concentrate on $\sqrt{\frac{M2}{M1} } = \frac{t2}{t1}$ . Notice how the molar mass and time are on the same position, recall effusion is when gas escapes from a container through a small hole. The time it takes it to leave depends on the molar mass. If the gas is heavy, like Xe, it would take a longer time (4.83 minutes). If it was light it would leave in less time, that gives us somewhat an idea what our element could be, we know that it's atleast an element before Xenon.

Let's plug everything in and solve for M2. I chose M2 to be the unknown here because it's easier to have it basically as a whole number already.

$\sqrt{\frac{M2}{131} } = \frac{2.29}{4.83}$

The square root is easier to deal with if you take it out in the first step, so let's remove it by squaring each side by 2, the opposite of square root essentially.

$(\sqrt{\frac{M2}{131} } )^2= (\frac{2.29}{4.83})^2$

${\frac{M2}{131} } = (0.47)^2$

${\frac{M2}{131} } = 0.22$

M2= 0.22 x 131

M2= 28.93 g/mol

8 0

3 years ago