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3 years ago

A composition of reflections over parallel lines is the same as a __________.

Chemistry

2 answers:

8090 [49]3 years ago

6 0

Answer : Option A) Translation

Explanation : A composition of reflections over parallel lines is the same as a Translation.

To identify if the composition of reflections over parallel lines are same as translation or not?

We can check using a picture of some shape in the plane. Place the picture on the right side of two vertical parallel. Now, we can see the reflected the shape over the nearest parallel line, then check the reflection over the other parallel line. We see that the shape winds up in the same orientation, like it was just shifted over to the right. Hence, it is translation.

Andrej [43]3 years ago

5 0

Answer:

rotation

Explanation:

reflected twice

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Match the term with its description.

Fynjy0 [20]

Answer:

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Explanation:

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3 years ago

Describe the relationship between hydrogen oxygen and water

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4 years ago

In a titration of 35.00 mL of 0.737 M H2SO4, __________ mL of a 0.827 M KOH solution is required for neutralization.

elena55 [62]

Answer:

In a titration of 35.00 mL of 0.737 M H₂SO₄, 62.4 mL of a 0.827 M KOH solution is required for neutralization.

Explanation:

The balanced reaction is

H₂SO₄ + 2 KOH ⇒ 2 H₂O + K₂SO₄

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) 1 mole of H₂SO₄ is neutralized with 2 moles of KOH.

The molarity M being the number of moles of solute that are dissolved in a given volume, expressed as:

$Molarity=\frac{number of moles}{volume}$

in units of $\frac{moles}{liter}$

then the number of moles can be calculated as:

number of moles= molarity* volume

You have acid H₂SO₄

35.00 mL= 0.035 L (being 1,000 mL= 1 L)
Molarity= 0.737 M

Then:

number of moles= 0.737 M* 0.035 L

number of moles= 0.0258

So you must neutralize 0.0258 moles of H₂SO₄. Now you can apply the following rule of three: if by stoichiometry 1 mole of H₂SO₄ are neutralized with 2 moles of KOH, 0.0258 moles of H₂SO₄ are neutralized with how many moles of KOH?

$moles of KOH=\frac{0.0258moles of H_{2} SO_{4}*2 moles of KOH }{1mole of H_{2} SO_{4}}$

moles of KOH= 0.0516

Then 0.0516 moles of KOH are needed. So you know:

Molarity= 0.827 M
number of moles= 0.0516
volume=?

Replacing in the definition of molarity:

$0.827 M=\frac{0.0516 moles}{volume}$

Solving:

$volume=\frac{0.0516 moles}{0.827 M}$

volume=0.0624 L= 62.4 mL

In a titration of 35.00 mL of 0.737 M H₂SO₄, 62.4 mL of a 0.827 M KOH solution is required for neutralization.

4 0

3 years ago

If the [H+] in a solution is 1 × 10–1 mol/L, what is the [OH–]? Show your work.

AVprozaik [17]

Hello!

The basic equations to solve this is
pH = -log[H+]
pOH = -log[OH-]
pH + pOH = 14
------------------------------------------------------------------------------------------------------
Find pH

pH = -log(1 * 10^-1)
pH = 1
------------------------------------------------------------------------------------------------------
Find pOH
1 + pOH = 14

pOH = 13
------------------------------------------------------------------------------------------------------
Find OH-
[OH-] = 10^(-pOH)
[OH-] = 1 * 10^-13mo/L

The answer is $[OH-] = 1 * 10^{-13} mol/L$

Hope this helps!

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4 years ago

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FrozenT [24]

it is a Chemical property

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3 years ago