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kotegsom [21]
3 years ago
10

A chemical buffer is a solution that increases pH changes when small quantities of an acid or base are added. True False

Chemistry
1 answer:
Alex3 years ago
7 0

Answer:

false

Explanation:

A buffer solution is an aqueous solution consisting of a mixture of a weak acid and its conjugate base, or vice versa. Its pH changes very little when a small amount of strong acid or base is added to it. A buffer protects from ph increasing.

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You are making a map of your state in your map scale 1 cm equals 10 km if two cities are 200 km apart how many centimeters apart
Agata [3.3K]
The answer is 20 centimeters
6 0
4 years ago
Could you guys help me with these questions pls. Thank you
lapo4ka [179]

Answer:

8.4) 27.1 cm³

8.5) 0.217 mol/dm³

Explanation:

Please see attached picture for full solution.

8.4) Since the results from titrations 3-5 are within 0.10cm³ from each other, these 3 results are concordant.

8.5) Find mol of sulfuric acid to find mol of NaOH using mole ratio. Equation has already been balanced for you, so mole ratio of NaOH to H2S04 is 2:1. To find concentration of NaOH, divide the number of moles by the volume in dm³ since the units for concentration is mol/dm³, which you can think of as the number of moles of NaOH in a dm³ of solution.

1 dm³ = 1000cm³

Thus, 1cm³= 1/1000 dm³

3 0
3 years ago
PLEASE HELP ME ASAP!! I NEED IT! NO LINKS
Natali [406]

Answer:

asexual cell

Explanation:there are mainly 2 types sexual and asexual i would say asexual because they grow without another parent and don't have a nucleus.

3 0
3 years ago
The list shows five processes that occur during digestion.
Kobotan [32]
I think the answer is B.
8 0
3 years ago
For the following reaction, 13.1 grams of glucose (C6H12O6) are allowed to react with 10.6 grams of oxygen gas. glucose (C6H12O6
mart [117]

Answer:

There will be formed 14.58 grams of CO2

O2 is the limiting reagent

There will remain 3.151 grams of glucose

Explanation:

Step 1: Data given

Mass of glucose = 13.1 grams

molar mass of glucose 180.156 g/mol

Mass of oxygen = 10.6grams

molar mass of oxygen = 32 g/mol

<u>Step 2</u>: The balanced equation

C6H12O6 + 6 O2 → 6H2O + 6CO2

<u>Step 3</u>: Calculate moles of glucose

Moles of glucose = mass glucose / molar mass glucose

Moles of glucose = 13.1 grams / 180.156 g/mol

Moles of glucose = 0.0727 moles

Step 4: Calculate moles of oxygen

Moles O2 = 10.6 grams / 32 g/mol

Moles O2 = 0.33125 moles

Step 5: Calculate the limiting reactant

For 1 mol of glucose , we need 6 moles of O2 to produce 6 moles of H2O and 6 moles of CO2

Oxygen is the limiting reactant. It will completely be consumed ( 0.33125 moles).

Glucose is in excess. There will be consumed 0.33125/6 = 0.05521 moles

There will remain 0.0727 - 0.05521 = 0.01749 moles

This is 0.01749 moles * 180.156 g/mol = 3.151 grams

Step 6: Calculate moles of CO2

For 1 mol of glucose , we need 6 moles of O2 to produce 6 moles of H2O and 6 moles of CO2

For 0.33125 moles of O2 we'll get 0.33125 moles of CO2 produced

Step 7: Calculate mass of CO2

Mass of CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.33125 moles * 44.01 g/mol

Mass CO2 = 14.58 grams

7 0
3 years ago
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