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2 years ago

What volume of a 0.295 M nitric acid solution is required to neutralize 24.8 mL of a 0.107 M potassium hydroxide solution

Chemistry

1 answer:

KengaRu [80]2 years ago

4 0

Answer: 9.00 mL

Explanation:

$M_{A}V_{A}=M_{B}V_{B}\\(0.295)V_{A}=(0.107)(24.8)\\V_{A}=\frac{(0.107)(24.8)}{(0.295)} \approx \boxed{9.00 \text{ mL}}$

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Seven
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Fluorine (F) and bromine (Br) are in the same group on the periodic table. How do atoms of these elements compare when they form

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Answer is: Both a fluorine atom and a bromine atom gain one electron, and both atoms become stable.

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Halogens need to gain one electron to have electron cofiguration like next to it noble gas.

Fluorine has atomic number 9, it means it has 9 protons and 9 electrons.

Fluorine tends to have eight electrons in outer shell like neon (noble gas) and gains one electron in chemical reaction.

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3 years ago

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What sequence of enzymes allows for the synthesis of glycogen from glucose-6-phosphate?

Nutka1998 [239]

Answer:

The answer is (e) : phosphoglucomutase, UDP-glucose pyrophosphorylase, glycogen synthase then amylo-(1,4-1,6)-transglycosylase.

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3 years ago

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3 years ago

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A sample of table sugar (sucrose, C12H22O11) has a mass of 1.202 g.

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Answer:

a) 0.003512 moles

b) Moles C= 0.04214 moles carbon

Moles H = 0.07726 moles hydrogen

Moles O = 0.03863 moles of oxygen

c) C atoms = 2.54 *10^22 carbon atoms

H atoms = 4.65 *10^22 hydrogen atoms

O atoms = 2.33 *10^22 oxygen atom

Explanation:

Step 1: Data given

Mass of sucrose = 1.202 grams

Molar mass of sucrose = 342.3 g/mol

Step 2: Calculate moles of sucrose

Moles sucrose = Mass sucrose / molar mass sucrose

Moles sucrose = 1.202 grams / 342.3 g/mol

Moles sucrose = 0.003512 moles

Step 3: Calculate moles of each element

For 1 mol of C12H22O11 we have 12 moles of carbon, 22 moles of hydrogen and 11 moles of oxygen

Moles C: 12*0.003512 = 0.04214 moles carbon

Moles H: 22* 0.003512 = 0.07726 moles hydrogen

Moles O: 11* 0.003512 = 0.03863 moles of oxygen

Step 4: Calculate the number of atoms

C atoms = 6.022 *10^23 / mol * 0.04214 moles = 2.54 *10^22 atoms carbon

H atoms = 6.022 * 10^23 / mol * 0.07726 moles = 4.65 *10^22 atoms H

O atoms = 6.022 * 10^23 / mol * 0.03863 moles = 2.33 *10^22 atoms O

0.003512 moles of sucrose containse 6.022 *10^23 * 0.003512 = 2.11 * 10^21 sucrose molecules

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3 years ago