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2 years ago

A) Of options (a) - (d) below, highlight the (unbalanced) chemical equation that represents

Chemistry

1 answer:

labwork [276]2 years ago

7 0

Answer:

c...............

Explanation:.............................yuuuuuuuuuuuuuuuuuuuuuwoopuuuuuuuu

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In a chemical process, you need to force a compound to bond with a halogen, causing it to lose hydrogen. Which type of reaction

Kazeer [188]

Answer is Halogenation because Halogenation is a type of substitution reaction in which a hydrogen atom is replaced by a halogen atom in a molecule.
The molecule looses its hydrogen atom as the halogen is introduced into the molecule. This sort of reaction is very common in organic chemistry. Many hydrocarbons can be halogenated in the presence of light.

5 0

2 years ago

Consider the following data on some weak acids and weak bases:

Juli2301 [7.4K]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

8 0

3 years ago

Which equation shows how to calculate how many grams (g) of KOH would be needed to fully react with 4 mol Mg(OH)2? The balanced

bazaltina [42]

Mole ratio:

MgCl₂ + 2 KOH = Mg(OH)₂ + 2 KCl

2 moles KOH ---------------- 1 mole Mg(OH)₂
moles KOH ------------------- 4 moles Mg(OH₂)

moles KOH = 4 x 2 / 1

= 8 moles of KOH

molar mass KOH = 56 g/mol

mass of KOH = n x mm

mass of KOH = 8 x 56

= 448 g of KOH

hope this helps!

3 0

3 years ago

Read 2 more answers

A learner was assigning oxidation numbers for different elements in the compounds OF2 and NaF. The learner assigned F an oxidati

77julia77 [94]

Answer: B. No, fluorine is always assigned an oxidation number of -1.

Explanation:

6 0

1 year ago

A solution is prepared at that is initially in benzoic acid , a weak acid with , and in sodium benzoate . Calculate the pH of th

Kisachek [45]

Answer:

$pH=4.1$

Explanation:

Hello,

In this case, for a concentration of 0.42 M of benzoic acid whose Ka is 6.3x10⁻⁵ in 0.33 M sodium benzoate, we use the Henderson-Hasselbach equation to compute the required pH:

$pH=pKa+log(\frac{[base]}{[acid]} )$

Whereas the concentration of the base is 0.33 M and the concentration of the acid is 0.42 M, thereby, we obtain:

$pH=-log(Ka)+log(\frac{[base]}{[acid]} )\\\\pH=-log(6.3x10^{-5})+log(\frac{0.33M}{0.42M} )\\\\pH=4.1$

Regards.

3 0

3 years ago