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3 years ago

Amoxicillin Suspension 125 mg/ 5 ml is 125 mg of Amoxicillin per 5 ml of suspension is an example of weight to ________.

1 answer:

8 0

Answer : Amoxicillin Suspension 125 mg/ 5 ml is 125 mg of Amoxicillin per 5 ml of suspension is an example of weight to volume.

Explanation :

Weight by volume (w/v) means that the mass of solute present in 100 mL volume of solution.

Weight by weight (w/w) means that the mass of solute present in 100 gram of solution.

Volume by volume (v/v) means that the volume of solute present in 100 mL volume of solution.

As per question, amoxicillin suspension is, 125 mg/ 5 ml that means 125 mg of Amoxicillin present in 5 mL of suspension. So, it is an example of weight to volume.

Hence, it is an example of weight to volume.

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In a reaction A+B--->C, reactant A has 5g of product C has 9g of product. How many grams should reactant B contain?

Rzqust [24]

According to the law of conservation of Mass:

In a chemical reaction mass can neither be created nor be destroyed

So, we can say that: Mass of A + Mass of B = Mass of C

In the given reaction,

One of the reactants weigh 5 grams and another one weighs x grams

The mass of the product of this reaction is 9 grams

<u>Mass of reactant B:</u>

Mass of A + Mass of B = Mass of C

5 + x = 9

x = 4 grams

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3 years ago

100 points! Report on an element

Luden [163]

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3 years ago

Read 2 more answers

Carbon dioxide dissolves in water to form carbonic acid, which is primarily dissolved CO2. Dissolved CO2 satisfies the equilibri

lorasvet [3.4K]

Explanation:

The reaction equation will be as follows.

$CO_{2}(aq) + H_{2}O \rightleftharpoons H^{+}(aq) + HCO^{-}_{3}(aq)$

Calculate the amount of $CO_{2}$ dissolved as follows.

$CO_{2}(aq) = K_{CO_{2}} \times P_{CO_{2}}$

It is given that $K_{CO_{2}}$ = 0.032 M/atm and $P_{CO_{2}}$ = $1.9 \times 10^{-4}$ atm.

Hence, $[CO_{2}]$ will be calculated as follows.

$[CO_{2}]$ = $K_{CO_{2}} \times P_{CO_{2}}$

= $0.032 M/atm \times 1.9 \times 10^{-4}atm$

= $0.0608 \times 10^{-4}$

or, = $0.608 \times 10^{-5}$

It is given that $K_{a} = 4.46 \times 10^{-7}$

As, $K_{a} = \frac{[H^{+}]^{2}}{[CO_{2}]}$

$4.46 \times 10^{-7} = \frac{[H^{+}]^{2}}{0.608 \times 10^{-5}}$

$[H^{+}]^{2}$ = $2.71 \times 10^{-12}$

$[H^{+}]$ = $1.64 \times 10^{-6}$

Since, we know that pH = $-log [H^{+}]$

So, pH = $-log (1.64 \times 10^{-6})$

= 5.7

Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.

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3 years ago

Which causes waves in the ocean to occur?

galina1969 [7]

Answer:

Explanation:

C. energy being transferred from wind

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3. 79,789kJ = (293k) (377kJ)

KatRina [158]

the anwser would most likely be j=0

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