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Levart [38]
3 years ago
7

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

Chemistry
1 answer:
kkurt [141]3 years ago
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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PLEASE ANSWER ASAP!!!!!!! 50 points if the answer is correct!
Ad libitum [116K]

5

 mL acetic acid

95

mL water

Explanation:

Since

5

%

of the vinegar, by volume, is acetic acid, and we have

100

mL of vinegar, we have

5

mL of vinegar.

Similarly, we have

100

%

−

5

%

=

95

%

being water, so we have

95

mL of water.

8 0
3 years ago
The formula of an iodate of lanthanum is la(io3)3. what is the formula of the sulfate of lanthanum with the same oxidation numbe
sammy [17]

Iodate has a charge of -1, this means its chemical symbol must be IO3(-1) while Lanthanum is La(+3). Therefore making the compound La(IO3)3.

However sulfate has a chemical symbol of SO4(-2), it has charge of -2, therefore the formula for lanthanum sulfate is:

<span>La2(SO4)<span>3</span></span>

8 0
3 years ago
Read 2 more answers
10 points
puteri [66]

Answer: Using more fossil fuels

Explanation: Burning fossil fuels releases Green House gasses into the atmosphere, such as carbon dioxide. The carbon dioxide in the atmosphere traps in heat, which causes global temperatures to increase.

3 0
3 years ago
Solid ammonium chloride, NH4Cl, is formed by the reaction of gaseous ammonia, NH3, and hydrogen chloride, HCl. NH3(g)+HCl(g)⟶NH4
Mashutka [201]

9.41 atm is the pressure in atmospheres of the gas remaining in the flask

<h3>What is the pressure in atmospheres?</h3>

The equation NH3(g) + HCl(g) ==> NH4Cl(s) is balanced.

Divide the moles of each reactant by its coefficient in the balanced equation, and the limiting reagent is identified as the one whose value is less. With the issue we now have...

6.44 g NH3 times 1 mol NH3/17 g equals 0.3688 moles of NH3 ( 1 = 0.3688)

HCl: 6.44 g of HCl times one mole of HCl every 36.5 g equals 0.1764 moles ( 1 = 0.1764). CONTROLLING REAGENT

NH4Cl will this reaction produce in grams

0.1764 moles of HCl multiplied by one mole of NH4Cl per mole of HCl results in 9.44 g of NH4Cl (3 sig. figs.)

the gas pressure, measured in atmospheres, that is still in the flask

NH3(g) plus HCl(g) results in NH4Cl (s)

0.3688......0.1764............0..........

Initial

-0.1764....-0.1764........+0.1764...Change

Equilibrium: 0.1924.......0...............+0.1924

There are 0.1924 moles of NH3 and no other gases in the flask. This is at a temperature of 25 °C (+273 = 298 °K) in a volume of 0.5 L. After that, we may determine the pressure by using the ideal gas law (P).

PV = nRT

P = nRT/V = 0.1924 mol, 0.0821 latm/mol, and 298 Kmol / 0.5 L

P = 9.41 atm

9.41 atm is the pressure in atmospheres of the gas remaining in the flask

To learn more about balanced equation refer to:

brainly.com/question/11904811

#SPJ1

7 0
1 year ago
Find the empirical formula of the compound ribose, a simple sugar often used as a nutritional supplement. A 14.229 g sample of r
MakcuM [25]

Answer:

CH2O

Explanation:

Firstly, we need to convert the masses of the elements to percentage compositions. This can be done by placing the mass of each element over the total mass multiplied by 100% . We can start with carbon.

C = 5.692/14.229 * 100 = 40%

O = 7.582/14.229 * 100 = 53.29%

H = 0.955/14.229 * 100 = 6.71%

We then proceed to divide each percentage composition by their atomic mass of 12, 16 and 1 respectively.

C = 40/12 = 3.333

O = 53.29/16 = 3.33

H = 6.71/2 = 6.71

Dividing by the smaller value which is 3.33

C = 3.33/3.33 = 1

O = 3.33/3.33= 1

H = 6.71/3.33 = 2

The empirical formula of the compound ribose is CH2O

6 0
3 years ago
Read 2 more answers
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