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Alborosie
3 years ago
9

What is the ratio strength of 100 ml containing a 1:50 povidone iodine solution diluted to 1000 ml? (answer must be numeric; no

units or commas; do not enter the 1 and the colon; enter only the number after the colon, as shown here with xs: 1:xxx.)?
Chemistry
1 answer:
guapka [62]3 years ago
4 0

We are given that:

100 mL total solution with 1:50 povidone iodine solution

 

This means that there is 1 mL of povidone iodine solution per 50 mL of total solution. Since we are given a total of 100 mL solution, therefore we have an initial amount of:

pure povidone iodine solution = 2 mL

 

This amount of pure povidone iodine solution is added or diluted (most perhaps with water) to make a total of 1000 mL total solution, therefore the new ratio is:

2:1000 povidone iodine solution

By dividing both sides by 2, this simplifies to

1:500 povidone iodine solution

 

Answer:

1:500

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Answer:

The number of protons is equal to the mass number of the element. Since an element always has a different number of protons, the mass can indicate how many neutrons are in an isotope. Atoms of the same element can have a different number of neutrons. There are three naturally-occurring isotopes of carbon.

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4 years ago
You know that 8.9 moles of solute particles are dissolved in that liquid solution. If the molarity of the solution is 25 M, how
pychu [463]

Answer:

V = 0.356 L

Explanation:

In this case, we need to use the following expression:

M = n/V (1)

Where:

M: molarity of solution (mol/L or M)

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V: Volume of solution (Liters)

From these expression, we can solve for V:

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5 0
3 years ago
(03.03 MC)
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2 years ago
What mass of HCL, in grams, is required to react with 0.610 g of al(oh)3 ?
kompoz [17]

Answer: 0.8541 grams of HCl will be required.

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\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of Al(OH)_3 = 0.610 g

Molar mass of Al(OH)_3 = 78 g/mol

\text{Number of moles}=\frac{0.610g}{78g/mol}

Number of moles of Al(OH)_3 = 0.0078 moles

The reaction between Al(OH)_3 and HCl is a type of neutralization reaction because here acid and base are reacting to form an salt and also releases water.

Chemical equation for the above reaction follows:

Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O

By Stoichiometry,

1 mole of  Al(OH)_3 reacts with 3 moles of HCl

So, 0.0078 moles of Al(OH)_3 will react with \frac{3}{1}\times 0.0078 = 0.0234 moles

Mass of HCl is calculated by using the mole formula, we get

Molar mass of HCl = 36.5 g/mol

Putting values in the equation, we get

0.0234moles=\frac{\text{Given mass}}{36.5g/mol}

Mass of HCl required will be = 0.8541 grams

3 0
3 years ago
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