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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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fiasKO [112]

Answer:

Explanation:every feet has 12in so divde, 67/12  so its 5ft 7/12

4 0
3 years ago
A 2.684-g sample of zinc oxide was reduced by hydrogen gas, resulting in 2. 156 g of pure zinc metal. Determine the empirical fo
Afina-wow [57]

The empirical formula of the initial zinc oxide is ZnO.

<h3>What is Empirical Formula?</h3>

The empirical formula of a compound represents the ratios of elements in a compound but not the actual numbers or arrangement of the atoms.

It is the lowest whole number ratio of the element in the compound.

<h3>How to find out the empirical formula?</h3>
  • Find out the given masses and molar masses of the elements

The molar mass of Zn = 65 gmol⁻¹

Given the mass of Zn = 2.156 g

The molar mass of Oxygen = 16 gmol⁻¹

The mass of Oxygen = Mass of a sample of zinc oxide - the mass of zinc metal

                                   = (2.684 - 2.156) g

                                   = 0.528 g

  • Find the number of moles of the elements in the compound

The number of moles is given by

n = \frac{m}{M}

where m = given mass and

M = Molar mass

Number of moles of Zinc = \frac{2.156}{65} = 0.033 moles

Number of moles of Oxygen =\frac{0.528}{16} = 0.033 moles

  • Find the simplest ratios of the elements in the compound. To find the ratios simply divide the number of moles by the lowest number of moles obtained.

Here, the number of moles is the same for both elements. Hence, the simplest ratio for Zn:O is 1:1.

Therefore, the empirical formula of zinc oxide is ZnO.

Learn more about the empirical formula:

brainly.com/question/1603500

#SPJ4

 

5 0
1 year ago
An interplanetary probe returns to Earth with soil samples. The atoms in the sample are sorted by their number of protons. A new
Leya [2.2K]

Answer is: the average atomic mass is 232.

ω₁ = 20% ÷ 100%.

ω₁ = 0.20.

ω₂ = 80% ÷ 100%.

ω₂ = 0.80.

Ar₁ = 120 (number of protons) + 120 (number of neutrons).

Ar₁ = 240.

Ar₂ = 120 + 110 .

Ar₂ = 230.

Average atomic mass of atoms of bolognium =  

Ar₁ · ω₁ + Ar₂ · ω₂.  

Average atomic mass of atoms of bolognium =  240 · 0.2 + 230 · 0.8.  

Average atomic mass of atoms of bolognium = 48 + 184.  

Average atomic mass of atoms of bolognium = 232.

4 0
3 years ago
If i have an 86 and i don’t do a 50 point assignment, what will my grade be in the class?
nexus9112 [7]
Well you are at a 86% so t<span>his is considered a "B" grade on an average grade scale. If you take 50 points we would need to know the Total point you could have got in that class to be able to see how much a percent was the assignment work and then take that percentage of the assignment and subtract it from the 86% to see where on the scale you would fall in after the assignment points were taken off.   Hope this helps :)
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7 0
3 years ago
(?)Li2O + (?)H2O → (?)LiOH
pav-90 [236]

Answer:

Li2O+H2O---->(2)LiOH

Explanation:

you have to balance the equation and not all the blanks have to be filled all the time but if it makes it easier for you in the first 2 question marks you can put a 1 which isnt necessary but if ur a visual person it will help.

hope this helps im litterally learning the same thing as u lol

6 0
3 years ago
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