Label each of the components in the equation below what type of reaction

To reduce 16 ounces of a 25% solution of antiseptic to a 10% solution, how much distilled water should a nurse add?

NemiM [27]

Let x = amt of distilled water
:
A simple equation
.25(16) = .10(x+16)
4 = .10x + 1.6

4 - 1.6 = .1x
2.4 = .1x
x = 2.4/0.1
x = 24 oz of distilled water
:
:
Prove this by seeing the amt of antiseptic is the same (only the % changes)
.25(16) = .10(24+16)
4 = .1(40)

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

8 0

3 years ago

Read 2 more answers

If we mix 25 grans if sodium oxide with a large amount of potassium chloride, how many grams of sodium chloride should be produc

amid [387]

Answer:

47.2 g

Explanation:

Data given:

mass of Sodium oxide (Na₂O) = 25 grams

mass of potassium chloride (KCl) = excess

amount of sodium chloride (NaCl) = ?

Solution:

First we look for reaction of sodium oxide with potassium chloride.

Na₂O + 2 KCl -----------> 2 NaCl + K₂O

As we know that potassium chloride is in excess amount and sodium oxide is 25 g so it means sodium oxide is a limiting reactant and amount of sodium chloride depends on the amount of sodium oxide.

So, now we will look for mole mole ration of Na₂O to NaCl.

Na₂O + 2 KCl -----------> 2 NaCl + K₂O

1 mol 2 mol

from above equation we come to know that 1 mole of Na₂O gives 2 moles of NaCl.

Now convert moles to mass

As we Know

Molar mass of Na₂O = 2(23) + 16 = 62 g/mol

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

So

Na₂O + 2 KCl -----------> 2 NaCl + K₂O

1 mol (62 g/mol) 2 mol (58.5 g/mol)

62 g 117 g

So, it means that 62 g of Na₂O produce 117 g of NaCl then how many grams of NaCl will be produce by 25 g of Na₂O

Apply unity formula

62 g of Na₂O ≅ 117 g of NaCl

25 g of Na₂O ≅ X g of NaCl

Do cross multiplication

X g of NaCl = 117 g x 25 g / 62 g

X g of NaCl = 47.2 g

So,

25 g of Na₂O gives 47.2 grams of NaCl.

4 0

3 years ago

What effects does the concentration of reactants have on the rate of a reaction ?

Harlamova29_29 [7]

Question:

What effects does the concentration of reactants have on the rate of a reaction?

Answer:

Reactant concentration. Increasing the concentration of one or more reactants will often increase the rate of reaction. This occurs because a higher concentration of a reactant will lead to more collisions of that reactant in a specific time period.

Increasing the concentration of reactants generally increases the rate of reaction because more of the reacting molecules or ions are present to form the reaction products. ... When concentrations are already high, a limit is often reached where increasing the concentration has little effect on the rate of reaction.

Hope this helps, have a good day. c;

5 0

3 years ago

one of the tallest dog breeds is the great dane which stands over a meter tall. one of the shortest is the pomeranian which stan

Gemiola [76]

We know that domestic dogs are descended from wolves. Exactly when and how this occurred is still being studied; many experts indicate that dogs were first domesticated about 12,000 years ago, but a few DNA studies indicate a much earlier date for the split between the wolf and the dog, perhaps as long ago as 130,000 years. Regardless of the exact timing of the divergence between wolves and dogs, one thing is true: both species display quite a bit of genetic diversity, particularly in size and coloration.In order to breed a dog of a specific size, all one has to do is choose parents with the appropriate variations. For instance, if you want to create a small dog breed, you would just breed the smallest dogs you have, then you would choose their smallest offspring to breed the next generation, and so on until you have reached the desired size.

8 0

3 years ago

Read 2 more answers

A gas fills a balloon at a temperature of 27 C and 101.325 kPa of pressure. What will the pressure of the balloon be if the gas

JulijaS [17]

135.1‬kPa

Explanation:

Given parameters:

T1 = 27°C

P1 = 101.325 kPa

T2 = 127°C

Unknown:

P2 = ?

Solution:

Using a derivative of the combined gas law where we assume that the gas has a constant volume, we can solve for the unknown.

At constant volume:

$\frac{P1}{T1} = \frac{P2}{T2}$

P1 is the initial pressure

T1 is the initial temperature

P2 is the final pressure

T2 is the final temperature

Take the given temperature to K

T1 = 27 + 273 = 300K

T2 = 127 + 273 = 400K

Input the variables:

$\frac{101.325}{300} = \frac{P2}{400}$

P2 = 135.1‬kPa

learn more:

Boyle's law brainly.com/question/8928288

#learnwithBrainly

3 0

3 years ago