3. Determine the uses of the following materials:

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

NemiM [27]

<span>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!
</span>

8 0

2 years ago

Read 2 more answers

Please show the work :(What is the estimated force applied to the box if the acceleration is .40 m/s2?

lara31 [8.8K]

Answer:

20N

Explanation:

Given parameters:

Force(N) Acceleration(m/s²)

10 0.2

? 0.4

Unknown:

The force applied when the acceleration is 0.4m/s²

Solution:

From newton's second law of motion;

Force = mass x acceleration

Since we are using the same box, let us find the mass of the box;

Force = mass x acceleration

10 = mass x 0.2

mass = $\frac{10}{0.2}$ = 50kg

Now,

The force in the second instance will be;

Force = 50 x 0.4 = 20N

8 0

2 years ago

Hi can someone please help me!

Ratling [72]

Answer:

a) 2

b) 2

c) 5

d) 5

e) 5

Explanation:

a) There is 1 Ag atom and 1 Cl atom. <em>When there's no subscript number next to an element, it means there is only one.</em>

b) There is 1 Ca atom and 1 O atom.

c) There are 3 Mg atoms (there's a subscript 3 next to Mg) and 2 N atoms.

d) There are 2 Al atoms and 3 O atoms.

e) There are 2 Sc atoms and 3 S atoms.

7 0

2 years ago

A student weighs an empty flask and stopper and finds the mass to be 55.844 g. She then adds about 5 mL of an unknown liquid and

Oduvanchick [21]

Answer :

(a) The pressure of the vapor in the flask in atm is, 0.989 atm

(b) The temperature of the vapor in the flask in Kelvin is, 372.7 K

The volume of the flask in liters is, 0.2481 L

(c) The mass of vapor present in the flask was, 0.257 g

(d) The number of moles of vapor present are 0.00802 mole.

(e) The mass of one mole of vapor is 32.0 g/mole

Explanation : Given,

Mass of empty flask and stopper = 55.844 g

Volume of liquid = 5 mL

Temperature = $99.7^oC$

Mass of flask and condensed vapor = 56.101 g

Volume of flask = 248.1 mL

Barometric pressure in the laboratory = 752 mmHg

(a) First we have to determine the pressure of the vapor in the flask in atm.

Pressure of the vapor in the flask = Barometric pressure in the laboratory = 752 mmHg

Conversion used :

$1atm=760mmHg$

or,

$1mmHg=\frac{1}{760}atm$

As, $1mmHg=\frac{1}{760}atm$

So, $752mmHg=\frac{752mmHg}{1mmHg}\times \frac{1}{760}atm=0.989atm$

Thus, the pressure of the vapor in the flask in atm is, 0.989 atm

(b) Now we have to determine the temperature of the vapor in the flask in Kelvin.

Conversion used :

$K=273+^oC$

As, $K=273+^oC$

So, $K=273+99.7=372.7$

Thus, the temperature of the vapor in the flask in Kelvin is, 372.7 K

Now we have to determine the volume of the flask in liters.

Conversion used :

1 L = 1000 mL

or,

1 mL = 0.001 L

As, 1 mL = 0.001 L

So, 248.1 mL = 248.1 × 0.001 L = 0.2481 L

Thus, the volume of the flask in liters is, 0.2481 L

(c) Now we have to determine the mass of vapor that was present in the flask.

Mass of flask and condensed vapor = 56.101 g

Mass of empty flask and stopper = 55.844 g

Mass of vapor in flask = Mass of flask and condensed vapor - Mass of empty flask and stopper

Mass of vapor in flask = 56.101 g - 55.844 g

Mass of vapor in flask = 0.257 g

Thus, the mass of vapor present in the flask was, 0.257 g

(d) Now we have to determine the number of moles of vapor present.

Using ideal gas equation:

PV = nRT

where,

P = Pressure of vapor = 0.989 atm

V = Volume of vapor = 0.2481 L

n = number of moles of vapor = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of vapor = 372.7 K

Putting values in above equation, we get:

$(0.989atm)\times 0.2481L=n\times (0.0821L.atm/mol.K)\times 372.7K\\\\n=0.00802mole$

Thus, the number of moles of vapor present are 0.00802 mole.

(e) Now we have to determine the mass of one mole of vapor.

$\text{Mass of one mole of vapor}=\frac{\text{Mass of vapor}}{\text{Moles of vapor}}$

$\text{Mass of one mole of vapor}=\frac{0.257g}{0.00802mole}=32.0g/mole$

Thus, the mass of one mole of vapor is 32.0 g/mole

8 0

3 years ago

This is for chemistry:/

Andreyy89

Answer:

Its the bottom one :) ......

5 0

3 years ago