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3 years ago

A scientist has two solutions, which she has labeled solution a and solution

1 answer:

6 0

To determine the amount of each solution that she should use, we do mass balance for the whole process. We let x be the amount the 65% salt and y the amount of the 90% salt. We do as follows:

The total mass should be equal to 150 oz.

x + y = 150

Doing a salt balance, we obtain the second equation:
.65x + .90y = 150(.85)

Solving for x and y, we obtain:

x = 30 oz
y = 120 oz

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0.0625 mol of an ideal gas is contained within a 1.0000 Lvolume container. Its pressure is 142,868 Pa and temperature is X K . W

aliya0001 [1]

<u>Answer:</u> The temperature of the ideal gas is $2.75\times 10^2K$

<u>Explanation:</u>

To calculate the temperature, we use the equation given by ideal gas equation:

$PV=nRT$

where,

P = Pressure of the gas = 142,868 Pa = 142.868 kPa (Conversion factor: 1 kPa = 1000 Pa)

V = Volume of gas = 1.0000 L

n = number of moles of ideal gas = 0.0625 moles

R = Gas constant = $8.31\text{L kPa }mol^{-1}K^{-1}$

T = temperature of the gas = ?

Putting values in above equation, we get:

$142.868kPa\times 1.0000=0.0625mol\times 8.31\text{L kPa }mol^{-1}K^{-1}\times T\\\\T=275K=2.75\times 10^2K$

Hence, the temperature of the ideal gas is $2.75\times 10^2K$

8 0

4 years ago

A chemistry student needs 45.0mL of pentane for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the stud

Masja [62]

The mass of pentane the student should weigh out is

The density of pentane is 0.626 gcm-3

To calculate the mass of pentane following expression is used,

(Density is defined as the mass divide by volume)

Density = mass / volume

mass of pentane = Density of pentane * Volume of pentane

mass of pentane = 0.626 gcm-3 * 45.0 mL

= 28.17 g

Here the unit of mass of pentane is g,

However the unit of density is gcm-3 and unit of volume is mL i.e. cm3

Hence, Mass = gcm-3 * cm3

Mass = g

The mass of pentane the student should weigh out is 28.17g

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5 0

2 years ago

How many protons and neutrons are in 119_Sn? O a. 50 n and 119 p O b. 50 p and 169 1 O c. 50 p and 69 O d. 50 n and 169 p O e. N

Anastaziya [24]

<u>Answer:</u> The given isotope of tin has 50 protons and 69 neutrons.

<u>Explanation:</u>

Atomic number is defined as the number of protons or number of electrons that are present in neutral atom. It is represented as Z.

Atomic number = Number of protons = Number of electrons

Mass number is defined as the sum of number of protons and number of neutrons. It is represented as A.

Mass number = Number of protons + Number of neutrons

We are given:

An isotope having representation $_{50}^{119}\textrm{Sn}$

Mass number of Sn = 19

Atomic number = 50

Number of neutrons = Mass number - Atomic number = 119 - 50 = 69

Hence, the given isotope of tin has 50 protons and 69 neutrons.

3 0

3 years ago

What type of mixture is throughly mixing powdered lemonade with water

vagabundo [1.1K]

Answer:

I believe it would be a solution I could be wrong tho

Explanation:

8 0

2 years ago

Read 2 more answers

A client who weighs 70 kg is receiving a solution of 0.9% sodium chloride (normal saline) 500 ml with dopamine 800 mg at 5 ml/ho

nordsb [41]

The mcg/kg/minute the client is receiving is 1.904 mcg/kg/min whose weight is 70 kg.

1 milligram (mg) = 1000 micrograms (mcg) or 0.001 grams (g) 1 g = 1000 mg 1 kilogram (kg) = 1000 g 1 kg = 2.2 pound (lb) 1 liter (L) = 1000 milliliters (mL)

The number of mcg/kg/minute = flow rate × concentration ÷ mass of client

Flow rate = 12 ml/hour = 12 ml/hour × 1 hr/60 min = 0.2 ml/min

Concentration = mass of dopamine/volume

where

mass of dopamine = 800 mg and

volume = 500 ml

Concentration =800 mg/500 ml

= 1.6 mg/ml

= 1.6 mg/ml × 1000 mcg/mg

= 1600 mcg/ml

mass of client = 70 kg

Calculating mcg/kg/minute

So, substituting the variables into the equation, we have

mcg/kg/minute = flow rate × concentration ÷ mass of client

mcg/kg/minute = 0.08 ml/min × 1600 mcg/ml ÷ 70 kg

mcg/kg/minute = 320 mcg/min ÷ 70 kg

mcg/kg/minute = 1.904 mcg/kg/min

Thus, the mcg/kg/minute the client is receiving is 1.904 mcg/kg/min

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7 0

2 years ago