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Lunna [17]
3 years ago
8

A student plans to use a density versus solution concentration standard curve to identify the sodium chloride concentration in a

n unknown salt water solution. The unknown should have a concentration between 3.0% and 6.0% by mass of sodium chloride. If the student prepares each standard by adding some mass of sodium chloride to 30.0 grams of Dl water, which of the following mass sets of sodium chloride will result in an appropriate standard curve? Select all that apply.
a. 0.50 g, 1.0 g, 1.5 g, 2.0 g, 2.5 g
b. 2.0 g, 4.0 g, 8.0 g, 10.0 g, 12.0g
c. 0.1 g, 0.2 g, 0.3 g, 0.4 g, 1.0 g
d. 0.1 g, 0.3 g, 0.6 g, 0.9 g, 1.2 g,
e. 2.0 g, 3.0 g, 4.0 g, 5.0 g. 6,0 g
f. 0.20 g, 0.40 g, 0.80 g, 1.0 g. 1.2 g
g. 0.1g,0.5 g, 1.0 g, 1.5 g, 2.0 g
Chemistry
1 answer:
Art [367]3 years ago
7 0

Answer:

a. 0.50 g, 1.0 g, 1.5 g, 2.0 g, 2.5 g;

g. 0.1 g, 0.5 g, 1.0 g, 1.5 g, 2.0 g

Explanation:

The percent mass is defined as a ratio between the mass of a solute and mass of a solution:

\omega = \frac{m_{solute}}{m_{solution}}\cdot 100\%

Since solution only consists of a solute and solvent, express its mass as:

m_{solution} = m_{solute} + m_{solvent}

Then:

\omega = \frac{m_{solute}}{m_{solution}}\cdot 100\%=\frac{m_{solute}}{m_{solute} + m_{solvent}}

Firstly, solve for how much mass is required to prepare 3.0 %. Let's say, we have x g of the solute:

0.03 = \frac{x}{30.0 + x}\therefore x = 0.03(30.0 + x)

x = 0.90 + 0.03x

0.97x = 0.90\therefore x = 0.93 g

Similarly, solve for 6.0 %, let's say, we have x g of the solute again:

0.06 = \frac{x}{30.0 + x}\therefore x = 0.06(30.0 + x)

x = 1.80 + 0.06x

0.94x = 1.80\therefore x = 1.91 g

Hence, masses should be in a range of 0.93 g to 1.91 g.

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In AG calculations, temperature is expressed in a. degrees Celsius. b. kelvins. c. degrees Fahrenheit, d. kilojoules.
schepotkina [342]

Answer:

The correct option is: a. degrees Celsius

Explanation:

The anion gap is the difference in the cations and anions in plasma, serum or urine, calculated from medical lab test results. It can be calculated by measuring the concentration of the anions or cations, which are expressed in millimoles/litre (mmol/L) or milliequivalents/liter (mEq/L).

The temperature in this test is expressed in degrees Celsius (°C).

6 0
3 years ago
CaC2(s) + 2H2O(l) --> Ca(OH)2(aq) + C2H2(g) In the reaction above, 0.5487 grams of calcium carbide are completely consumed to
jasenka [17]

Answer:

239.7mL

Explanation:

Using the general gas equation;

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas constant (0.0821 Latm/molK)

T = temperature (K)

The balanced chemical equation in this question is as follows:

CaC2(s) + 2H2O(l) --> Ca(OH)2(aq) + C2H2(g)

From the equation, 1 mole of CaC2 produces 1 mole of ethylene gas, C2H2.

Using mole = mass/molar mass

Molar mass of CaC2 = 40 + 12(2)

= 40 + 24

= 64g/mol

mole = 0.5487/64

mole = 0.00857mol of CaC2

Hence, 0.00857mol of CaC2 produced 0.00857mol of C2H2

Based on the information provided, n = 0.00857mol, T = 43°C = 43 + 273 = 316K, p = 0.926 atm

PV = nRT

V = nRT/P

V = 0.00857 × 0.0821 × 316/0.926

V = 0.222/0.926

V = 0.2397L

In mL, volume = 0.2397 × 1000

= 239.7mL

3 0
3 years ago
A family pool holds 10,000 gallons of water. How many cubic centimeters is this?
Ierofanga [76]

Answer:

3.785412e+7cm³

Explanation:

Have a nice day.

3 0
2 years ago
A sample of a gas has volume 78.5ml at 318.15 K.. What Volume at will the sample occupy at 273. 15 K when the pressure is held c
DochEvi [55]

Charles’ Law

V₁/T₁=V₂/T₂

78.5/318.15=V₂/273.15

V₂=67.4 ml

7 0
1 year ago
what is the percent by mass of NaHCO3 in a solution containing 20.0 g of NaHCO3 dissolved in 600.0 ml of water
Ne4ueva [31]
%(NaHCO3)= ((mass NaHCO3)/(mass NaHCO3 + mass water))*100%

m=Volume*Density
Density of water =1 g/ml
m(water) = Volume(water)*Density(water) = 600.0 ml * 1g/ml=600g water

%(NaHCO3)= ((20.0 g)/(20.0 g + 600 g))*100%=0.0323*100%=32.3%
7 0
3 years ago
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