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dusya [7]
2 years ago
12

Copy the lists of measurements shown below. Pay close attention to the units that follow each number. List 1: 150 mL 11 mL 200 m

L List 2: 2 mL 801 mL 27 cm3 List 3: 1 L 999 mL 998 cm3
a. Cross out the smallest volume in each list. b. Circle the largest volume in each list.
Chemistry
2 answers:
stich3 [128]2 years ago
7 0

a) cross out 11mL from list 1, 2 mL from list 2, and 998 cm3 from list 3.

b) circle 200 mL from list 1, 801 mL from list 2, and 1 L from list 3.

zlopas [31]2 years ago
5 0
Cm^3 and mL are equals. 1000 mL = 1L. :) hope that helps you out!!
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How many grams of sodium oxide can be produced when 55.3 g Na react with 64.3 g O2?. . Unbalanced equation: Na + O2 → Na2O. . Sh
GenaCL600 [577]
 <span>this is a limiting reagent problem. 

first, balance the equation 
4Na+ O2 ---> 2Na2O 

use both the mass of Na and mass of O2 to figure out how much possible Na2O you could make. 
start with Na and go to grams of Na2O 

55.3 gNa x (1molNa/23.0gNa) x (2 molNa2O/4 molNa) x (62.0gNa2O/1molNa2O) = 75.5 gNa2O 

do the same with O2 

64.3 gO2 x (1 molO2/32.0gO2) x (2 molNa2O/1 mol O2) x (62.0gNa2O/1molNa2O) = 249.2 g Na2O 

now you must pick the least amount of Na2O for the one that you actually get in the reaction. This is because you have to have both reacts still present for a reaction to occur. So after the Na runs out when it makes 75.5 gNa2O with O2, the reaction stops. 

So, the mass of sodium oxide is 

75.5 g</span>
3 0
3 years ago
Read 2 more answers
171 g of sucrose ( MW of 342, melting point 186 oC, boiling point very high, and vapor pressure is negligible) is dissolved in o
Fantom [35]

The complete question is as follows: 171 g of sucrose ( MW of 342, melting point 186 oC, boiling point very high, and vapor pressure is negligible) is dissolved in one liter of water at 25 oC. At 25 oC the vapor pressure of water is 24 mmHg. Which value is closest to the vapor pressure (VP) of this solution at 25 oC?

a. 16mm Hg

b. 24mm Hg  

c. 20mm Hg  

d. 12mm Hg

Answer: The vapor pressure (VP) of this solution at 25^{o}C is closest to the value 24 mm Hg.

Explanation:

Given: Mass of sucrose = 171 g

Mass of water = 1 L = 1000 g

Vapor pressure of water = 24 mm Hg

As moles is the mass of substance divided by its molar mass. Hence, moles of water (molar mass = 18.02 g) is calculated as follows.

Moles = \frac{mass}{molar mass}\\= \frac{1000 g}{18.02 g/mol}\\= 55.49 mol

Similarly, moles of sucrose (molar mass = 342 g/mol) is as follows.

Moles = \frac{mass}{molar mass}\\= \frac{171 g}{342 g/mol}\\= 0.5 mol

Total moles = 55.49 + 0.5 mol = 55.99 mol

Mole fraction of water is as follows.

Mole fraction = \frac{moles of water}{total moles}\\= \frac{55.49}{55.99}\\= 0.99

Formula used to calculate vapor pressure of the solution is as follows.

P_{i} = P^{o}_{i} \times \chi_{i}

where,

P_{i} = vapor pressure of component i over the solution

P^{o}_{i} = vapor pressure of pure component i

\chi_{i} = mole fraction of i

Substitute the values into above formula to calculate vapor pressure of water as follows.

P_{i} = P^{o}_{i} \times \chi_{i}\\= 24 mm Hg \times 0.99\\= 23.76 \\or 24 mm Hg\\

Thus, we can conclude that the vapor pressure (VP) of this solution at 25^{o}C is closest to the value 24 mm Hg.

4 0
3 years ago
Which statement correctly compares sound and light waves?
Mrac [35]

Answer:

Light waves carry energy parallel to the motion of the wave, while sound waves carry energy perpendicu

Explanation:

5 0
3 years ago
The type of music represents which variable?
gayaneshka [121]

Answer:

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3 0
3 years ago
Read 2 more answers
First, get about 20 mL of the quinine stock solution in a clean beaker to prepare 250 mL of about 2 ppm quinine in 0.05 M H2SO4.
iris [78.8K]

Answer:

Volume of acid, Va=250mL; Volume of quinine,Vb=20mL; Molarity of acid, Ma=0.05M.

Molar mass of acid= H2+S+O4= 2+32+(16X4)= 2+32+64=98g

Concentration of acid, Ca= Molar mass of acid/ Ma =98/0.05=1960g/mol

Explanation: To calculate concentration of quinine, Cb is as follow

Va*Ca=Vb*Cb

∴ Cb=Va*Ca/Vb =250*1960/20 =24500g/mol

5 0
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