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matrenka [14]
3 years ago
15

Como se calcula la relación volumen-volumen?

Chemistry
1 answer:
Reil [10]3 years ago
3 0

Answer:

yes

Explanation:

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Sally does the work in 2.3 hours and Pete does 2.5
Yuri [45]
Pete because 2.5 x 2.3 is faster
5 0
3 years ago
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What is the difference between boiling point and normal boiling point?​
Softa [21]

Answer:

The boiling point is the temperature at which the vapor pressure equals the pressure of gas.

The normal boiling point is the temperature at which the vapor pressure equals one atmosphere

Explanation:

5 0
3 years ago
If i have 340mL of a 1.5 M NaBr solution, What will the concentration be for 1000mL?
Gekata [30.6K]

Answer:

0.51M

Explanation:

Given parameters:

Initial volume of NaBr = 340mL

Initial molarity  = 1.5M

Final volume  = 1000mL

Unknown:

Final molarity = ?

Solution;

This is a dilution problem whereas the concentration of a compound changes from one to another.

In this kind of problem, we must establish that the number of moles still remains the same.

    number of moles initially before diluting = number of moles after dilution

Number of moles  = Molarity x volume

Let us find the number of moles;

          Number of moles  = initial volume x initial molarity

Convert mL to dm³;

                  1000mL  = 1dm³

                     340mL gives \frac{340}{1000}   = 0.34dm³

Number of moles  = initial volume x initial molarity  = 0.34 x 1.5 = 0.51moles

Now to find the new molarity/concentration;

               Final molarity  = \frac{number of moles}{Volume}    = \frac{0.51}{1}    = 0.51M

We can see a massive drop in molarity this is due to dilution of the initial concentration.

6 0
3 years ago
What is the percent yield of ferrous sulfide if the actual yield is 220.0 g and the theoretical yield is 275.6 g
Anika [276]
To get the percent yield, we will use this formula: ((Actual Yield)/(Theoretical Yield)) * 100% Values given: actual yield is 220.0 g theoretical yield is 275.6 g Now, let us substitute the values given. (220.0 grams)/(275.6 grams) = 0.7983 Then, to get the percentage, multiply the quotient by 100. 0.7983 (100) = 79.83% Among the choices, the most plausible answer is 79.8% <span>
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4 0
3 years ago
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How many grams of argon would it take to fill a large light bulb with a volume of 0.745 L at STP?
EastWind [94]

Answer:

Mass = 1.33 g

Explanation:

Given data:

Mass of argon required = ?

Volume of bulb = 0.745 L

Temperature and pressure = standard

Solution:

We will calculate the number of moles of argon first.

Formula:

PV = nRT

R = general gas constant = 0.0821 atm.L/mol.K

By putting values,

1 atm ×0.745 L = n × 0.0821 atm.L/mol.K× 273.15 K

0.745 atm. L = n × 22.43 atm.L/mol

n = 0.745 atm. L / 22.43 atm.L/mol

n = 0.0332 mol

Mass of argon:

Mass = number of moles × molar mass

Mass = 0.0332 mol × 39.95 g/mol

Mass = 1.33 g

8 0
3 years ago
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