Why is atomic radius not a well-defined quantity?

Suppose the concentration of the standard stock solution is 50.04 ppm and that you delivered 4.60 mL of this standard stock solu

12345 [234]

Answer:

9.21 ppm

Explanation:

Considering

$concentration_{working\ solution}\times Volume_{working\ solution}=concentration_{stock\ solution}\times Volume_{stock\ solution}$

Given that:

$concentration_{working\ solution}=?$

$Volume_{working\ solution}=25.00mL$

$Volume_{stock\ solution}=4.60mL$

$concentration_{stock\ solution}=50.04 ppm$

So,

$concentration_{working\ solution}\times 25.00mL=50.04 ppm\times 4.60mL$

$concentration_{working\ solution}=\frac{50.04\times 4.60}{25.00}\ ppm=9.21\ ppm$

4 0

3 years ago

An aqueous KNO3 solution is made using 76.6 g of KNO3 diluted to a total solution volume of 1.84 L. (Assume a density of 1.05 g/

defon

Answer:

The answer to your question is Molarity = 0.41

Explanation:

Data

mass of KNO₃ = 76.6 g

volume = 1.84 l

density = 1.05 g/ml

Process

1.- Calculate the molecular mass of KNO₃

molecular mass = 39 + 14 + (16 x 3) = 101 g

2.- Calculate the number of moles

101 g of KNO₃ --------------- 1 mol

76.6 g of KNO₃ ------------ x

x = (76.6 x 1) / 101

x = 0.76 moles

3.- Calculate molarity

Molarity = $\frac{number of moles}{volume}$

Substitution

Molarity = $\frac{0.76}{1.84}$

Result

Molarity = 0.41

7 0

3 years ago

The acetylene tank contains 35.0 mol C2H2, and the oxygen tank contains 84.0 mol O2.

harkovskaia [24]

Answer:- As per the question is asked, 35.0 moles of acetylene gives 70 moles of carbon dioxide but if we solve the problem using the limiting reactant which is oxygen then 67.2 moles of carbon dioxide will form.

Solution:- The balanced equation for the combustion of acetylene is:

$2C_2H_2(g)+5O_2(g)\rightarrow 4CO_2(g)+2H_2O(g)$

From the balanced equation, two moles of acetylene gives four moles of carbon dioxide. Using dimensional analysis we could show the calculations for the formation of carbon dioxide by the combustion of 35.0 moles of acetylene.

$35.0molC_2H_2(\frac{4molCO_2}{2molC_2H_2})$

= $70molCO_2$

The next part is, how we choose 35.0 moles of acetylene and not 84.0 moles of oxygen.

From balanced equation, there is 2:5 mol ratio between acetylene and oxygen. Let's calculate the moles of oxygen required to react completely with 35.0 moles of acetylene.

$35.0molC_2H_2(\frac{5molO_2}{2molC_2H_2})$

= $87.5molO_2$

Calculations shows that 87.5 moles of oxygen are required to react completely with 35.0 moles of acetylene. Since only 84.0 moles of oxygen are available, the limiting reactant is oxygen, so 35.0 moles of acetylene will not react completely as it is excess reactant.

So, the theoretical yield should be calculated using 84.0 moles of oxygen as:

$84.0molO_2(\frac{4molO_2}{5molO_2})$