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3 years ago

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A solution is prepared by mixing 38.0L of pyrimidine and 442.0L of water. What is the percent volume of pyrimidine in this solut

ion?

1 answer:

klasskru [66]3 years ago

0 0

I have the same question and cant still answer it so I need the answers

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What is the number of moles of solute in 250 mL of a 0.4 M solution?

mafiozo [28]

Answer:

0,1 mol

Explanation:

We know that the formula of concentration is C= moles of solute/ volume

0,4 M= moles of solute/ 250 mL

Convert mL to L 250 mL =0,25 L

0,4 M x 0,25 L= moles of solute

0,1 moles= moles of solute

3 0

3 years ago

Marking anybody who got it the brainliest

bija089 [108]

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2 years ago

In a lab experiment 80.0 g of ammonia [NH3] and 120 g of oxygen are placed in a reaction vessel. At the end of the reaction 72.2

valentinak56 [21]

The percent yield of the reaction : 89.14%

<h3>Further explanation</h3>

Reaction of Ammonia and Oxygen in a lab :

<em>4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)</em>

mass NH₃ = 80 g

mol NH₃ (MW=17 g/mol):

$\dfrac{80}{17}=4.706$

mass O₂ = 120 g

mol O₂(MW=32 g/mol) :

$\tt \dfrac{120}{32}=3.75$

Mol ratio of reactants(to find limiting reatants) :

$\tt \dfrac{4.706}{4}\div \dfrac{3.75}{5}=1.1765\div 0.75\rightarrow O_2~limiting~reactant(smaller~ratio)$

mol of H₂O based on O₂ as limiting reactants :

mol H₂O :

$\tt \dfrac{6}{5}\times 3.75=4.5$

mass H₂O :

4.5 x 18 g/mol = 81 g

The percent yield :

$\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{72.2}{81}\times 100\%=89.14\%$

6 0

3 years ago

The ________ ion is represented by the electron configuration [ar]3d2.

Tresset [83]

The Ti 2+ ions is represented by electron configuration (Ar)3d2. Titanium is in atomic number 22 and its electronic configuration is (Ar)3d2 4s2. Titanium loss two electron that is 4s2 electrons hence the electronic configuration ( Ar)3d2. 4s2 is the valence electron hence it the one which is lost to form Ti2+

3 0

3 years ago

Read 2 more answers

The reaction A(B) = 2B(g) has an equilibrium constant of K = 0.045. What is the equilibrium constant for the reaction B(g) =1/2A

Mamont248 [21]

Answer:

The $K_c$ for the reaction $B(g) = \frac{1}{2}A$ will be 4.69.

Explanation:

The given equation is A(B) = 2B(g)

to evaluate equilibrium constant for $B(g) = \frac{1}{2}A$

$K_c=[B]^2[A]$

= 0.045

The reverse will be $2B\leftrightharpoons A$

Then, $K_c = \frac{[A]}{[B]^2}$

= $\frac{1}{0.045}$

= $22m^{-1}$

The equilibrium constant for $B(g) = \frac{1}{2}A$ will be

$K_c = \sqrt{K_c}$

$=\sqrt{22}$

= 4.69

Therefore, $K_c$ for the reaction $B(g) = \frac{1}{2}A$ will be 4.69.

5 0

3 years ago