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devlian [24]
3 years ago
9

What volume is needed to make a 2.45 M solution of KCl using 0.50 mol of KCl

Chemistry
1 answer:
Ivan3 years ago
4 0

Answer:

Explanation:

91.4

grams

Explanation:

C

=

m

o

l

v

o

l

u

m

e

2.45

M

=

m

o

l

0.5

L

2.45

M

⋅

0.5

L

=

m

o

l

m

o

l

=

1.225

Convert no. of moles to grams using the atomic mass of K + Cl

1.225

m

o

l

⋅

(

39.1

+

35.5

)

g

m

o

l

1.225

m

o

l

⋅

74.6

g

m

o

l

=

1.225

⋅

74.6

g

=

91.4

g

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Analysis of skunk spray yields a molecule with 44.77% c, 7.46% h and 47.76% s by mass. what is the empirical formula for this mo
stiv31 [10]
Suppose we have 100 gr of the substance. Then by weight, it would contain 44.77 gr of C, 7.46 gr of H and 47.76 gr of S. We need to look up the atomic weights of these atoms; M_H=1, M_C=12, M_S=32. The following formula holds (where n are the moles of the substance, M its molecular mass and m its mass): n=m/M. Substituting the known quantities for each element, we get that the substance has 3.73 moles of C, 7.46 moles of H and 1.49 moles of S. In the empirical formula for the molecule, all atoms appear an integer amout of times. Hence, for every mole of Sulfur, we have 2.5 moles of C and 5 moles of H (by taking the moles ratios). Thus, for every 2 moles of sulfur, we have 5 moles of C and 10 moles of H. Now that all the coefficients are integer, we have arrived at an empirical formula for the skunk spray agent: C_5H_{10}S_2
4 0
4 years ago
Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.7 g of ethane is m
Bond [772]

Answer:

m_{H_2O}=4.86gH_2O

Explanation:

Hello,

In this case, the described chemical reaction is:

C_2H_6+\frac{7}{2} O_2\rightarrow 2CO_2+3H_2O

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

n_{C_2H_6}=2.7gC_2H_6*\frac{1molC_2H_6}{30gC_2H_6} =0.09molC_2H_6

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

n_{C_2H_6}^{consumed\ by \ O_2}=13.0gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6}{\frac{7}{2} molO_2}=0.116molC_2H_6

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

m_{H_2O}=0.09molC_2H_6*\frac{3molH_2O}{1molC_2H_6} *\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=4.86gH_2O

Best regards.

3 0
3 years ago
The equilibrium constant (Kp) for the decomposition of phosphorus pentachloride (PCl5) to phosphorus trichloride (PCl3) and mole
Oksanka [162]

Answer: The partial pressure of Cl_2 is 1.86 atm

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_c

The given balanced equilibrium reaction is,

                            PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)

Pressure at eqm.     0.973 atm                 0.548atm      x atm

The expression for equilibrium constant for this reaction will be,

K_c=\frac{(p_{PCl_3}\times (p_{Cl_2})}{(p_{PCl_5})}

Now put all the given values in this expression, we get :

1.05=\frac{(0.548)\times (x)}{(0.973)}

By solving the term 'x', we get :

x = 1.86 atm

Thus, the partial pressure of Cl_2 is 1.86 atm

4 0
4 years ago
In which of the following reactions is chlorine (Cl) oxidized? Br2 + 2Cl− → Cl2 + 2Br− Cl2 + 2e− → 2Cl− 2ClO3− + 12H+ → Cl2 + 6H
Tomtit [17]

2CIO3 is the answer to your question

7 0
3 years ago
Read 2 more answers
Help and explain step by step if you can!
Leona [35]

Answer:

the third one

Explanation:

if you look at it mathematically its the one that makes the most sense

4 0
3 years ago
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