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NeTakaya
4 years ago
14

Solve the following mole-mass problem. Refer to the periodic table. How much mass does 0.50 mole of copper contain?

Chemistry
2 answers:
andre [41]4 years ago
7 0
It is 31.75g of copper is contained in .50 mole of copper.
s344n2d4d5 [400]4 years ago
4 0

0.50 moles of copper is equal to 32 g .

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Which of the following shows correctly an ion pair and ionic compound the two ions from
zubka84 [21]

Answer:

what are the choices?

Explanation:

4 0
3 years ago
What is the molar concentration of 35 mL of H2SO4 that neutralizes 25 mL of 0.320M NaOH
stira [4]
V ( H2SO4) = 35 mL / 1000 => 0.035 L

M ( H2SO4) = ?

V ( NaOH ) = 25 mL / 1000 => 0.025 L 

M ( NaOH ) = 0.320 M

number of moles NaOH:

n = M x V

n = 0.025 x 0.320 => 0.008 moles of NaOH

Mole ratio:

<span>2 NaOH + H2SO4 = Na2SO4 + 2 H2O
</span>
2 moles NaOH ---------------------- 1 mole H2SO4
0.008 moles moles NaOH ---------- ??

0.008 x 1 / 2 => 0.004 moles of H2SO4 :

Therefore:

M ( H2SO4) = n / V

M = 0.004 /  0.035

= 0.114 M

hope this helps!



6 0
4 years ago
A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.
irga5000 [103]

Answer:

1. pH = 1.23.

2. H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

pH=pKa+log(\frac{[base]}{[acid]} )

Whereas the pKa is:

pKa=-log(Ka)=-log(5.90x10^{-2})=1.23

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Which is also shown in net ionic notation.

Best regards!

4 0
3 years ago
What should the temperature of the detergent water be in a 3 compartment sink
kumpel [21]
It should be At least 110°F
3 0
3 years ago
Read 2 more answers
what modifications to either or both of the half-cell reactions would kim need to make if she wanted to use them to produce equa
kompoz [17]

We have to add the both half cell equations and eliminate the number of electrons lost/gained.

<h3>What modification must Kim make to the equations?</h3>

The term redox reaction is a type of reaction that occurs when an electron  is lost or gained in a reaction system. We can see that in this reaction, zinc looses two electron which are gained by copper.

If we want to obtain the equation 4.9 which is the overall equation of the redox reaction from the various half cell equations then we have to add the both half cell equations and eliminate the number of electrons lost/gained.

Learn kore about redox reaction:brainly.com/question/13293425

#SPJ1

7 0
2 years ago
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