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3 years ago

in order to dilute 1.0L of a 6.00M solution of sodium hydroxide to 0.500M concentration, how much water (in mL) must you add?

Chemistry

2 answers:

Shtirlitz [24]3 years ago

8 0

To solve this we use the equation,

M1V1 = M2V2

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

6.00 M ( 1.0 L ) = ( 0.500 M ) V2

V2 = 12 L

Therefore, approximately 11 L of water should be added to the 1 L of 6.00 M solution. Hope this answers the question. Have a nice day.

quester [9]3 years ago

5 0

Another M1V1=M2V2
V1=1.0L 
M1=6.00M 
M2=0.500M 
V2=?? 
(6.00M)(1.0L)=(0.500M)V2 
Solve for V2 you get 12L or 12000mL

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Calculate the average atomic mass of an element containing 39.9% of atoms with a mass of 69 amu and 60.1% of atoms with a mass o

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Answer:

70.202 amu

Explanation:

From the question given above, the following data were obtained:

Isotope A:

Abundance (A%) = 39.9%

Mass of A = 69 amu

Isotope B:

Abundance (B%) = 60.1%

Mass of B = 71 amu

Average atomic mass =?

The average atomic mass of the element can be obtained as follow:

Average atomic mass = [(Mass of A × A%)/100] + (Mass of B × B%)/100]

= [(69 × 39.9)/100] + [(71 × 60.1)/100]

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4 0

3 years ago

A solution of permanganate is standardized by titration with oxalic acid, . To react completely with mol of oxalic acid required

Zielflug [23.3K]

Answer:

$M=0.120M$

Explanation:

Hello,

In this case, the undergone chemical reaction is:

$MnO_4^-(aq)+H_2C_2O_4(aq)\rightarrow Mn^{+2}+CO_2$

In such a way, the acidic redox balance turns out:

$(Mn^{+7}O_4)^-+5e^-+8H^+\rightarrow Mn^{+2}+4H_2O\\H_2C_2O_4\rightarrow2CO_2+2H^++2e^-$

Which leads to the total balanced equation as follows:

$2(MnO_4)^-(aq)+6H^+(aq)+5H_2C_2O_4(aq)\rightarrow2Mn^{+2}(aq)+8H_2O(l)+10CO_2(g)$

Thus, as the mass of oxalic acid is not given, one could suppose a value of 1 g (which you can modify based on the actual statement) in order to compute the oxalic acid moles as shwon below:

$1gH_2C_2O_4*\frac{1molH_2C_2O_4}{90.04gH_2C_2O_4} *\frac{2mol(MnO_4)^-}{5molH_2C_2O_4} =0.00444mol(MnO_4)^-$

Whereby the molality results:

$M=\frac{0.00444mol(MnO_4)^-}{0.03702L} =0.120M$

Remember you can modify the oxalic acid mass as you desire.

Best regards.

5 0

4 years ago