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3 years ago

If HCr2O7 is titrated with Ca(OH)2, what is the balanced chemical equation of this neutralization reaction?

Chemistry

1 answer:

Harrizon [31]3 years ago

6 0

Answer:

2HCr2O7(aq) + Ca(OH)2(aq) --> 2H2O(l) + Ca(Cr2O7)2

Explanation:

A neutralization reaction produces water and a salt. Since Ca(OH)2 has two hydroxides, two water molecules are produced, which means you also need two molecules of the acid. Having two molecules of the acid also ensures that you have two dichromates on both sides.

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marysya [2.9K]

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3 years ago

The chart below lists data on four different projects designed to restore a wetlands habitat destroyed by human activity and a r

snow_lady [41]

Answer:

Project 3.

Explanation:

Project 3's anticipated cost is 12 to 17 million dollars. It is a lower anticipated cost than Project 2 and Project 4, but higher than Project 1 by one million dollars at maximum cost anticipation. Additionally, the percentage of wildlife to benefit is 70-80%, which is second to the most wildlife to benefit which is Project 4 at 75-80%.

And finally, for community support for Project 3 - the chart lists it as high. This outclasses Project 2 and Project 4, but balances with Project 1. However, Project 1 costs 13 to 16 million dollars and only benefits 15-25% of wildlife.

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3 years ago

At the end of mitosis the cells produced are identical to each other is it True or False

My name is Ann [436]

True:

The result is two genetically identical daughter nuclei.

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3 years ago

If acetic acid is the only acid that vinegar contains (ka=1.8×10−5), calculate the concentration of acetic acid in the vinegar.

kicyunya [14]

Ethanoic (Acetic) acid is a weak acid and do not dissociate fully. Therefore its equilibrium state has to be considered here.

$CH_{3}COOH \ \textless \ ---\ \textgreater \ H^{+} + CH_{3}COO^{-}$

In this case pH value of the solution is necessary to calculate the concentration but it's not given here so pH = 2.88 (looked it up)

pH = 2.88 ==> $[H^{+}]$ = $10^{-2.88}$ = 0.001 $moldm^{-3}$

The change in Concentration Δ $[CH_{3}COOH]$ = 0.001 $moldm^{-3}$

  CH3COOH H+ CH3COOH
Initial   $moldm^{-3}$      $x$ 0 0

Change $moldm^{-3}$ -0.001 +0.001 +0.001

Equilibrium $moldm^{-3}$ $x$ - 0.001 0.001 0.001


Since the $k_{a}$ value is so small, the assumption
$[CH_{3}COOH]_{initial} = [CH_{3}COOH]_{equilibrium}$ can be made.

$k_{a} = [tex]= 1.8*10^{-5} = \frac{[H^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]} = \frac{0.001^{2}}{x}$

Solve for x to get the required concentration.

note: 1.)Since you need the answer in 2SF don&t round up values in the middle of the calculation like I've done here.

2.) The ICE (Initial, Change, Equilibrium) table may come in handy if you are new to problems of this kind

Hope this helps!

8 0

3 years ago

Group I element are very active.give reason

RUDIKE [14]

Answer:

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Explanation:

7 0

3 years ago