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lutik1710 [3]
3 years ago
5

The chemical standard for the quantity of a mole is based on which substance? a.hydrogen-1 b.helium-2 c.oxygen-16 d.carbon-12

Chemistry
1 answer:
jekas [21]3 years ago
5 0
D.carbon-12 is the answer.......
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Stronger acids dissociate completes and produce more ions.
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I want to use a galvanic cell to power a 60-watt light bulb. Complete the following steps to determine how long the galvanic cel
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Explanation:

On dissolution of chromium(II) sulfate is water 2 ions of chromium(III) ion and and 3 sulfate ions are formed.

The balanced chemical equations of the dissolution of chromium (II) sulfate:

Cr_2(SO_4)_3(aq)\rightarrow 2Cr^{3+}(aq)+3SO_4^{2-}(aq)

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The balanced chemical equations of the dissolution of gold(III) nitrate:

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6 0
3 years ago
What is the mole ratio of D to A in the generic chemical reaction? 4A + B --> C + D​
storchak [24]
<h3>Answer:</h3>

Mole ratio of D to A is 1 : 4

<h3>Explanation:</h3>

We are given the generic chemical equation;

4A + B → C + D​

We are supposed to determine the mole ratio of D to A

What is mole ratio?

  • Mole ratio is the ratio of the number of moles of reactants or products in a chemical reaction.
  • We determine the mole ratio using the coefficients of reactants or products in question.

For example;

  • In the equation, 4A + B → C + D​, the coefficient of A is 4 while the coefficient of D is 1.
  • This means, 4 moles of A reacts with 1 mole of b to produce 1 mole of C and 1 mole of D
  • Thus, mole ratio of D to A  is 1 : 4
6 0
3 years ago
1. In any nuclear reaction, nuclei are transformed in some way.
Grace [21]
<span>1. In any nuclear reaction, nuclei are transformed in some way.

True

2. Two fissionable substances commonly used in nuclear reactors and nuclear weapons include uranium-235 and plutonium-239.

True

3. Fission reactions can be controlled; thus, they are used to create energy and electricity. But fusion reactions are uncontrollable and require too much energy for economical use.

True

4. Fission reactions occur at such high temperatures in the Sun that the Sun appears to glow.

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Hope this answers the question. Have a nice day.</span>
4 0
3 years ago
Read 2 more answers
A buffer solution is prepared by placing 5.86 grams of sodium nitrite and 32.6 mL of a 4.90 M nitrous acid solution into a 500.0
N76 [4]

This question is incomplete, the complete question is;

A buffer solution is prepared by placing 5.86 grams of sodium nitrite and 32.6 mL of a 4.90 M nitrous acid solution into a 500.0 mL volumetric flask and diluting to the calibration mark. If 10.97 mL of a 1.63 M solution of potassium hydroxide is added to the buffer, what is the final pH? The Ka for nitrous acid = 4.6 × 10⁻⁴.

Answer:

the final pH is 3.187

Explanation:

Given the data in the question;

Initial moles of HNO2 = 32.6/1000 × 4.90 = 0.15974 mol

Initial moles of NO2- = mass/molar mass = 5.86/68.995 =  0.0849336 mol

Moles of KOH added = 10.97/1000 × 1.63  = 0.0178811 mol

so

HN02 + KOH → NO2- + H2O

moles of HNO2 = 0.15974 - 0.0178811 = 0.1418589 mol

Moles of NO2- = 0.0849336 + 0.0178811  =  0.1028147 mol

Now,

pH = pka + log( [NO2-]/[HNO2])

pH = -log ka + log( moles of NO2- / moles of HNO2 )

we substitute

pH = -log( 4.6 × 10⁻⁴ ) + log( 0.1028147  / 0.1418589  )

pH = -log( 4.6 × 10⁻⁴ ) + log( 0.724767 )

pH =  3.337242 + (-0.1398 )

pH = 3.187

Therefore, the final pH is 3.187

8 0
3 years ago
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