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r-ruslan [8.4K]
3 years ago
12

How many grams of calcium cyanide (Ca(CN)2) are contained in 0.79 mol of calcium cyanide?

Chemistry
1 answer:
Allushta [10]3 years ago
3 0
I dont know but do you know da wae brudda?


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What is the atomic number and name of an element that has 15 protons in its nucleus?
just olya [345]

Answer:

phosphorus.

Explanation:

The atomic number of phosphorus is 15 so the protons will be 15.

4 0
3 years ago
Read 2 more answers
78.9 + 890.43 - 21 = 9.5 x 10^2
maria [59]

948 or 9.48 x 10^2

There are two sets of rules for significant figures

• One set for addition and subtraction

• Another set for multiplication and division

You used the set for multiplication and division.

This problem involves addition and subtraction, and the rule is

The number of places after the decimal point in the answer must be <em>no greater than the number of decimal places in every term</em> in the sum.

Thus, we have

78.9

+890.43

-21.

= 948.33

The "21" term has the fewest digits after the decimal point (none), so the answer must have no digits after the decimal point.

To the correct answer is 948 = 9.48 x 10^2. It has three significant figures.

8 0
3 years ago
The order of components in a typical flame atomic absorption spectrometer is
Zinaida [17]

The order of components in a typical flame atomic absorption spectrometer is hollow cathode lamp--flame--monochromator--detector

<u>Explanation:</u>

  • The hollow cathode lamp practices a cathode created of the element of interest with a low internal pressure of inert gas.
  • Remove scattered light of other wavelengths from the flame. AAS flame includes aiming at first the fuel than the oxidant and then lighting the flame with the instrument's auto-ignition system. Applying flame Ddtroy any analyte ions and breakdown complexes.
  • The process  of the monochromator is to divide analytical lines photons moving through the flame
  • Photomultiplier tube (PMT)  as the detector the PMT determines the intensity of photons of the analytical line exiting the monochromator.
4 0
3 years ago
(a) Write the balanced neutralization reaction that occurs between H2SO4 and KOH in aqueous solution. Phases are optional. (b) S
Sunny_sXe [5.5K]

These are two questions and two answers

Answer:

    Question 1:

  • <u>H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l)</u>

    Question 2:

  • <u>0.201 M</u>

Explanation:

<u>Question 1:</u>

The<em> neutralization</em> reaction that occurs between H₂SO₄ and KOH is an acid-base reaction.

The products of an acid-base reaction are salt and water.

This is the sketch of such neutralization reaction:

1) <u>Word equation:</u>

  • sulfuric acid + potassium hydroxide → potassium sulfate + water

                 ↑                               ↑                              ↑                       ↑

               acid                          base                        salt                   water

<u>2) Skeleton equation (unbalanced)</u>

  • H₂SO₄ + KOH → K₂SO₄ + H₂O

<u>#) Balanced chemical equation (including phases)</u>

  • H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l) ← answer

<u>Question 2:</u>

<u>1) Mol ratio:</u>

Using the stoichiometric coefficients of the balanced chemical equation you get the mol ratio:

  • 1 mol H₂SO₄ (aq) : 2 mol KOH (aq) : 1 mol K₂SO₄ (aq) : mol 2H₂O (l)

<u>2) Moles of H₂SO₄:</u>

  • V = 0.750 liter
  • M = 0.480 mol/liter
  • M = n/V ⇒ n = M×V = 0.480 mol/liter × 0.750 liter = 0.360 mol

<u>3) Moles of KOH:</u>

  • V = 0.700 liter
  • M = 0.290 mol/liter
  • M = n/V ⇒ n = M × V = 0.290 mol/liter × 0.700 liter = 0.203 mol

<u>4) Determine the limiting reagent:</u>

a) Stoichiometric ratio:

   1 mol H₂SO₄ / 2 mol NaOH = 0.500 mol H₂SO4 / mol NaOH

b) Actual ratio:

   0.360 mol H₂SO4 / 0.203 mol NaOH = 1.77 mol H₂SO₄ / mol NaOH

Since hte actual ratio of H₂SO₄  is greater than the stoichiometric ratio, you conclude that H₂SO₄ is in excess.

<u>5) Amount of H₂SO₄ that reacts:</u>

  • Since, KOH is the limiting reactant, using 0.203 mol KOH and the stoichiometryc ratio 1 mol H₂SO₄ / 2 mol KOH, you get:

         x / 0.203 mol KOH = 1 mol H₂SO₄ / 2 mol KOH ⇒

         x = 0.203 / 2 = 0.0677 mol of H₂SO₄

<u>6) Concentration of H₂SO₄ remaining:</u>

  • Initial amount - amount that reacted = 0.360 mol - 0.0677 mol = 0.292 mol

  • Total volume = 0.700 liter + 0.750 liter = 1.450 liter

  • Concetration = M

        M = n / V = 0.292 mol / 1.450 liter = 0.201 M ← answer

6 0
3 years ago
Why is computational chemistry important?​
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Ydhegeywyehdvxhdhdishhshs shshsjssisp
4 0
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