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AleksAgata [21]
3 years ago
7

1. A dining hall had a total of 25 tables-some long rectangular tables and some round

Chemistry
1 answer:
TEA [102]3 years ago
7 0

x = 20 long tables

y = 5 round table

Explanation:

We have the following system of equations:

x + y = 25

8x + 6y = 190

From the first equation we have:

x = 25 - y

And we replace x in the second equation:

8(25 - y) + 6y = 190

200 - 8y + 6y = 190

200 - 2y = 190

200 - 190 = 2y

10 = 2y

y = 5

Now we insert the value of y in the next equation:

x = 25 - y

x = 25 - 5

x = 20

Learn more about:

system of equations

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The amount of I−3(aq) in a solution can be determined by titration with a solution containing a known concentration of S2O2−3(aq
Pavlova-9 [17]

Answer : The molarity of I_3^- in the solution is, 0.128 M

Explanation :

The given balanced chemical reaction is,

2S_2O_2^{-3}(aq)+I_3(aq)\rightarrow S_4O_2^{-6}(aq)+3I^-(aq)

First we have to calculate the moles of Na_2S_2O_3.

\text{Moles of }Na_2S_2O_3=\text{Molarity of }Na_2S_2O_3\times \text{Volume of solution}

\text{Moles of }Na_2S_2O_3=0.260mole/L\times 0.0296L=0.007696mole

Conversion used : (1 L = 1000 ml)

Now we have to calculate the moles of I_3^-.

From the balanced chemical reaction, we conclude that

As, 2 moles of S_2O_2^{-3} react with 1 mole of I_3^-

So, 0.007696 moles of S_2O_2^{-3} react with \frac{0.007696}{2}=0.003848 mole of I_3^-

The moles of I_3^- = 0.003848 mole

Now we have to calculate the molarity of I_3^-.

\text{Molarity of }I_3^-=\frac{\text{Moles of }I_3^-}{\text{Volume of solution}}

Now put all the given values in this formula, we get:

\text{Molarity of }I_3^-=\frac{0.003848mole}{0.03L}=0.128mole/L=0.128M

Therefore, the molarity of I_3^- in the solution is, 0.128 M

8 0
3 years ago
Ionic character of lih
adoni [48]
I believe LiH becomes HF 
8 0
4 years ago
Consider the exothermic reaction
alisha [4.7K]

Answer:

The answer to your question is -2855 J

Explanation:

Reaction

                     2C₂H₆  +  7O₂   ⇒   4CO₂  +  6H₂O

Formula

Heat of reaction = ΔHrxn = ΣΔHrxn products - ΣΔHrxn reactants

Substitution

ΔHrxn = { 4(-393.5) + 6(-241.8)} - {2(-84.7) + 7(0)}

ΔHrxn = {-1574 -1450.8} - {-169.4}

ΔHrxn = -3024.8 + 169.4

ΔHrxn = -2855.4 J

4 0
3 years ago
Experiment #1A melting point of an old sample of Naphthalene was completed and a melting range of 77-83 oC was observed and reco
White raven [17]

Answer:

%error = 0.32%

Explanation:

Let's answer both questions, by parts.

1. Percentage error:

In this case, I do not have the video, but I do have the reported melting point of naphtalene which is 80.26 °C.

The expression to calculate the percentage error is the following:

%Error = absolute error / actual percentage. (1)

And the absolute error is:

Abs error = actual value - experimental value  (2)

But the experimental value is a range, so we just have to get a average of that:

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Now the absolute error:

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Finally the %error:

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2. Meaning of melting point range and %error

The melting point range just means that the sample of naphtalene has impurities, and when a sample of any compound has impurities, melting point tends to be low. However, this decrease of temperature is a wider range. But usually a range of just 5° C means that compound has little traces of impurities but it can still be used for reactions.

The %error means that the impurities of the sample are really low, so the sample is practically pure with little traces of impurities.

3 0
3 years ago
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Maslowich

Answer:

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At standard room temperature and pressure, most of the elements are solids.

Just a few of the elements are liquids and gases at this temperature.

The periodic table is made up of metals, metalloids and non-metals. Most of these substances are actually solid.

Some non-metals are gaseous at standard room temperature.

8 0
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