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Serjik [45]
3 years ago
8

Is c10H8 a conductor

Chemistry
1 answer:
Lunna [17]3 years ago
5 0

Answer:

No

Explanation:

Napthalene cannot conduct electricity

You might be interested in
15.1 L N2 at 25 °C and 125 kPa and 44.3 L O2 at 25 °C and 125 kPa were transferred to a tank with a volume of 6.25 L. What is th
Y_Kistochka [10]

Answer:

The total pressure of the mixture in the tank of volume 6.25 litres at 51°C  is 1291.85 kPa.

Explanation:

For N2,

                Pressure(P₁)=125 kPa

                  Volume(V₁)=15·1 L

                Temperature (T₁)=25°C=25+273 K=298 K

Similarly, for Oxygen,

                   Pressure(P₂)= 125 kPa

                   Volume(V₂)= 44.3 L

                  Temperature(T₂)=25°C= 298 K

Then, for the mixture,

              Volumeof the mixture( V)= 6.25 L

                                     Pressure(P)=?

                Temperature (T)= 51°C = 51+273 K=324 K

Then, By Combined gas laws,

                                 \frac{P_{1} V_{1} }{T_{1} } +\frac{P_{2} V_{2} }{T_{2} } =\frac{PV}{T}

                      or, \frac{15.1*125}{298} +\frac{44.3*125}{298} =\frac{P*6.25}{324}

                     or, 6.34+18.58=\frac{P*6.25}{324}

                     or, P=\frac{24.92*324}{6.25}

                        ∴P=1291.85 kPa

So the total pressure of the mixture in the tank of volume 6.25 litres at 51°C  is 1291.85 kPa.

3 0
3 years ago
Write the fraction of the mass of kcl produced from 1 g of k2c03​
Minchanka [31]

Mass of KCl= 1.08 g

<h3>Further explanation</h3>

Given

1 g of K₂CO₃

Required

Mass of KCl

Solution

Reaction

K₂CO₃ +2HCl ⇒ 2KCl +H₂O + CO₂

mol of K₂CO₃(MW=138 g/mol) :

= 1 g : 138 g/mol

= 0.00725

From the equation, mol ratio K₂CO₃ : KCl = 1 : 2, so mol KCl :

= 2/1 x mol K₂CO₃

= 2/1 x 0.00725

= 0.0145

Mass of KCl(MW=74.5 g/mol) :

= mol x MW

= 0.0145 x 74.5

= 1.08 g

6 0
3 years ago
Classify each of the observed changes according to whether or not they are likely to represent evidence that a chemical reaction
Mama L [17]

Answer:

a solution color becoming less intense  due to dilution- is not an evidence of a chemical reaction

bubbles (gas formation) - evidence of a chemical reaction

explosion or fire - evidence of a chemical reaction

changes in color- evidence of a chemical reaction

precipitation- evidence of a chemical reaction

changes in temperature - evidence of a chemical reaction

a solid liquifying - is not an evidence of a chemical reaction

solution colors mixing - is not an evidence of a chemical reaction

Explanation:

A chemical change is not easily reversible and yields new substances. It is often accompanied by a loss or gain of heat.

In the answer section, i have shown some evidences that lead us to conclude that a chemical reaction has taken place. The occurrence of a chemical change often goes with the formation of new substances as earlier stated and any of these signs may accompany the process.

For instance, when a metal is dropped in dilute acid solution, bubble of hydrogen gas indicates that a chemical reaction has taken place.

8 0
3 years ago
Cans of regular soda have volumes with a mean of 14.5 oz and a standard deviation of 0.18 oz. What is the maximum "usual" value
tangare [24]

Answer:

14.68 oz

Explanation:

The mean volume is the average volume of all the regular soda cans put together; 14.5 oz. The standard deviation 0.18 oz is the maximum deviation from the mean. It can be 14.5-0.18= 14.32 oz which is the usual minimum or 14.5 + 0.18=14.68 oz which is the usual maximum

4 0
3 years ago
If you added 45,000 calories to water that was at 25 degrees C, and the ending temperature was 35 degrees C, how much water did
user100 [1]

<u>Answer:</u>

<em>4.5 L water we have in litres (L).</em>

<em><u></u></em>

<u>Explanation:</u>

Q=m\times c \times \Delta T

where

\Delta T = Final T - Initial T

Q is the heat energy in calories

c is the specific heat capacity (for water 1.0  cal/(g℃))  

m is the mass of water

Plugging in the values  

\\$45000 \mathrm{cal}=m \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times\left(35^{\circ} \mathrm{C}-25^{\circ} \mathrm{C}\right)$\\\\$45000 \mathrm{cal}=m \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times 10^{\circ} \mathrm{C}$\\\\$m=\frac{45000 \mathrm{cal}}{1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times 10^{\circ} \mathrm{C}}$\\\\$m=4500 \mathrm{g}$\\\\Density of water $=\frac{\text { mass }}{\text { volume }}$

So,

Volume of water = mass/density

\\\\=\frac{4500 \mathrm{g}}{\frac{1.09}{\mathrm{mL}}}=4500 \mathrm{mL}$$

=4.5 L (Answer)

6 0
3 years ago
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