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g100num [7]
3 years ago
13

The legs and the wings of an insect are attached to the ________. a. head b. thorax c. abdomen d. antennae

Chemistry
2 answers:
AveGali [126]3 years ago
7 0
<span>The legs and the wings of an insect are attached to the ________.


The correct answer is:
</span>b. thorax<span>
</span>
FrozenT [24]3 years ago
4 0
Legs and wings are attached to the thorax!
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The sum of pH and pOH for any aqueous solution equals .
Vsevolod [243]

Answer:

pH is calculated by taking negative logarithm of hydrogen ion concentration. Acids have pH ranging from 1 to 6.9, bases have pH ranging from 7.1 to 14 and neutral solutions have pH equal to 7. Thus the sum of pH and pOH is 14

Explanation:

6 0
3 years ago
2NaCl → 2Na +Cl2<br> What reaction is this
babunello [35]

Answer:

Decomposition reaction

Explanation:

A single reactant breaking down to form 2 or more products is decomposition

5 0
2 years ago
If 16.29 grams of Na2SO4 is mixed with 3.697 grams of C and allowed to react according to the balanced equation: Na2SO4(aq) + 4
abruzzese [7]

<u>Answer:</u> The limiting reagent in the given chemical reaction is carbon metal.

<u>Explanation:</u>

Excess reagent is defined as the reagent which is present in large amount in a chemical reaction.

Limiting reagent is defined as the reagent which is present in small amount in a chemical reaction. Formation of product depends on the limiting reagent.

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

  • <u>For sodium sulfate:</u>

Given mass of sodium sulfate = 16.29 g

Molar mass of sodium sulfate = 142 g/mol

Putting values in equation 1, we get:

\text{Moles of sodium sulfate}=\frac{16.29g}{142g/mol}=0.115mol

  • <u>For carbon:</u>

Given mass of carbon = 3.697 g

Molar mass of carbon = 12 g/mol

Putting values in equation 1, we get:

\text{Moles of carbon}=\frac{3.697g}{12g/mol}=0.31mol

For the given chemical reaction:

Na_2SO_4(aq.)+4C(s)\rightarrow Na_2S(s)+4CO(g)

By Stoichiometry of the reaction:

4 moles of carbon reacts with 1 mole of sodium sulfate

So, 0.31 moles of carbon will react with = \frac{1}{4}\times 0.31=0.0775mol of sodium sulfate

As, given amount of sodium sulfate is more than the required amount. So, it is considered as an excess reagent.

Thus, carbon metal is considered as a limiting reagent because it limits the formation of product.

Hence, the limiting reagent in the given chemical reaction is carbon metal.

4 0
3 years ago
Why does the temperature of the water stay the same when it melts and boils?
krok68 [10]

The temperature stays the same when a solid changes to a liquid because energy is required to break the forces between particles of water therefore changing the state of matter and separating the particles away from each other.

When a liquid boils, the energy is needed by the particles to escape the surface of the liquid and boil. Instead of raising the temperature, the energy goes into the particles' kinetic energy store so it has enough speed to escape the surface of the liquid.

5 0
3 years ago
In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in
RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 m^{3}

Density of water = 1000 kg/m^{3}

Therefore,  mass of water = Density × Volume

                       = 1000 kg/m^{3} \times 0.25 m^{3}

                       = 250 kg  

Initial Temperature of water (T_{1}) = 20^{o}C

Final temperature of water = 140^{o}C

Heat of vaporization of water (dH_{v}) at 140^{o}C  is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point (140^{o}C) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

                           = 62.5 (kg) × 2133 (kJ/kg)

                           = 133.3 \times 10^{3} kJ

All this heat was supplied in 60 min = 60(min)  × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = \frac{133.3 \times 10^{3}kJ}{3600 s} = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from 20^{o}C to 140^{o}C of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

                      = 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

                     = 125520 kJ  

Time required = Heat required / Power rating

                       = \frac{125520}{37}

                       = 3392 sec

Time required to raise the temperature from 20^{o}C to 140^{o}C of 0.25 m^{3} water is calculated as follows.

                    \frac{3392 sec}{60 sec/min}

                     = 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0
3 years ago
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