How many kilometers are in a 7.21. mm

What are the products of the following combustion reaction?<br><br> 2C2H2(g) + 5O2(g) →

kakasveta [241]

A combustion reaction of an will generally produce CO2 and H20 -- carbon dioxide and water and/or an oxide

looking at the combustion material C2H2, you know that the end products will be CO2 and H20, so the question is how much of each will you get

well, look at the total amount of carbon atoms, 2 C2, which means a total of 4 carbon atoms in this reaction, since only CO2 has carbon atoms, that means there must be 4 CO2 as an end product and 4 CO2 will use up 4 of 5 O2 molecule leaving only 1 O2 molecule for the H2 reaction.

now O2 has a total of 2 oxygen molecules whereas H20 has only a single oxygen molecule, hence the end product must have 2 H20

check that the H atoms balance out on both sides

4 0

3 years ago

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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

What are the answers to the blanks to make this equation balanced

Eddi Din [679]

1Al+6NaOH==>2 $Na_{3}$ Al $O_{3}$ +3 $H_{2}$

Hope this helps!

5 0

3 years ago

0.00105 liters is the same as how many cm and ml

Ainat [17]

The answer is 1.05 cubic centimeters and 1.05 mL (1 cubic centimeter is equal to 1 mL)

8 0

3 years ago

Why doesnt iron burst into flames when it rusts

Katarina [22]

Rust does not have the properties to catch onto flames. However, if you light it on fire, then it would probably catch in flames but not instantly and will not continue to burn unless you have soaked it in oil or flammable object or substance. :) Hope this helps!

3 0

3 years ago