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Svetach [21]
2 years ago
7

What is the mass of 11.3 mole of argon?

Chemistry
2 answers:
zheka24 [161]2 years ago
7 0

Answer:

They will determine the mass of various common elements and compounds and convert this data into values of moles, atoms and molecules.

Explanation:

MAXImum [283]2 years ago
6 0
The answer is 451.4124 argon
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If the equation: FeCl3+O2-->Fe2O3+Cl2, how many moles of chlorine gas can be produced if 4 moles of FeCl3 react with 4 moles
BlackZzzverrR [31]

Answer:

6 moles of Cl2

Explanation:

First, the equation has to be balanced, which makes it 4 FeCl3 + 3 O2 --> 2 Fe2O3 + 6 Cl2

Using this information, we can see that one mole of O2 will not be present in the reaction. Since four moles of FeCl3 are needed to react in the equation, which would produce six moles of Cl2, and only four moles of FeCl3 are present, six moles of Cl2 would be produced.

4 0
2 years ago
Define a fluid very sort answer ​
Ganezh [65]

Answer:

A fluid is a medium that has a defined mass and volume, but no fixed shape, at a constant temperature and pressure. This may include gases, liquids, plasmas, and to some extent plastic solids. A fluid can flow and deform, preventing it from carrying loads in a static equilibrium.  A fluid is always compressible and internal frictional forces always occur due to the viscosity of the fluid.

4 0
3 years ago
Who proposed the "Law of Octaves" for organizing events?
djyliett [7]

b

Explanation:

February 7, 1863, was the day John Newlands published a paper outlining what would be known as “The Law of Octaves”. Newlands discovered if he ordered the known elements by increasing atomic weight, the chemical properties of the elements would be similar for every eighth group

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
Give an example of a change that the ecosystem was able to recover from and return to equilibrium.
Tanzania [10]

Answer:

The Fox is at the top of the food chain; so, the other animals could increase in population and the fox could return

4 0
2 years ago
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