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Oxana [17]
3 years ago
13

3. What detrmines the identity of an atom?​

Chemistry
1 answer:
Brrunno [24]3 years ago
5 0
The number of protons in one atom of an element determines the atom's identity, and the number of electrons determines its electrical charge. The atomic number tells you the number of protons in one atom of an element. ... The atomic mass of an element is the total number of protons and neutrons in the nucleus of the atom.
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How many moles of hypomanganous acid. H3 MnO4, are contained in 22.912 g?
gregori [183]

Answer:

0.188mol

Explanation:

Using the formula;

mole = mass/molar mass

Molar mass of hypomanganous acid. (H3MnO4) = 1(3) + 55 + 16(4)

= 3 + 55 + 64

= 122g/mol

According to this question, there are 22.912g of H3MnO4

mole = 22.912g ÷ 122g/mol

mole = 0.188mol

6 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
A wetland being drained so a housing development can be built is an example of ____________.
Reil [10]

Answer:

a now give me the points

Explanation:

5 0
3 years ago
Please help me ASAP I’ll mark Brainly
mixer [17]

Answer:

cell

chloroplast and cell wall

nucleus

life processes

cell membrane

shape and size

vacuole

Hope it helps

5 0
3 years ago
What acid occurs naturally in the body and performs important regulatory functions
gtnhenbr [62]

Answer: carbonic acid

8 0
3 years ago
Read 2 more answers
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