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Iteru [2.4K]
2 years ago
15

WOKWOEKWWKEOWKEEAKKOWAEKWOKSWS

Chemistry
2 answers:
4vir4ik [10]2 years ago
8 0

Answer:

do ju gwana b ma frind

hiExplanation:

jfiififjrijfirfjrijfirfirfjri

CaHeK987 [17]2 years ago
3 0

Answer: independent varible

Explanation:

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All of the following are benefits of recycling except?
IgorLugansk [536]
Recycling Isn’t Always Cost Effective.
High Up Front Cost
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Recycling Sites Are Commonly Unsafe.
6 0
3 years ago
How many moles are in 110 grams of nahco3
yuradex [85]
Molar mass NaHCO₃ = 23 + 1 + 12 + 16 x 3 = 84 g/mol

1 mole ---------- 84 g
? mole ---------- 110 g

moles NaHCO₃ = 110 . 1 / 84

moles NaHCO₃ = 110 / 84

= 1.309 moles

hope this helps!
4 0
3 years ago
Help help help I will give brainliest
vampirchik [111]

Answer:

its b alkyne

Explanation:

brinliest po plsss

8 0
3 years ago
Read 2 more answers
Can you check this for me?
Aliun [14]
Hi, your answer is correct.
3 0
4 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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