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Tcecarenko [31]
3 years ago
8

a company has designed a new product and tested the porotype which is the next step in product development

Chemistry
1 answer:
Masja [62]3 years ago
6 0
I’d say market and distribution would be the next step
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A balloon contains 0.15 moles of co2 gas. how many molecules of co2 are in the balloon?
Sauron [17]
1- One mole is = 6.02 x 10^23 of anything, So one mole of atoms is 6.02x10^23.

2- when the balloon contains 0.15 moles of Co2 gas so:

the no.of molecules of Co2 = 0.15 x 6.02x 10^23

                                              = 9.0 x 10^22
5 0
3 years ago
Read 2 more answers
Why is it important to control the reaction rate of the chemical process
xxTIMURxx [149]
Mg+2HCl=MgCl_2+H_{2(g)}

The reaction creates dihydrogen, hence if it's uncontrolled it could lead to potentially dangerous amounts of gas being released at once.
8 0
3 years ago
Give the direction of the reaction, if K >> 1. Give the direction of the reaction, if K >> 1. The forward reaction i
anygoal [31]

Answer:

A. for K>>1 you can say that the reaction is nearly irreversible so the forward direction is favored. (Products formation)

B. When the temperature rises the equilibrium is going to change but to know how is going to change you have to take into account the kind of reaction. For endothermic reactions (the reverse reaction is favored) and for exothermic reactions (the forward reaction is favored)

Explanation:

A. The equilibrium constant K is defined as

K=\frac{Products}{reagents}

In any case  

aA +Bb  equilibrium Cd +dD

where K is:

K= \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}

[] is molar concentration.

If K>>> 1 it means that the molar concentration of products is a lot bigger that the molar concentration of reagents, so the forward reaction is favored.

B. The relation between K and temperature is given by the Van't Hoff equation

ln(\frac{K_{1}}{K_{2}})=\frac{-delta H^{o}}{R}*(\frac{1}{T_{1}}-\frac{1}{T_{2}})

Where: H is reaction enthalpy, R is the gas constant and T temperature.  

Clearing the equation for K_{2} we get:

K_{2}=\frac{K_{1}}{e^{\frac{-deltaH^{o}}{R}*(\frac{1}{T_{1}} -\frac{1}{T_{2}})}}

Here we can study two cases: when delta H^{o} is positive (exothermic reactions) and when is negative (endothermic reactions)

For exothermic reactions when we increase the temperature the denominator in the equation would have a negative exponent so K_{2} is greater that K_{1} and the forward reaction is favored.

When we have an endothermic reaction we will have a positive exponent so K_{2} will be less than K_{1} the forward reactions is not favored.  

{e^{\frac{-deltaH^{o}}{R}*(\frac{1}{T_{1}} -\frac{1}{T_{2}})}}

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
Help help help plsss
Olin [163]

Answer:

4.58×10^3

3.72×10^7

Explanation:

3.8×10^2 =380

+ 4.2×10^3=4200

= 4580 = 4.58×10^3

8 0
2 years ago
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