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boyakko [2]
3 years ago
12

After creating a prototype, which of the following would not be an appropriate next step in the design process of a new technolo

gy?
A. Test product safety.
B. Troubleshoot problems.
C. Obtain a patent.
D. Research the need.
Chemistry
1 answer:
Ivanshal [37]3 years ago
4 0
I would say D but i am not entirely sure 
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Determine % yield if a student obtains 45 g of product in an experiment and the theoretical amount is determined to be 50 g. *
ycow [4]

Answer:

90%

Explanation:

Percentage yield = ?

Theoretical yield = 50g

Actual yield = 45g

To calculate the percentage yield of a compound, we'll have to use the formula of percentage yield which is the ratio between the actual yield to theoretical multiplied by 100

Percentage yield = (actual yield / theoretical yield) × 100

Percentage yield = (45 / 50) × 100

Percentage yield = 0.9 × 100

Percentage yield = 90%

The percentage yield of the substance is 90%

6 0
3 years ago
What is the significance of enzymes in maintaining homeostasis in living organisms ?
GalinKa [24]
 Enzymes are needed for metabolic pathways in the body, respiration, digestion and other important life processes. When enzymes function properly, homeostasis is maintained. However, if an enzyme is lacking or has an incorrect shape due to genetic mutation, this can lead to disease within an organism.
7 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
After creating a prototype, which of the following would not be an appropriate next step in the design process of a new technolo
Ivanshal [37]
I would say D but i am not entirely sure 
4 0
3 years ago
`Let's just consider the concept of stoichiometry without any confusing chemicals. Here is a generic equation: 2A + 6B ? 3C. If
lesya [120]

Answer:

Option D. 5.5

Explanation:

The equation is this:

2A + 6B  ⇒  3C

With the amounts that we were given, let's determine which is the <em>limting reactant</em>

2 A reacts with 6 B

4 A will react with ( 4 .6)/2 = 12B

I have 11 B, so the limiting is B

6 B react with 2 A

11 B will react with (11 .2 )/6 =3.66 A

I have 4 A, so A is the excess.

6 B produce 3 C

11 B will produce ( 11 .3)/6 = 5.5C

7 0
3 years ago
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