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elena-s [515]
3 years ago
11

Which of the following is an example of incomplete dominance?

Chemistry
1 answer:
tatuchka [14]3 years ago
4 0
The example of the incomplete dominance is B. A cross between a black mouse and a white mouse produces a gray mouse.

In incomplete dominance both colors mix and produce another color.

Hope this helps :)
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What must be changed, temperature or heat energy, during condensation explain in 2-3 sentences
Nastasia [14]

Answer:

Change occur in heat energy.

Explanation:

Change occur in the heat energy of molecules during condensation process. The molecules of gas releases its heat energy and converted into liquid state. With this heat energy, they are active and escape from each other but when the heat energy is removed from them, the attractive forces between these gas molecules are formed which convert them into liquid state so change in heat energy occur in the condensation process.

4 0
3 years ago
6 Which substances have atoms of the same element but different molecular structures?
Bezzdna [24]
Well you want the answer with the same letter in it. A.k.a 2. This means they have the same element in them
O2 (Diatomic oxygen that we breathe) and O3 (Ozone that would be deadly to breathe) have different structures.
So the answer is 2
8 0
3 years ago
Martian periodic table
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3 0
3 years ago
A 0.150-kg sample of a metal alloy is heated at 540 Celsius an then plunged into a 0.400-kg of water at 10.0 Celsius, which is c
Zarrin [17]

Answer:

C_{alloy}=0.497\frac{J}{g\°C}

Explanation:

Hello there!

In this case, according to this calorimetry problem on equilibrium temperature, it is possible for us to infer that the heat released by the metal allow is absorbed by the water for us to write:

Q_{allow}=-(Q_{water}+Q_{Al})

Thus, by writing the aforementioned in terms of mass, specific heat and temperature, we have:

m_{alloy}C_{alloy}(T_{eq}-T_{alloy})=-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})

Then, we solve for specific heat of the metallic alloy to obtain:

C_{alloy}=\frac{-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})}{m_{alloy}(T_{eq}-T_{alloy})}

Thereby, we plug in the given data to obtain:

C_{alloy}=\frac{-(400g*4.184\frac{J}{g\°C} (30.5\°C-10.0\°C)+200g*0.900\frac{J}{g\°C}(30.5\°C-10.0\°C)}{150g(30.5\°C-540\°C)} \\\\C_{alloy}=0.497\frac{J}{g\°C}

Regards!

3 0
3 years ago
Which sentence best describes an igneous intrusion
Ulleksa [173]
What are the sentences? 
It would help to see the sentences.
5 0
3 years ago
Read 2 more answers
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