The number of pounds present in 2.72 kg of Mastiff puppy is 5.996 pounds
<h3>What is weight?</h3>
Weight can be defined as the gravitation pull on a body. or it can also be defined as the product of the mass of a body and its acceleration due to gravity.
<h3>Units of weight</h3>
weight has several units and they include
- Newton
- pounds
- ounces
- Tons.
These units can be converted.
From the question,
If
But,
Then,
Therefore,
- 2.72 kg of Mastiff = (2720/453.6) pounds of average Mastiff = 5.996 pounds
Hence, The number of pounds present in 2.72 kg of Mastiff puppy is 5.996 pounds.
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Wheres the diagram
We have to see the diagram to answer your question
Answer:
The temperature of the solute/solvent without any external effect would decrease.
Explanation:
As the bonding between the solute particles is really strong, therefore a large amount of energy is required to overcome these forces. So that the new bonding between the solute and solvent is created.
In order to achieve this, there will be a lot of energy required and that is through the heating process. So the solution will require energy so the solute will dissolve fully either by provision of external force i.e stirring or by heating.
Explanation:
This problem is asking to predict the pressure in the container at a temperature of 1,135 K with no apparent background; however, in similar problems we can be given a graph having the pressure on the y-axis and the temperature on the x-axis and a trendline such as on the attached file, which leads to a pressure of 21.2 atm by using the given equation and considering the following:
<h3>Graph analysis.</h3>
In chemistry, experiments can be studied, modelled and quantified by using graphs in which we have both a dependent and independent variable; the former on the y-axis and the latter on the x-axis.
In addition, when data is recorded and graphed, one can use different computational tools to obtain a trendline and thus, attempt to find either the dependent or independent value depending on the requirement.
In this case, since the provided trendline by the graph and the program it was put in is y = 0.017x+1.940, we understand y stands for pressure and x for temperature so that we can extrapolate this equation even beyond the plotted points, which is this case.
In such a way, we can plug in the given temperature to obtain the required pressure as shown below:
y = 0.017 ( 1,135 ) + 1.940
y = 21.2
Answer that is in atm according to the units on the y-axis:
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