<span>1. Fill a beaker or graduated cylinder with enough water to completely immerse the sphere in. 2. Record the baseline initial measurement. 3. Drop the sphere in. 4 <span>Record final measurement.</span></span>
Answer : (C) Hafnium is the most likely identity of the given substance.
Solution : Given,
Mass of given substance (m) = 46.9 g
Volume of given substance (V) = 3.5 
First, find the Density of given substance.
Formula used :
Now,put all the values in this formula, we get
= 13.4 g/
So, we conclude that the density of given substance (13.4 g/) is approximately equal to the density of Mercury and Hafnium (13.53 and 13.31 g/ respectively).
According to the question the substance is solid at room temperature but Mercury is liquid at room temperature. So, Mercury is not identical to the given substance.
Another element i.e, Hafnium is the element whose density is approximately equal to the given substance and also solid at room temperature. And we know that the melting point of solid is high.
So, Hafnium is the most likely element which is the identity of the given substance.
Wind abrades rock by sandblasting, this is the process in which wind causes the
The mass would be 28. The mass of an atom equals the number of neutrons and protons added together. In this case, they do not say the protons. But we must use our inside knowledge to know that the number of protons equals the number of electrons. Thus, your number of protons would be 13
Hello!
To find the amount of energy need to raise the temperature of 125 grams of water from 25.0° C to 35.0° C, we will need to use the formula: q = mcΔt.
In this formula, q is the heat absorbed, m is the mass, c is the specific heat, and Δt is the change in temperature, which is found by final temperature minus the initial temperature.
Firstly, we can find the change in temperature. We are given the initial temperature, which is 25.0° C and the final temperature, which is 35.0° C. It is found by subtract the final temperature from the initial temperature.
35.0° C - 25.0° C = 10.0° C
We are also given the specific heat and the grams of water. With that, we can substitute the given values into the equation and multiply.
q = 125 g × 4.184 J/g °C × 10.0° C
q = 523 J/°C × 10.0° C
q = 5230 J
Therefore, it will take 5230 joules (J) to raise the temperature of the water.
