Answer:
a) 5
b) 4
c) 3
d) 3
e) 4
Explanation:
I use only one rule when the decimal is present, meaning you can see the decimal (as is the case with all of these).
When the decimal is Present, start counting sig figs from the Pacific (left) side of the number beginning with the first non-zero digit and count all the way to the end.
So, for example, in "a", the first non-zero digit starting from the left is 1, then continue counting all the way to the right side.
For "c", the first non-zero digit is the left most 4 (skip the first 4 zeros), then count all the way to the right side.
The 3 parts are
1) A five carbon ribose sugar
2) A Phosphate molecule
3) The four nitrogenous bases
I hope that's help !
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:
Learn more about trendlines: brainly.com/question/13298479
Answer:
V₁ = 208.3 mL
Explanation:
Given data:
Initial molarity of HCl = 6.0 M
Final volume = 500 mL
Final molarity = 2.5 M
Volume of initial solution required = ?
Solution:
Formula:
M₁V₁ = M₂V₂
Now we will put the values in formula.
6.0 M × V₁ = 2.5 M ×500 mL
6.0 M × V₁ = 1250 M.mL
V₁ = 1250 M.mL / 6.0 M
V₁ = 208.3 mL
Energy added = mass*specific heat* rise in temperature.
= 5 * (60-50) * 4.18
= 50*4.18
= 209 J (answer).