0.25 mols SO₂ x 64.058 g SO₂/ 1 mol SO₂ = 16.0145 g SO₂
molar mass of SO₂: 64.058 g
answer: 16 grams of SO₂ (2 sig figs)
check the question to see if its asked for a specific unit for mass (grams or kilograms, if they asked for kiligrams then convert 16 grams to kilograms by dividing it by 1000)
Answer:
the answer is Fungi
Explanation:
it makes its own food and doesn't move from place to place that why this is the answer
<h3>
Answer:</h3>
1.85 M
<h3>
Explanation:</h3>
<u>We are given;</u>
- Number of moles as 0.50 mol
- Volume of the solution is 270 ml
But, 1000 mL = 1 L
- Thus, volume of the solution is 0.27 L
We are required to calculate the molarity of the solution;
- Molarity refers to the concentration of a solution in moles per liter.
- It is calculated by dividing number of moles with the volume.
Molarity = Moles ÷ Volume
In this case;
Molarity = 0.50 moles ÷ 0.27 L
= 1.85 Mol/L or 1.85 M
Therefore, molarity of the solution is 1.85 M
This question is asking for a method for the determination of the freezing point in a solution that does not have a noticeable transition in the cooling curve, which is basically based on a linear fit method.
The first step, would be to understand that when the transition is well-defined as the one on the attached file, we can just identify the temperature by just reading the value on the graph, at the time the slope has a pronounced change. For instance, on the attached, the transition occurs after about 43 seconds and the freezing point will be about 4 °C.
However, when we cannot identify a pronounced change in the slope, it will be necessary to use a linear fit method (such as minimum squares) to figure out the equation for each segmented line having a significantly different slope and then equal them so that we can numerically solve for the intercept.
As an example, imagine two of the segmented lines have the following equations after applying the linear fit method:

First of all, we equal them to find the x-value, in this case the time at which the freezing point takes place:

Next, we plug it in in any of the trendlines to obtain the freezing point as the y-value:

This means the freezing point takes place after 7.72 second of cooling and is about 1.84 °C. Now you can replicate it for any not well-defined cooling curve.
Learn more:
A diagram of the composition of air would be the air's chemical formula and it's structure.