Answer:
The sample will be heated to 808.5 Kelvin
Explanation:
Step 1: Data given
Volume before heating = 2.00L
Temperature before heating = 35.0°C = 308 K
Volume after heating = 5.25 L
Pressure is constant
Step 2: Calculate temperature
V1 / T1 = V2 /T2
⇒ V1 = the initial volume = 2.00 L
⇒ T1 = the initial temperature = 308 K
⇒ V2 = the final volume = 5.25 L
⇒ T2 = The final temperature = TO BE DETERMINED
2.00L / 308.0 = 5.25L / T2
T2 = 5.25/(2.00/308.0)
T2 = 808.5 K
The sample will be heated to 808.5 Kelvin
Yes, it mixes it and has vitamins in the tea.
The way you calculate the empirical formula is to firstly assume 100g. To find each elements moles you take each elements percentage listed, times it by one mole and divide it by its atomic mass. (ex: moles of K =55.3g x 1 mole/39.1g, therefore there is 1.41432225 moles of Potassium) Once you’ve completed this for every element you list each elements symbol beside it’s number of moles and divide by the smallest number because it can only go into its self once. After you’ve done this, you’ve found your empirical formula, which is the simplest whole number ratio of atoms in a compound. I’ve added an example of a empirical question I completed last semester :)
<h3><u>Answer;</u></h3>
Molarity = 0.25 M
<h3><u>Explanation;</u></h3>
Molarity is given by moles/Liter.
First we find moles:
Number of moles = Mass /molar mass
= (10.7g NH4Cl)/(53.5g/mol NH4Cl)
= 0.200 moles NH4Cl
Then we convert to liters:
= (800mL)*(1L/1000mL) = 0.800L
Therefore; molarity = 0.2moles/0.8L
= 0.25M
Answer:
2)

3)

Explanation:
Hello!
2) In this case, since magnesium sulfate is MgSO₄, we can see how magnesium weights 24.305 g/mol, sulfur 32.06 g/mol and oxygen 64.00 g/mol as there is one atom of magnesium as well as sulfur but four oxygen atoms for a total of g/mol; thus the percent compositions are:

3) In this case, although the element seems to contain Ag and O, we infer its molecular formula is Ag₂O; thus, since we have two silver atoms weighing 215.74 g/mol and one oxygen atom weighing 16.00 g/mol for a total of 231.74 g/mol, we obtain the following percent compositions:

Best regards!