Answer:
0.00984211
Explanation:
At his favorite restaurant, Dylan orders a soda with his dinner. The ice in his drink has a mass of 6 g and a volume of 20 cm3. What is the density of the ice? Explain the steps you took to solve this problem, showing your work.
A mixture of propane and butane is burned with pure oxygen. The combustion products contain 46.7 mole% H2O. After all the water is removed from the products, the residual gas contains 69.0 mole% CO2 and the balance O2 a. What is the mole percent of propane in the fuel?
Answer:
28%
Explanation:
Basically, all o did was write the equations, balance it and solve for them. Also, at the place I stared, I used simultaneous equation to solve it. Multiplying by 8 and also 3.
It's a pretty straightforward question.
At the final step that's missing, I Did
(y)C3H8 = 2.8 / ( 2.8 + 7.1)
(y)C3H8 = 0.28
lab
table b:eyedrop of blue water in cold water
table c:eyedrop of red water in hot water
table d:see how temperature changes on white + black paper with light shining on them
plz help
brainliest for 1st correct answer
Answer:
lol, I just did this assignment, cold water
0:30: slowly expand
1:00: still expanding
1:30: expanding is continued
2:00: still expanding slowly
2:30: expanding slowly
3:00: expanded twice the size from the first time
hot water
0:30: expanding quickly
1:00: still expanding
1:30: starts to reach the surface and still expanding
2:00: almost all the of the dye reached the top
2:30: the dye reached the top and is starting to split
3:00: the dye split appart and is moving down again
radiation
white paper
0:30: at a low temp. at around 3.3 C
1:00: still at a low temp. around 4.4 C
1:30: slowly rising temp. around 5.5 C
2:00: still slowly rising temp. around 7.2 C
2:30: still at a low temp. around 8.3 C
3:00: ends at a low temp. around 10 C
black paper
0:30: at a resonable temp. at around 6.6 C
1:00: temp. rises quicker at around 13.3 C
1:30: temp. stil rising quickly at around 17.7 C
2:00: temp. is getting hotter at around 21.6 C
2:30: temp. is still getting hotter at around 26.1 C
3:00: temp. ends at a hot temp. at around 30 C
Explanation:
If the compound below is oxidized, the resulting product is ___. Select one: a. methane and propane b. butanal c. butanoic acid d. butane
The complete question is
If the compound below is oxidized, the resulting product is ___.
the compound given is Butanal
Select one:
a. methane and propane
b. butanal
c. butanoic acid
d. butane
Answer:
C. Butanoic Acid
Explanation:
When Butanal is oxidized using reagents like KMnO4, Tollens reagent etc.
When Aldehyde is oxidized corresponding Carboxylic Acid is produced
Butanal → Butanoic Acid
Why do we need to identify matter?
Answer:
It's important for scientists to know the properties of matter because all things are made up of matter. Each type of matter has different physical characteristics and scientists need to know and understand these characteristics to make calculations. ... The main phases of matter are solid, liquid, and gas
Explanation:
^^
C. If an electron is added to the outer shell, how will this affect the charge and volume of the atom?
if an electron is added on the outer shell
Answer:
An atom that gains or loses an electron becomes an ion. If it gains a negative electron, it becomes a negative ion. If it loses an electron it becomes a positive ion.
Answer:
when an atom gains an electron, it gains a negative charge . it's volume increases as well due to the increase in repulsion between the electrons
What is the pH after two solutions are mixed: 1. A 0.155 M H2SO4 solution with a volume of 35.0 mL and 2. A 0.225 M solution of KOH with a volume of 35.0 mL. The volumes are additive.
Answer:
[tex]pH=1.67[/tex]
Explanation:
Hello.
In this case, since the reaction between the sulfuric acid and the potassium hydroxide is:
[tex]2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O[/tex]
We can see a 2:1 mole ratio between them, thus, we compute the moles of acid and base that are present in the solution by using the employed volume and its concentration:
[tex]n_{H_2SO_4}=0.155mol/L*0.0350L=0.005425molH_2SO_4\\\\n_{KOH}=0.225mol/L*0.0350L=0.007875molKOH\\[/tex]
Next, we use the 2:1 mole ratio to compute the consumed moles of acid:
[tex]n_{H_2SO_4}^{consumed}=0.007875molKOH*\frac{1molH_2SO_4}{2molKOH}=0.0039375molH_2SO_4[/tex]
It means that the leftover of sulfuric acid is:
[tex]n_{H_2SO_4}^{leftover}=0.005425-0.0039375=0.001488molH_2SO_4[/tex]
Therefore, since sulfuric acid is a strong acid, the pH is defined in terms of its concentration which is actually equal to the concentration of hydrogen ions in solution, which is computed considering the total additive volume:
[tex][H^+]=[H_2SO_4]=\frac{0.001488mol}{0.070L}=0.02125M[/tex]
So the pH turns out:
[tex]pH=-log(0.02125)\\\\pH=1.67[/tex]
Best regards.