Well it would move 10,000cm so converting that into km would be it would move 0.1km
Answer: There are 0.006 moles of acid in the flask.
Explanation:
Given:
= 21.35 mL,
= 0.150 M
= 25.0 mL,
= ?
Formula used to calculate molarity of
is as follows.

Substitute the values into above formula as follows.

As molarity is the number of moles of a substance present in a liter of solution.
Total volume of solution = 
= 21.35 mL + 25.0 mL
= 46.36 mL (1 mL = 0.001 L)
= 0.04636 L
Therefore, moles of acid required are calculated as follows.

Thus, we can conclude that there are 0.006 moles of acid in the flask.
Answer:
3 different elements are present in the compound.
During photosynthesis, 5 moles of water are needed to produce 150 grams of glucose. The correct option is D.
<h3>What is photosynthesis?</h3>
It is the process by which green plants and some other organisms use sunlight to synthesize nutrients from carbon dioxide and water.
- Step 1: Write the balanced equation for photosynthesis.
6 CO₂ + 6 H₂O ⇒ C₆H₁₂O₆ + 6 O₂
- Step 2: Convert 150 g of C₆H₁₂O₆ to moles.
The molar mass of C₆H₁₂O₆ is 180.16 g/mol.
150 g × 1 mol/180.16 g = 0.833 mol
- Step 3: Calculate the moles of water required to form 0.833 moles of C₆H₁₂O₆
The molar ratio of H₂O to C₆H₁₂O₆ is 6:1.
0.833 mol C₆H₁₂O₆ × 6 mol H₂O/1 mol C₆H₁₂O₆ = 5.00 mol H₂O
During photosynthesis, 5 moles of water are needed to produce 150 grams of glucose. The correct option is D.
Learn more about photosynthesis here: brainly.com/question/3529377
#SPJ1
Answer:
The pH of the solution is 5.31.
Explanation:
Let "
is the dissociation of weak acid - HCN.
The dissociation reaction of HCN is as follows.

Initial C 0 0
Equilibrium c(1-
) c
c
Dissociation constant = 

In this case weak acids
is very small so, (1-
) is taken as 1.


From the given the concentration = 0.050 M
Substitute the given value.

![[H_{3}O^{+}]=c\alpha](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%5D%3Dc%5Calpha)
![[H_{3}O^{+}]=0.05\times 9.8\times 10^{-4}= 4.9\times10^{-6}](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%5D%3D0.05%5Ctimes%209.8%5Ctimes%2010%5E%7B-4%7D%3D%204.9%5Ctimes10%5E%7B-6%7D)
![pH= -log[H_{3}O^{+}]](https://tex.z-dn.net/?f=pH%3D%20-log%5BH_%7B3%7DO%5E%7B%2B%7D%5D)
![=-log[4.9\times10^{-6}]](https://tex.z-dn.net/?f=%3D-log%5B4.9%5Ctimes10%5E%7B-6%7D%5D)

Therefore, The pH of the solution is 5.31.