Thank you for posting your math problem here. To convert 3.9x10^5mg to dg the answer is <span>3.9 x 10^3 dg. Below is the solution:
Solution:
</span><span>1mg=0.01dg
</span><span> dg= 3.9 X 10^5mg
</span>dg = <span>(3.9 X 10^5) x 0.01
dg = </span><span>3.9 x 10^3 </span>
Answer:
9.63 L.
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
So the consumed amounts of hydrochloric acid and bromine are the same to the beginning based on:
In such a way, the yielded moles of hydrobromic acid and chlorine are:
Thus, the volume of the sample, after the reaction is the same as no change in the total moles is evidenced, that is 9.63L.
Best regards.
There are different formula you need to keep in mind when solving for [OH-]
Given that pH = 6.10
pH + pOH = 14
6.10 + pOH = 14
pOH = 7.9
[OH-] = 10^(-pOH)
[OH-] = 10^(-7.9)
[OH-] = 0.000000013
[OH-] = 1.3 x 10^-8
<h2>
<u>Answer: [OH-] = 1.3 x 10^-8</u></h2>
Answer: an invisible line around which an object rotates, or spins.
Explanation: //Give thanks(and or Brainliest) if helpful (≧▽≦)//
<h3>
Answer:</h3>
2.125 g
<h3>
Explanation:</h3>
We have;
- Mass of NaBr sample is 11.97 g
- % composition by mass of Na in the sample is 22.34%
We are required to determine the mass of 9.51 g of a NaBr sample.
- Based on the law of of constant composition, a given sample of a compound will always contain the sample percentage composition of a given element.
In this case,
- A sample of 11.97 g of NaBr contains 22.34% of Na by mass
A sample of 9.51 g of NaBr will also contain 22.345 of Na by mass
% composition of an element by mass = (Mass of element ÷ mass of the compound) × 100
Mass of the element = (% composition of an element × mass of the compound) ÷ 100
Therefore;
Mass of sodium = (22.34% × 9.51 g) ÷ 100
= 2.125 g
Thus, the mass of sodium in 9.51 g of NaBr is 2.125 g
