Answer: 44g
Explanation: The formular for finding Moles is ;
Moles = Mass / Molar Mass or Formular Mass.
Base on this question; Moles = 10, Mass = 440g, and Formular Mass = ?
Making 'Formular Mass', subject of the formular; we thus have;
Formular mass = Mass / Moles = 440/ 10 = 44g
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Answer:

Explanation:
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In this case, when we want to balance chemical reactions such as in this case, the idea is to equal to number of atoms of each element at each side of the equation according to the lay of conservation of mass, just as shown below:

Because we have four phosphorous and ten oxygen atoms at each side.
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Answer:
pH = 5.54
Explanation:
The pH of a buffer solution is given by the <em>Henderson-Hasselbach (H-H) equation</em>:
- pH = pKa + log
![\frac{[CH_3COO^-]}{[CH_3COOH]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_3COO%5E-%5D%7D%7B%5BCH_3COOH%5D%7D)
For acetic acid, pKa = 4.75.
We <u>calculate the original number of moles for acetic acid and acetate</u>, using the <em>given concentrations and volume</em>:
- CH₃COO⁻ ⇒ 0.377 M * 0.250 L = 0.0942 mol CH₃COO⁻
- CH₃COOH ⇒ 0.345 M * 0.250 L = 0.0862 mol CH₃COOH
The number of CH₃COO⁻ moles will increase with the added moles of KOH while the number of CH₃COOH moles will decrease by the same amount.
Now we use the H-H equation to <u>calculate the new pH</u>, by using the <em>new concentrations</em>:
- pH = 4.75 + log
= 5.54
This is an acid – base reaction and this always result a salt and water
in a neutralization reaction. <span>
The salt that is formed will be calcium bromide (calcium
is located in group 2 so calcium bromide has a formula of CaBr2)
so essentially we got:
HBr + Ca(OH)2 ------> CaBr2 + H2O </span>
balancing the elements: <span>
<span>2HBr(aq) + Ca(OH)2(aq) --------> CaBr2(aq) +
2H2O(l)</span></span>