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MAXImum [283]
3 years ago
8

If one wants 100g of Ag, how much ag2o is needed?

Chemistry
1 answer:
KonstantinChe [14]3 years ago
3 0

Mass Ag₂O needed = 107.416 g

<h3>Further explanation</h3>

Given

100 g Ag

Required

mass of Ag2O

Solution

Proust stated the Comparative Law that compounds are formed from elements with the same Mass Comparison so that the compound has a fixed composition of elements

mass of Ag in Ag2O :

= ((2 x Ar Ag)/molar mass Ag2O )x mass Ag2O

\tt mass~Ag=\dfrac{2.Ar~Ag}{MW~Ag_2O}\times mass~Ag_2O

Ar Ag : 107,8682 g/mol

MW Ag₂O = 231,735 g/mol

Input the value :

mass Ag₂O=(mass Ag x MW Ag₂O) : (2 x Ar Ag)

mass Ag₂O = (100 g x 231,735 g/mol) : ( 2 x 107,8682 g/mol)

mass Ag₂O = 107.416 g

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cestrela7 [59]
In 1 mol of CH3OH, you have 4 H-atoms (because 3 H-atoms are attached to the C-atom, and one H-atom in the OH group). That means in 0.500 mol of CH3OH, you have 2 H-atoms since it is halved. And then we have Avogadro's constant: 6.02 * 1023.

The question asks for how many hydrogen atoms there are in 0.500 mol CH3OH. Using the numbers that we have (Avogadro's constant and no. of H-atoms), the answer of the question will be something like:

<span>H-atoms in CH3OH = 2 * 6.02 * </span>1023<span> = ~1.2 * 10</span>24

 


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3 years ago
Lt takes 4 hr 39 min for a 2.00-mg sample of radium-230 to decay to 0.25 mg. what is the half-life of radium-230?
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Data:

Co = 2.00 mg
C = 0.25 mg
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Time conversion: 4 hr 39 min = 4.65 hr

1) Replace the data in the equation to find k

C = Co { e ^ (-kt) } => C / Co = e ^ (-kt) => -kt = ln { C / Co} => kt = ln {Co / C}

=> k = ln {Co / C} / t =  ln {2.00mg / 0.25mg} / 4.65 hr = 0.44719

2) Use C / Co  = 1/2 to find the hallf-life

C / Co = e ^ (-kt) => -kt = ln (C / Co)

=> -kt = ln (1/2) => kt = ln(2) => t = ln (2) / k

t = ln(2) / 0.44719 = 1.55 hr.

Answer: 1.55 hr
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Answer:

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Target equation is 4C(s) + 5H2(g) = C4H10

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