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allsm [11]
3 years ago
12

How many moles of disulfur decafluoride are present in 3.99 x 104 molecules of this compound? moles

Chemistry
1 answer:
arsen [322]3 years ago
6 0

<u>Answer:</u> The number of moles of disulfur decafluoride is 6.62\times 10^{-20}

<u>Explanation:</u>

We are given:

Number of disulfur decafluoride molecules = 3.99\times 10^4

According to mole concept:

6.022\times 10^{23} number of molecules are contained in 1 mole of a compound.

So, 3.99\times 10^4 number of molecules will be contained in = \frac{1mol}{6.022\times 10^{23}}\times 3.99\times 10^{4}=6.62\times 10^{-20}mol of disulfur decafluoride.

Hence, the number of moles of disulfur decafluoride is 6.62\times 10^{-20}

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3 years ago
A compound is 52.0% zinc 9.6% carbon and 38.4% oxygen. Calculate the empirical formula of the compound.
skad [1K]

Answer:

So a compound is 52% Zinc(Zn), 9.6% Carbon(C), and 38.4% Oxygen (O). Let’s first start off by assuming that we have 100 g of this compound. This means that we have 52 g of Zinc, 9.6 g of Carbon, and 38.4 g of Oxygen.Zinc = 65.38 g/molCarbon = 12 g/molOxygen = 16 g/molThis means we have:52 g of Zn(1 mol Zn/65.38 g of Zn) ≈0.8 mol of Zn.9.6 g of C(1 mol C/12 g of C) = 0.8 mol of C38.4 g of O(1 mol of O/16 g of O) = 2.4 mol of O.

Explanation:

What we want to do next is divide each element by the common factor of all of them, which is 0.8. In most cases, you divide each element by the element with the least amount of moles. After we divide each by 0.8, you’ll notice you have 1 Zn, 1 C, and 3 O. This gives you the empirical formula of ZnCO3, or Zinc Carbonate.

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2 years ago
The temperature of the water was measured at different depths of a pond.
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A. Dependent variable
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4 years ago
4. DBearded waste of Co-60 must be stored until it is no longer radioactive. Cobalt-60
Bingel [31]

464 g radioisotope was present when the sample was put in storage

<h3>Further explanation</h3>

Given

Sample waste of Co-60 = 14.5 g

26.5 years in storage

Required

Initial sample

Solution

General formulas used in decay:  

\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}

t = duration of decay  

t 1/2 = half-life  

N₀ = the number of initial radioactive atoms  

Nt = the number of radioactive atoms left after decaying during T time  

Half-life of Co-60 = 5.3 years

Input the value :

\tt 14.5=No.\dfrac{1}{2}^{26.5/5.3}\\\\14.5=No.\dfrac{1}{2}^5\\\\No=\boxed{\bold{464~g}}

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3 years ago
What are the 3 reactants necessary for photosynthesis
Brut [27]

Answer:

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7 0
3 years ago
Read 2 more answers
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