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TiliK225 [7]
4 years ago
13

7.26 of a hydrate, Cu(NO3)2.xH2O, formed 2.4 g copper(II) oxide.

Chemistry
1 answer:
goldfiish [28.3K]4 years ago
6 0

Number of moles= mass/ molar mass

Or n=m/MM

n = number of moles

m = mass

MM = molar mass

1) n CuO = 2.4g / 79.54g/mol = 0.03 mol CuO

2) n Cu(NO3)2.xH2O = 7.26 g / 205.6 = 0.035 moles of Cu(NO3)2.xH2O

3) 205.6 g

Cu = 63.5 g

N = 14g

O =16g

H= 1 g

63.5+ (14+(16*3))*2+1*2+16 =205.6 g

4) yes is 188g

5) I don’t know, I assume was 1

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What is the mass of 1.7 × 1023 atoms of zinc (Zn)?
Rudiy27

Number of moles is defined as the ratio of given mass in g to the molar mass.

The mathematical formula is:

Number of moles = \frac{given mass in g}{molar mass}   (1)

Number of zinc atoms is equal to 1.7\times10^{23}, by Avogadro number, number of moles can be calculated.

As, 1 mol=6.022\times10^{23} atoms, hence,

1.7\times10^{23}\times\frac{1 mol}{6.022\times10^{23} }

= 0.2822 mol

Now, from formula (1), calculate mass in g (molar mass of zinc = 65.4 g/mol)

0.2822 mol = \frac{mass in g}{65.4 g/mol}

mass in g  = {0.2822 mol \times 65.4 g/mol

= 18.45588 g

Thus, by rounding off the above number, it will come near about 19 g approximately.

Hence, option (C) is the correct answer.



 



8 0
4 years ago
A fishing line sinker contains 0.650 moles of lead. How many atoms of lead are in the sinker? Show all of your work, including s
34kurt

Answer:

3.91x10²³ atoms of lead

Explanation:

In chemistry, a mole of a substance is defined as 6.022x10²³ particles that could be atoms, molecules, ions, etc.

As you can see, in the problem, you have 0.650moles of lead in a fishing line sinker, the present atoms are:

0.650mol Pb \frac{6.022x10^{23}atoms}{1mol} =<em> 3.91x10²³ atoms of lead</em>

7 0
3 years ago
True or False: As the ball rises from point 1 to point 3, it slows down
quester [9]
True, it would slow down.
7 0
3 years ago
Consider a certain type of nucleus that has a half-life of 32 min. calculate the percent of original sample of nuclides remainin
Irina18 [472]
t1/2 = ln 2 / λ = 0.693 / λ
Where t1/2 is the half life of the element and λ is decay constant.

32 = 0.693 / λ 
λ   = 0.693 / 32          (1) 

Nt = Nο eΛ(-λt)          (2)

Where Nt is atoms at t time, λ is decay constant and t is the time taken.
t = 1.9 hours = 1.9 x 60 min

From (1) and (2),


Nt = Nο e⁻Λ(0.693/32)*1.9*60
Nt =  0.085Nο 

Percentage = (Nt/Nο) x 100%
                   = (0.085Nο/Nο) x 100%
                   = 8.5%

Hence, Percentage of remaining atoms with the original sample is 8.5%

8 0
4 years ago
Urgent help will be much appreciated
Scorpion4ik [409]

Answer:

b c d a

Explanation:

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6 0
3 years ago
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