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Tresset [83]
2 years ago
13

How many molecules are in 3.6 grams of NaCl?

Chemistry
1 answer:
raketka [301]2 years ago
6 0

Answer:

\boxed{\text{None}}

Explanation:

There are no molecules in NaCl, because it consists only of ions.

However, we can calculate the number of formula units (FU) of NaCl.

Step 1. Calculate the moles of NaCl

\text{No. of moles}=\text{3.6 g NaCl}\times \dfrac{\text{1 mol NaCl}}{\text{63.54 g NaCl}} = \text{0.0567 mol NaCl}

Step 2. Convert moles to formula units

\text{FU} = \text{0.0567 mol NaCl} \times \dfrac{6.022 \times 10^{23}\text{ FU NaCl}}{\text{1 mol NaCl}}\\\\= 3.4 \times 10^{22} \text{ FU NaCl}

There are \boxed{3.4 \times 10^{22} \text{ FU NaCl}} in 3.6 g of NaCl.

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A sample of an unknown substance has a mass of .158 kg if 2,510 J of heat is required to heat the substance from 32°C to 61°C wh
ziro4ka [17]

Answer: 0.548J/g°C

Explanation:

Q = s × m × DeltaT

Q = Heat (J)

S = Specific Heat Capacity

M = mass (g)

DeltaT = Change in temperature (°C)

0.158Kg x 1000 = 158g

2.510J = s x 158g x (61°C-32°C)

2.510J/(158g x 29°C) = s

S = 0.54779.... J/g°C

S = 0.548 J/g°C

7 0
2 years ago
Question 4: Which members of an ecosystem are part of the energy flow?
Vanyuwa [196]

Answer:

The answer is A (number 1)

5 0
2 years ago
Select the NMR spectrum that corresponds best to p-bromoaniline. (see Hint for the structure.) The selected tab will be highligh
lana [24]

Answer:

The NMR spectrum that corresponds best to p-bromoaniline  is the one that is attached in the image below.

Explanation:

For the p-bromoaniline 3 types of hydrogen are observed. The first signal that appears at 3.7 ppm would be from the hydrogens of the NH2 group, the hydrogens in ortho position with respect to the NH2 group give a double at approximately 6.54 ppm, and finally the characteristic 7.21 ppm signal is observed for the hydrogens in meta position with with respect to the NH2 group.

7 0
2 years ago
If a penny is made of 3.11 grams of copper, how many atoms of copper are in the penny
Pie

Answer:

2.94x10²² atoms of Cu

Explanation:

We must work with NA to solve this, where NA is the number of Avogadro, number of particles of 1 mol of anything.

Molar mass Cu = 63.55 g/mol

Mass / Molar mass = Mol → 3.11 g / 63.55 g/m = 0.0489 moles

1 mol  of Cu has 6.02x10²³ atoms of Cu

0.0489 moles of Cu, will have (0.0489  .NA)/ 1 = 2.94x10²² atoms of Cu

8 0
2 years ago
Hafnium has six naturally occurring isotopes: 0.16% of 174Hf, with an atomic weight of 173.940 amu; 5.26% of 176Hf, with an atom
ser-zykov [4K]

Answer:

177.277amu

Explanation:

the total occuring isotopes for Hafnium is =6.

First isotope had an atomic weight of 173.940amu

Second isotope =175.941amu

Third isotope =176.943amu

Fourth isotope=177.944amu

Fifth isotope. =178.946amu

sixth isotope .179.947amu

<em>Avera</em><em>ge</em><em> </em><em>ato</em><em>mic</em><em> </em><em>wei</em><em>ght</em><em> </em><em>of</em><em> </em><em>Haf</em><em>nium</em><em>=</em><em> </em><em>sum</em><em> </em><em>of</em><em> </em><em>all</em><em> </em><em>the </em><em>atomi</em><em>c</em><em> </em><em>weights</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>iso</em><em>topes</em><em>/</em><em> </em><em>Tota</em><em>l</em><em> </em><em>occu</em><em>ring</em><em> </em><em>isotopes</em>

Thus, 173.940amu+175.941amu+176.943amu+177.944amu+178.946amu+179.947amu.= 1063.661amu

Average atomic weight= 1063.661amu /6 = 177.2768333amu

= 177.277amu to 3 decimal places.

6 0
2 years ago
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