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3 years ago

How many atoms of chlorine are there in 16.5 g of iron (iii) chloride?

1 answer:

7 0

Fecl3 has a RFM of 56 + (35.5 * 3) = 162.5

so find out its moles 16.5 ÷ 162 5 = aprox. 0.1M

then there is a 1 : 3 ratio so 0.1 * 3 = 0.3

then times that by the avogado constant which is 6.02×10^23

0.3 × 6.02×10^23 = ans

hope that helps

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Find the total number of atoms in a sample of cocaine hydrochloride, c17h22clno4, of mass 23.0 mg .

Nataliya [291]

From the periodic table:
mass of carbon = 12 grams
mass of hydrogen = 1 gram
mass of nitrogen = 14 grams
mass of oxygen = 16 grams
mass of chlorine = 35.5 grams
Therefore,
molar mass of <span>c17h22clno4 = 17(12) + 22(1) + 35.5 + 14 + 4(16) = 339.5 grams

number of moles = mass / molar mass
number of moles = (23*10^-3) / (339.5)
number of moles = 6.77 * 10^-5 moles

number of atoms = number of moles * Avogadro's number
number of atoms = 6.77*10^-5 * 6.022*10^-23
number of atoms = 4.079 * 10^-27 atoms</span>

3 0

3 years ago

Find the mass of one atom of uranium-235. Recall that the mass in atomic mass units is equal to the mass in grams of one mole of

stiv31 [10]

Answer:

$3.90*10^{-25}kg/atom$

Explanation:

The molar mass of uranium-235 is 235 g/mol. So one mole of uranium-235 has a mass of 235 g. Put differently 6.022×10^23 atoms of uranium-235 have a mass of 235 g. Knowing that, how can we use that to find the mass of one atom?

mass of one atom = $\frac{235 g}{1mol} *\frac{1 mol}{6.022*10^{23}atoms } *\frac{1kg}{1000g}= 3.90*10^{-25}kg/atom$

7 0

2 years ago

The principal ingredient of glass is ______________________.

Lilit [14]

The principal ingredient of glass is quartz sand (SiO₂).

4 0

3 years ago

A sample of xenon occupies a volume of 736 mL at 2.02 atm and 1 °C. If the volume is changed to 416 mL and the temperature is ch

kozerog [31]

Answer:

$\large \boxed{\text{4.63 atm}}$

Explanation:

To solve this problem, we can use the Combined Gas Laws:

$\dfrac{p_{1}V_{1} }{n_{1}T_{1}} = \dfrac{p_{2}V_{2} }{n_{2}T_{2}}$

Data:

p₁ = 2.02 atm; V₁ = 736 mL; n₁ = n₁; T₁ = 1 °C

p₂ = ?; V₂ = 416 mL; n₂ = n₁; T₂ = 82 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = ( 1 + 273.15) K = 274.15 K

T₂ = (82 + 273.15) K = 355.15 K

(b) Calculate the new pressure

$\begin{array}{rcl}\dfrac{p_{1}V_{1}}{n_{1} T_{1}} & = & \dfrac{p_{2}V_{2}}{n_{2} T_{2}}\\\\\dfrac{\text{2.02 atm}\times \text{736 mL}}{n _{1}\times \text{274.15 K}} & = &\dfrac{p_{2}\times \text{416 mL}}{n _{1}\times \text{355.15 K}}\\\\\text{5.423 atm} & = &1.171{p_{2}}\\p_{2} & = & \dfrac{\text{5.423 atm}}{1.171}\\\\ & = & \textbf{4.63 atm} \\\end{array}\\\text{The new pressure will be $\large \boxed{\textbf{4.63 atm}}$}}$

6 0

2 years ago

Complete the balanced equation: Cu(NO3)2 + Na2CO3 > ?

deff fn [24]

I think its CuCO3+NaNO3 but idk if its right

7 0

2 years ago