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2 years ago

Please help make sure its correct thanks

Chemistry

1 answer:

Wewaii [24]2 years ago

5 0

The empirical formula of metal iodide : CoI₃(Cobalt(III) Iodide)

<h3>Further explanation</h3>

13.02 g sample of Cobalt , then mol Co(MW=58.933 g/mol) :

$\tt mol=\dfrac{mass}{Ar}=\\\\mol=\dfrac{13.02~g}{58.933}\\\\mol=0.221$

Mass of metal iodide formed : 97.12 g, so mass of Iodine :

$\tt =mass~metal~iodide-mass~Cobalt\\\\=97.12-13.02\\\\=84.1~g$

Then mol iodine (MW=126.9045 g/mol) :

$\tt \dfrac{84.1}{126.9045}=0.663$

mol ratio of Cobalt and Iodine in the compound :

$\tt 0.221\div 0.663=1\div 3$

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Lead has a density of 11.4 g/cm^3. What is the density in kilograms per cubic meter?

Vikentia [17]

The density in kg/m³ = 1.14 x 10⁴

<h3>Further explanation </h3>

Density is a quantity derived from the mass and volume

Density is the ratio of mass per unit volume

With the same mass, the volume of objects that have a high density will be smaller than objects with a smaller type of density

The unit of density can be expressed in g/cm³ or kg/m³

Density formula:

$\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}$

ρ = density , g/cm³ or kg/m³

m = mass , g or kg

v = volume , cm³ or m³

A density of Lead : ρ = 11.4 g/cm³

the density in kg/m³ :

$\tt 11.4~\dfrac{g}{cm^3}\times \dfrac{kg}{10^3~g}\times \dfrac{cm^3}{10^{-6}~m^3}=\boxed{\bold{1.14\times 10^4~\dfrac{kg}{m^3}}}$

8 0

3 years ago

Calculate the molarity of a salt solution with a volume of 0.250L that contains 0.70 mol of NaCl. (SHOW WORK)

Kazeer [188]

Answer:

= 0.28M

Explanation:

data:

volume = 0.250 L

= 0.250dm^3 ( 1litre = 1dm^3)

moles = 0.70 moles

Solution:

molarity = $\frac{no. of moles}{volume in dm^3}$

= 0.70 / 0.250

molarity = 0.28 M

5 0

1 year ago

How many atoms are in the 3.5 moles of arsenic atoms?

Dmitry_Shevchenko [17]

Number of atom in one mole = 6.022 × 10²³
So, in 3.5 moles, it would be: = 6.022 × 10²³ * 3.5 = 2.1 × 10²⁴

In short, Your Answer would be 2.1 × 10²⁴ atoms

Hope this helps!

6 0

3 years ago

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Maurinko [17]

Answer:

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2 years ago

Calculate the specific heat in J/(g·ºC) of an unknown substance if a 2.50-g sample releases 12.0 cal as its temperature changes

Degger [83]

Answer:

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