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ivanzaharov [21]
3 years ago
6

Metal carbonates decompose to the metal oxide and CO2 on heating according to this general equation. Mx(CO3)is) → My(s) + yCO2(g

) You heat 0.0900 g of a white, solid carbonate of a Group 2A metal and find that the evolved CO2 has a pressure of 69.8 mm Hg in a 285 mL flask at 25 °C. Determine the molar mass of the metal carbonate.
Chemistry
1 answer:
Mrrafil [7]3 years ago
4 0

Answer:

84.11 g/mol

Explanation:

A metal from group 2A will form the cation M²⁺, and the ion carbonate is CO₃²⁻, so the metal carbonate must be: MCO₃, and the reaction:

MCO₃(s) → MO(s) + CO₂(g)

For the stoichiometry of the reaction, 1 mol of MCO₃(s) will produce 1 mol of CO₂. Using the ideal gas law, it's possible to calculate the number of moles of CO₂:

PV = nRT , where P is the pressure, V is the volume(0.285 L), R is the gas constant (62.36 mmHg*L/mol*K), n is the number of moles, and T is the temperature (25 + 273 = 298 K).

69.8*0.285 = n*62.36*298

18583.28n = 19.893

n = 0.00107 mol

So, the number of moles of the metal carbonate is 0.00107. The molar mass is the mass divided by the number of moles:

0.0900/0.00107 = 84.11 g/mol

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A package of aluminum foil contains 50. ft2 of foil, which weighs approximately 8.5 oz . aluminum has a density of 2.70 g/cm3. w
Serhud [2]
Hello!

We have the following data:

Area (A) = 50 square feet
Mass (m) = 8.5 ounces
Density (d) = 2.70 g/cm³
Volume (V) = ?
Thickness (T) =? (in mm)

To move on, we must transform the area of 50 ft² in cm², let's see:

1 ft² ------- 929,0304 cm²
50 ft² ----- A

A = 50*929,0304

\boxed{A = 46451,52\:cm^2}\Longleftarrow(Area)

In the same way, we will convert the mass of 8.5 oz in grams, see:

1 oz -------- 28,3495 g
8,5 oz ------- m

m = 8,5*28,3495

\boxed{m = 240,97075\:g}\Longleftarrow(mass)

Knowing that the density is 2.70 g/cm³ and the mass is 240.97075 g, we will find the volume, applying the data in the density formula we have:

d =  \dfrac{m}{V} \to V =  \dfrac{m}{d}

V =  \dfrac{240,97075\:\diagup\!\!\!\!\!g}{2,70\:\diagup\!\!\!\!\!g/cm^3}

V = 89,24842593... \to \boxed{V \approx 89,25\:cm^3}

The statement wants to find the thickness of the packaging, for this we have some important data, such as: V (volume) = 89,25 cm³ and Area (A) = 46451,52 cm² and T (thickness) =? (in mm)

In the calculations of Costs in Surface Treatment of a part within the flat geometry, we will use the following formula:

V (volume) = A (Area) * T (Thickness)

89,25\:cm^3 = 46451,52\:cm^2\:*\:T

46451,52\:cm^2*T = 89,25\:cm^3

T =  \dfrac{89,25\:cm^3}{46451,52\:cm^2}

T = 0,001921358009...\:cm

We will convert to millimeters, going through a decimal place on the right

T = 0,01921358009..\:mm

\boxed{\boxed{T \approx 0,0192\:mm\:or\:T\approx \:1,92*10^{-2}\:mm}}\end{array}}\Longleftarrow(thickness)\qquad\checkmark

Hope this helps! :))





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Read 2 more answers
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According to Newton's first law of motion, it takes an unbalanced force to move an object at rest.

I hope this helps :)

3 0
3 years ago
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