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Shalnov [3]
3 years ago
5

How many moles is 0.025 g of NaCO3? (Show Your Work)

Chemistry
1 answer:
poizon [28]3 years ago
4 0

mol of Na2CO3 = 2.36 x 10⁻⁴

<h3>Further explanation</h3>

Given

Mass : 0.025 g of Na2CO3

Required

moles

Solution

The mole is the number of particles contained in a substance

1 mol = 6.02.10²³

Moles can also be determined from the amount of substance mass and its molar mass  :

mol = mass : molar mass

mass = mol x molar mass

Input the value :

mol = mass : MW Na2CO3

mol = 0.025 g : 106 g/mol

mol = 2.36 x 10⁻⁴

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Heat is a measure of the flow of thermal energy from one object or substance to another. ... Thermal energy typically flows from a warmer material to a cooler material. Generally, when thermal energy is transferred to a material, the motion of its particles speeds up and its temperature increases.

Explanation:

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The acetylene tank contains 35.0 mol C2H2, and the oxygen tank contains 84.0 mol O2.
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Answer:- As per the question is asked, 35.0 moles of acetylene gives 70 moles of carbon dioxide but if we solve the problem using the limiting reactant which is oxygen then 67.2 moles of carbon dioxide will form.

Solution:- The balanced equation for the combustion of acetylene is:

2C_2H_2(g)+5O_2(g)\rightarrow 4CO_2(g)+2H_2O(g)

From the balanced equation, two moles of acetylene gives four moles of carbon dioxide. Using dimensional analysis we could show the calculations for the formation of carbon dioxide by the combustion of 35.0 moles of acetylene.

35.0molC_2H_2(\frac{4molCO_2}{2molC_2H_2})

= 70molCO_2

The next part is, how we choose 35.0 moles of acetylene and not 84.0 moles of oxygen.

From balanced equation, there is 2:5 mol ratio between acetylene and oxygen. Let's calculate the moles of oxygen required to react completely with 35.0 moles of acetylene.

35.0molC_2H_2(\frac{5molO_2}{2molC_2H_2})

= 87.5molO_2

Calculations shows that 87.5 moles of oxygen are required to react completely with 35.0 moles of acetylene. Since only 84.0 moles of oxygen are available, the limiting reactant is oxygen, so 35.0 moles of acetylene will not react completely as it is excess reactant.

So, the theoretical yield should be calculated using 84.0 moles of oxygen as:

84.0molO_2(\frac{4molO_2}{5molO_2})

= 67.2molCO_2

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The gauge pressure inside a vessel is ‐40kPa, at an elevation of 5000m. a) What is the absolute pressure? b) At this elevation,
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Answer:

a) Pabs = 48960 KPa

b) T = 433.332 °C

Explanation:

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∴ d = 1000 Kg/m³

∴ g = 9.8 m/s²

∴ h = 5000 m

∴ P gauge = - 40 KPa * ( 1000 Pa / KPa ) = - 40000 Pa; Pa≡Kg/m*s²

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∴ P = 48960 KPa

⇒ ( 48960 KPa / 0.61094 ) = exp ( 17.625T / (T+ 243.04))

⇒ 80138.803 = exp ( 17.625T / ( T + 243.04))

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⇒ 2744.289 = 17.625T - 11.292T

⇒ 2744.289 = 6.333T

⇒ T = 433.332 °C

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