How many moles are in 4.6 x 10^19 atoms? *#27 is the question

A 0.08024 M solution of NaOH was used to titrate a solution of unknown concentration of HCl. A 32.08 mL sample of the HCl soluti

iris [78.8K]

Answer:

0.08024

Explanation:

4 0

2 years ago

The critical point of carbon dioxide is 304 k and 73 atm. under which conditions is carbon dioxide a liquid? i. 303 k and 73 atm

Sergeeva-Olga [200]

Answer is: a) I only.
Above critical temperature of CO₂, a gas cannot be liquefied no matter how much pressure is applied. Temperature and pressure above its critical point is called supercritical fluid and this is intermediate between gaseous and liquid states.

7 0

2 years ago

Write a balanced chemical equation, including physical state symbols, for the decomposition of solid calcium carbonate (CaCO_3)

nadya68 [22]

Answer:

CaCO₃(s) → CaO(s) + CO₂(g)

Explanation:

The decomposition reaction always make two compounds from one.

The products always have simpler chemical structure, originated from a determined compound. This can happens spontaneously or by a third party.

A notable example of decomposition is hydrolysis. As for example the case of water, which decomposes and generates oxygen and hydrogen gas

2H₂O (l) → 2 H₂ (g) + O₂ (g)

In this case, the calium carbonate decomposes into CaO and CO₂

These two, are the products of the decomposition.

Of course, the unique reactant is the Calcium Carbonate

The balanced equation is:

CaCO₃(s) → CaO(s) + CO₂(g)

6 0

2 years ago

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Calculate the density of a solid in g/cm3 if it weighs 38.3 kg and has a volume of 0.00463 m3.

RideAnS [48]

Answer: D=8.27 g/cm³

Explanation:

Density is mass/volume. Mass is in grams and volume is in liters. In this case, the problem wants our volume to be in cm³. All we need to do is to make some conversions to convert kg/m³ to g/cm³.

$D=\frac{38.3kg}{0.00463m^3} *\frac{(1m)^3}{(100cm)^3} *\frac{1000g}{1kg}$

With this equation, the m³ and kg cancel out, and we are left with g/cm³.

D=8.27 g/cm³

3 0

2 years ago

A 0.5438 g of a C.H.O. compound was combusted in air to make 1.039 g of CO2 and 0.6369 g H20. What is the empirical formula? Bal

goblinko [34]

Answer:

C₃H₅O₂

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

Explanation:

The reaction can be expressed as:

CₓHₓOₓ + nO₂ → CO₂ + H₂O

Under the assumption that there was a total combustion, all of the carbon in the reactant was combusted into CO₂, so the mass of C contained in the C.H.O. compound is the same mass of C contained in 1.039 g of CO₂:

$1.039gCO_{2}*\frac{1molCO_{2}}{44gCO_{2}} *\frac{1molC}{1molCO_{2}} *\frac{12gC}{1molC} =0.2834gC$

All of the hydrogens atoms in the compound ended up becoming H₂O, so the mass of H contained in the C.H.O. compound is the same mass of H contained in 0.6369 g of H₂O:

$0.6369g*\frac{1molH_{2}O}{18gH_{2}O} *\frac{1molH}{1molH_{2}O} *\frac{1gH}{1molH} =0.0354gH$

Because the compound is composed only by C, H and O, the mass of O in the compound can be calculated by substraction:

0.5438 g Compound - 0.2834 g C - 0.0354 g H = 0.2250 g O

In order to determine the empirical formula, we calculate the moles of each component:

mol C = 0.2834 g C ÷ 12 g/mol = 0.0236 mol C

mol H = 0.0354 g H ÷ 1 g/mol = 0.0354 mol H

mol O = 0.2250 g O ÷ 16 g/mol = 0.0141 mol O

Then we divide those values by the lowest one:

0.0236 mol C ÷ 0.0141 = 1.67

0.0354 mol H ÷ 0.0141 = 2.51

0.0141 mol O ÷ 0.0141 = 1

If we multiply those values by 2, we're left with the empirical formula C₃H₅O₂.

The reaction is:

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

8 0

2 years ago