1.25 x 1025 molecules of glucose (C H202) is how many moles of glucose?

For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts.

Nikolay [14]

Explanation:

Molar mass of $KClO_{3}$ = 39.1 + 35.5 + 3(16.0) = 122.6 g

Molar mass of KCl = 39.1 + 35.5 = 74.6 g

Molar mass of $O_{2}$ = 32.0 g

According to the equation, 2 moles of $KClO_{3}$ reacts to give 3 moles of oxygen.

Therefore, 2 (122.6) = 245.2 g of $KClO_{3}$ will give 3 (32.0) = 96.0 g of oxygen. Thus, 245.2 g of $KClO_{3}$ gives 96.0 g of oxygen.

(a) Calculate the amount of oxygen given by 2.72 g of $KClO_{3}$ as follows.

$\frac {96g}{245.2g} \times 2.72 g = 1.06 g$ of $O_{2}$

(b) Calculate the amount of oxygen given by 0.361 g of $KClO_{3}$ as follows.

$\frac {96g}{245.2g} \times 0.361 g = 0.141 g$ of $O_{2}$

c) Calculate the amount of oxygen given by 83.6 kg $KClO_{3}$ as follows.

$\frac {96g}{245.2g} \times 83.6 g = 32.731 kg$ of $O_{2}$

Convert kg into grams as follows.

$\frac{32.731 kg}{1 kg} \times 1000 g$ = 32731 g of $O_{2}$

(d) Calculate the amount of oxygen given by 22.5 mg of $KClO_{3}$ as follows.

$\frac {96g}{245.2g} \times 22.5 mg = 8.79 mg$

Convert mg into grams as follows.

$\frac{8.79 mg}{1 mg}\times 10^{-3} g = 8.79 \times 10^{-3} g$ of $O_{2}$

8 0

4 years ago

A balloon occupies a volume of 2.00 l at 40.0oc. how much volume will it occupy at 30.0oc?

hodyreva [135]

Charles' law says "at a constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature".

V α T

Where V is the volume and T is the temperature in Kelvin of the gas. We can use this for two situations as,
V₁/T₁ = V₂/T₂

V₁ = 2.00 L
T₁ = 40.0 ⁰C = 313 K
V₂ = ?
T₂ = 30.0 ⁰C = 303 K

By applying the formula,
2.00 L / 313 K = V₂ / 303 K
V₂ = (2.00 L / 313 K) x 303 K
V₂ = 1.94 L

Hence, the volume of the balloon at 30.0 ⁰C is 1.94 L

7 0

4 years ago

At 35 C, a sample of gas has a volume of 256 ml and a pressure of 720.torr. What would the volume

Natalija [7]

Answer: Volume would be 196.15 mL if the temperature were changed to $22^{o}C$ and the pressure to 1.25 atmospheres.

Explanation:

Given: $T_{1} = 35^{o}C = (35 + 273) K = 308 K$ , $V_{1}$ = 256 mL,

$P_{1}$ = 720 torr (1 torr = 0.00131579 atm) = 0.947368 atm

$T_{1} = 22^{o}C = (22 + 273) K = 295 K$ , $P_{2} = 1.25 atm$

Formula used to calculate volume is as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

Substitute the values into above formula as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{1 atm \times 256 mL}{308 K} = \frac{1.25 atm \times V_{2}}{295 K}\\V_{2} = 196.15 mL$

Thus, we can conclude that the volume would be 196.15 mL if the temperature were changed to $22^{o}C$ and the pressure to 1.25 atmospheres.

4 0

3 years ago

A 25.00 mL sample containing an unknown amount of Al3+ and Pb2+ required 17.14 mL of 0.04907 M EDTA to reach the end point. A 50

dolphi86 [110]

Answer:

pAl³⁺ = 1,699

pPb²⁺ = 1,866

Explanation:

In this problem, the first titration with EDTA gives the moles of Al³⁺ and Pb²⁺, these moles are:

Al³⁺ + Pb²⁺ in 25,00mL = 0,04907M×0,01714L = 8,411x10⁻⁴ moles of EDTA≡ moles of Al³⁺ + Pb²⁺.

The molar concentration is 8,411x10⁻⁴ moles/0,02500L = 0,0336M Al³⁺+Pb²⁺.

In the second part each Al³⁺ reacts with F⁻ to form AlF₃. Thus, you will have in solution just Pb²⁺.

The moles added of EDTA are:

0,02500L×0,04907M = 1,227x10⁻³ moles of EDTA

The moles of EDTA in excess that react with Mn²⁺ are:

0,02064M × 0,0265L = 5,470x10⁻⁴ moles of Mn²⁺≡ moles of EDTA

That means that moles of EDTA that reacted with Pb²⁺ are:

1,227x10⁻³ moles - 5,470x10⁻⁴ moles = 6,800x10⁻⁴ moles of EDTA ≡ moles of Pb²⁺.

The molar concentration of Pb²⁺ is:

6,800x10⁻⁴mol/0,0500L = 0,0136 M Pb²⁺

Thus, molar concentration of Al³⁺ is:

0,0336M Al³⁺+Pb²⁺ - 0,0136 M Pb²⁺ = 0,0200M Al³⁺

pM is -log[M], thus pAl³⁺ and pPb²⁺ are:

pAl³⁺ = 1,699

pPb²⁺ = 1,866

I hope it helps!

3 0

3 years ago

Nother metal phosphate is iron phosphate. it will behave similar to calcium phosphate in an acid solution. what is the net ionic

Serhud [2]

Iron phosphate when in acidic solution would dissociate into ions namely the iron ions and the phosphate ions. Furthermore, the phosphate ion would react to the hydronium ions forming HPO4^2-. To determine the net ionic equation we do as follows:

FePO4 <---------> Fe3+ + (PO4)3-
(PO4)3- + H3O+ <--------------> HPO4^2- + H2O

Adding the two equations would yield to:

FePO4(s) + H3O+(aq) ⇌ Fe^3+(aq) + HPO4^2−(aq) + H2O(l)

3 0

3 years ago