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ArbitrLikvidat [17]
3 years ago
13

Apply the Law of Conservation of Mass to the following problem: During a combustion reaction, 12.2 grams of methane reacts with

14 g of oxygen. The reaction produces carbon dioxide and water. If 20 grams of water are produced, how many grams of carbon dioxide are produced?
Chemistry
1 answer:
Ratling [72]3 years ago
5 0

Answer:

5.8 g of carbon dioxide are produced

Explanation:

The Law of Conservation of Mass states that the mass of the reactants must equal the mass of the products in all chemical reactions.

This is the chemical reaction (combustion)

 CH₄     +   2O₂    →    CO₂   +    2H₂O

12.2 g        14 g             x             20g

Mass in reactants = 12.2 g + 14 g = 26.2 g

Mass in products =  x + 20 g

26.2 g = x + 20g

26.2 g - 20g = x

5.8 g = x

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A box with a mass of 3.5 kg is pushed along a table. If the net force on the box is 35 Newtons, what is the acceleration of the
Vesnalui [34]

Answer:

<h2>10 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

m is the mass

f is the force

From the question

mass = 3.5 kg

force = 35 N

We have

a =  \frac{35}{3.5}  \\  = 10

We have the final answer as

<h3>10 m/s²</h3>

Hope this helps you

8 0
3 years ago
Silver occurs in trace amounts in some ores of lead, and lead can displace silver from solution: Pb(s) + 2Ag+ (aq) LaTeX: \longr
VikaD [51]

Answer : The value of \Delta G^o and K is, -180 kJ/mol and 3.6\times 10^{31}

Explanation :

The balanced cell reaction will be,

Pb(s)+2Ag^+(aq)\rightarrow Pb^{2+}(aq)+2Ag(g)

The half-cell reactions are:

Oxidation reaction (anode) : Pb(s)\rightarrow Pb^{2+}(aq)+2e^-

Reduction reaction (cathode) : 2Ag^+(aq)+2e^-\rightarrow 2Ag(g)

Relationship between standard Gibbs free energy and standard electrode potential follows:

\Delta G^o=-nFE^o_{cell}

where,

\Delta G^o = standard Gibbs free energy

F = Faraday constant = 96500 C

n = number of electrons in oxidation-reduction reaction = 2

E^o_{cell} = standard electrode potential of the cell = 0.93 V

Now put all the given values in the above formula, we get:

\Delta G^o=-2\times 96500\times 0.93

\Delta G^o=-179490J/mol=-179.49kJ/mol\approx -180kJ/mol

Now we have to calculate the value of 'K'.

\Delta G^o=-RT\ln K

where,

\Delta G_^o =  standard Gibbs free energy  = -180 kJ/mol

R = gas constant = 8.314\times 10^{-3}kJ/mole.K

T = temperature = 298 K

K = equilibrium constant = ?

Now put all the given values in the above formula 1, we get:

-180kJ/mol=-(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln K

K=3.6\times 10^{31}

Therefore, the value of \Delta G^o and K is, -180 kJ/mol and 3.6\times 10^{31}

5 0
3 years ago
Why are scientists worried that climate change will cause these toxic algae blooms to become more frequent?
ipn [44]

Answer: Eutrophication is the enhancement of the growth of algae in the water body.

Explanation:

The scientists are worried for the climate change as if the climate changed to prolonged rainy then the frequent raining can remove toxic chemicals from the agricultural sites, landfills, industries, and from other locations and deposit them to the water body (river, lakes, ponds, and others). The deposition of the salts of nitrogen, phosphorus, and sulfur promotes the growth of algae in the water body. This leads to reduction in the concentration of oxygen in the water body. This is called eutrophication. The lack of oxygen can lead to mortality of aquatic animals.

4 0
2 years ago
"How much NH4Cl, when present in 2.00 liters of 0.200 M ammonia, will give a solution with pH = 8.20? For NH3, Kb = 1.8 x 10-5"
Andru [333]

Answer:

245.66g of NH₄Cl is the mass we need to add to obtain the desire pH

Explanation:

The mixture of NH3/NH4Cl produce a buffer. We can find the pH of a buffer using H-H equation:

pH = pKa + log [A⁻] / [HA]

<em>Where [A⁻] is the molar concentration of the base, NH₃, and [HA] molar concentration of the acid, NH₄⁺. This molar concentration can be taken as the moles of each chemical</em>

<em />

First, we need to find pKa of NH₃ using Kb. Then, the moles of NH₃ and finally replace these values in H-H equation to solve moles of NH₄Cl we need to obtain the desire pH.

  • <em>pKa NH₃/NH₄⁺</em>

pKb = - log Kb

pKb = -log 1.8x10⁻⁵ = <em>4.74</em>

pKa = 14 - pKb

pKa = 14 - 4.74

pKa = 9.26

  • <em>Moles NH₃</em>

<em>2.00L ₓ (0.200mol NH₃ / L) = 0.400 moles NH₃</em>

  • <em>H-H equation:</em>

pH = pKa + log [NH₃] / [NH₄Cl]

8.20 = 9.26 + log [0.400 moles] / [NH₄Cl]

-1.06 =  log [0.400 moles] / [NH₄Cl]

0.0087 =  [0.400 moles] / [NH₄Cl]

[NH₄Cl] = 0.400 moles / 0.0087

[NH₄Cl] = 4.59 moles of NH₄Cl we need to add to original solution to obtain a pH of 8.20. In grams (Using molar mass NH₄Cl=53.491g/mol):

4.59 moles NH₄Cl ₓ (53.491g / mol) =

<h3>245.66g of NH₄Cl is the mass we need to add to obtain the desire pH</h3>

<em />

3 0
4 years ago
Can I have a gamergirl
Tom [10]

Answer:

no

Explanation:

3 0
3 years ago
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