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Soloha48 [4]
3 years ago
15

Chemistry questions below! (●'◡'●)

Chemistry
1 answer:
bearhunter [10]3 years ago
4 0

(1) Islands are always surrounded by bodies of water, in this case oceans. These islands tend to have more 'stable climates' and 'smaller temperature fluctuations' due to the specific heat of water, which is quite high, 4180 J/kg C. The higher the specific heat, the more the energy needed to raise the temperature of the substance. Land requires less energy to raise it's temperature than water due to it's lower specific heat. Thus, islands tend to have more stable temperatures than regions surrounded by land.

(2) Based on the conditions provided, the thermochemical equation for the reaction should be:  

<u>2NaHCO₃ (s) ---> Na₂CO₃ (s) + H₂O (g) + O₂ (g)</u>

This is a decomposition reaction, as it explains that the baking soda breaks down into the products. All we had left to do is balance the equation, adding 2 as the coefficient of the reactant.

(3) Let's rewrite the given equations:

First chemical reaction: H₂(g) + F₂(g)  ---> 2HF(g), ΔH₁ = –537 kJ.

Second chemical reaction: C(s) + 2F₂(g) ---> CF₄(g), ΔH₂ = –680 kJ

Third chemical reaction: 2C(s) + 2H₂(g) ---> C₂H₄(g), ΔH₃ = +52.3 kJ.

Fourth chemical reaction: C₂H₄(g) + 6F₂(g) ---> 2CF₄(g) + 4HF(g), ΔH₄ = ?

Now the 'fourth chemical reaction' is the one in which we need to determine the enthalpy:

ΔH₄ = 2(ΔH₁) + 2(ΔH₂) - ΔH₃,

ΔH₄ = 2(–537) + 2(–680) - 52.3 =<u> -2486.3 kJ</u>

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The density of the hydrocarbon in part (a) is 2.0 g l-1 at 50°c and 0.948atm. (i) calculate the molar mass of the hydrocarbon. (
Vedmedyk [2.9K]

1. Answer;

=56 g/mol

Explanation and solution;

PV = nRT

nRT= mass/molar mass (RT)

molar mass = (mass/V ) × (RT/P)

                   = Density × (RT/P)

Molar mass = 2.0 g/L × (0.0821 × 323 K)/0.948 atm

Molar mass = 56 g/mol


2. Answer;

Molecular mass is C4H8

Explanation;

Empirical mass × n = molar mass

Empirical mass for CH2 = 14 g/mol

Therefore;

56 g/mol = 14 g/mol × n

   n = 4

The molecular formula= 4(CH2)

    = C4H8


7 0
2 years ago
Argon, which comprises almost 1 percent of the atmosphere, is approximately 27 times more abundant than CO2 but does not contrib
kykrilka [37]

B. Argon's vibrational energy is not excited by infrared radiation.

Explanation:

The property of carbon dioxide to get excited by infrared electromagnetic radiation is what qualifies it as a greenhouse gas. When infrared from the earth's surface is reflecting back to space, some of the radiation is absorbed and remitted by the carbon-dioxide molecules in the atmosphere. This causes a phenomenon called greenhouse effect that causes the atmosphere to be relatively warmer. The more the carbon dioxide molecules the more the greenhouse effect.

6 0
3 years ago
What percentage of a radioactive species would be found as daughter material after six half–lives?
Novosadov [1.4K]
100%.....50%.....25%......12.5%......6.25%......3.125%......1.5625%
...........1............2...........3..............4.................5................6..................

After six half-lives would be found 1.5625% of readioactive species.
3 0
3 years ago
How many mL of 2M stock solution would I use to prepare 0.500 L of 0.1 M NaCl?
nordsb [41]

Answer:

25 mL

Explanation:

Step 1: Given data

  • Concentration of the concentrated solution (C₁): 2 M
  • Volume of the concentrated solution (V₁): ?
  • Concentration of the diluted solution (C₂): 0.1 M
  • Volume of the diluted solution (V₂): 0.500 L

Step 2: Calculate the volume of the concentrated NaCl solution

We will use the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂ / C₁

V₁ = 0.1 M × 0.500 L / 2 M

V₁ = 0.025 L = 25 mL

4 0
3 years ago
A tank contains 90 kg of salt and 2000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain
sashaice [31]

Answer:

a) 90 kg

b) 68.4 kg

c) 0 kg/L

Explanation:

Mass balance:

-w=\frac{dm}{dt}

w is the mass flow

m is the mass of salt

-v*C=\frac{dm}{dt}

v is the volume flow

C is the concentration

C=\frac{m}{V+(6-3)*L/min*t}

-v*\frac{m}{V+(6-3)*L/min*t}=\frac{dm}{dt}

-3*L/min*\frac{m}{2000L+(3)*L/min*t}=\frac{dm}{dt}

-3*L/min*\frac{dt}{2000L+(3)*L/min*t}=\frac{dm}{m}

-3*L/min*\int_{0}^{t}\frac{dt}{2000L+(3)*L/min*t}=\int_{90kg}^{m}\frac{dm}{m}

-[ln(2000L+3*L/min*t)-ln(2000L)]=ln(m)-ln(90kg)

-ln[(2000L+3*L/min*t)/2000L]=ln(m/90kg)

m=90kg*[2000L/(2000L+3*L/min*t)]

a) Initially: t=0

m=90kg*[2000L/(2000L+3*L/min*0)]=90kg

b) t=210 min (3.5 hr)

m=90kg*[2000L/(2000L+3*L/min*210min)]=68.4kg

c) If time trends to infinity the  division trends to 0 and, therefore, m trends to 0. So, the concentration at infinit time is 0 kg/L.

6 0
3 years ago
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