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Karolina [17]
1 year ago
6

7. When solid carbon reacts with oxygen gas to produce carbon dioxide gas, woul the triangle H value be on the reactant or produ

ct side of the equation?
Chemistry
1 answer:
garik1379 [7]1 year ago
7 0

When solid carbon reacts with oxygen gas to produce carbon dioxide gas. the deltaH (enthalpy change ) value is negative .DeltaH would be on the product side of the equation.

<h3>What is enthalpy change? </h3>

In a thermodynamic system, energy is measured by enthalpy. Enthalpy is a measure of a system's overall heat content and is equal to the system's internal energy plus the sum of its volume and pressure.

Knowing whether q is endothermic or exothermic allows one to characterise the relationship between q and H. An endothermic reaction is one that absorbs heat and demonstrates that heat from the environment is used in the reaction, hence q>0 (positive). For the aforementioned equation, under constant pressure and temperature, if q is positive, then H will also be positive. In a similar manner, heat is transferred to the environment when it is released during an exothermic reaction. Thus, q=0 (negative). Therefore, if q is negative, H will also be negative.

Learn more about enthalpy change here :

brainly.com/question/1445358

#SPJ13

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Black_prince [1.1K]
Im confused here. Add the rest of the question for me to help you
5 0
2 years ago
I’m so lost please help
Vadim26 [7]

d= 8/10 so yah that was a guess

8 0
3 years ago
How many molecules are in 7.62 L of CH4, at 87.5°C and 722 torr
pickupchik [31]

Answer: There are 1.469 \times 10^{23} molecules present in 7.62 L of CH_4 at 87.5^{o}C and 722 torr.

Explanation:

Given : Volume = 7.62 L

Temperature = 87.5^{o}C = (87.5 + 273) K = 360.5 K

Pressure = 722 torr

1 torr = 0.00131579

Converting torr into atm as follows.

722 torr = 722 torr \times \frac{0.00131579 atm}{1 torr}\\= 0.95 atm

Therefore, using the ideal gas equation the number of moles are calculated as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

PV = nRT\\0.95 atm \times 7.62 L = n \times 0.0821 L atm/mol K \times 360.5 K\\n = \frac{0.95 atm \times 7.62 L}{0.0821 L atm/mol K \times 360.5 K}\\= \frac{7.239}{29.59705}\\= 0.244 mol

According to the mole concept, 1 mole of every substance contains 6.022 \times 10^{23} atoms. Hence, number of atoms or molecules present in 0.244 mol are calculated as follows.

0.244 mol \times 6.022 \times 10^{23}\\= 1.469 \times 10^{23}

Thus, we can conclude that there are 1.469 \times 10^{23} molecules present in 7.62 L of CH_4 at 87.5^{o}C and 722 torr.

5 0
3 years ago
Metric conversion ( convert )
Nikitich [7]

Answer:

13. 2g

14. 5000mL

15. 104,000m

16. 160mm

17. 5600000mg

18. 10hs

19. 0.250km

20. 1daL

Explanation:

13. 1000milligram (mg) = 1gram (g)

Hence, 2000mg = 2000/1000

= 2g

14. 1 litre (L) = 1000millilitre (mL)

Hence, 5L = 5 × 1000

= 5000mL.

15. 1kilometre (km) = 1000metre (m)

Hence, 104km = 104 × 1000

= 104,000m

16. 1 centimetre (cm) = 10millimeters (mm)

Hence, 16cm = 16 × 10

= 160mm

17. 1kilogram (kg) = 1000000 milligram (mg)

Hence, 5.6kg = 5.6 × 1000000

= 5600000mg

18. 1 second (s) = 0.01 hectosecond (hs)

Hence, 1000s = 1000 × 0.01

= 10hs

19. 1000metre (m) = 1kilometre (km)

Hence, 250m = 250/1000

= 0.250km

20. 1 centiliter (cl) = 0.001 Decaliter (daL)

Hence, 1000cl = 1000 × 0.001

= 1daL

3 0
3 years ago
Which of the following substance is being reduced in the following reaction.? Cu (s) + 2AgNO3 (aq) + 2Ag (s) + Cu(NO3)2 (aq) A)
nikklg [1K]

<u>Answer:</u> The correct answer is Option A.

<u>Explanation:</u>

Oxidation reaction is defined as the reaction in which an atom looses its electrons. Here, oxidation state of the atom increases.

X\rightarrow X^{n+}+ne^-

Reduction reaction is defined as the reaction in which an atom gains electrons. Here, the oxidation state of the atom decreases.

X^{n+}+ne^-\rightarrow X

For the given chemical reaction:

Cu(s)+2AgNO_3(aq.)\rightarrow 2Ag(s)+Cu(NO_3)_2(aq.)

The half reactions for the above reaction are:

Oxidation half reaction: Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-

Reduction half reaction:  2Ag^+(aq.)+2e^-\rightarrow 2Ag(s)

From the above reactions, copper is loosing its electrons. Thus, it is getting oxidized.

Silver ion is gaining electrons and thus is getting reduced.

Hence, the correct answer is Option A.

3 0
3 years ago
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