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wolverine [178]
3 years ago
12

Why dose earth keep spinning

Chemistry
1 answer:
vichka [17]3 years ago
5 0
Because it tilts on its axis
You might be interested in
Lord kelvin described the concept of absolute zero temperature and the laws relating the change in thermal energy during chemica
kondaur [170]
Lord Kelvin, were he alive today, would be considered a Thermochemist. Thermochemistry is interested in the role of heat in chemical reactions. This includes the role of heat both as a biproduct of chemical reactions and a facilitator.

Kelvin's description of absolute zero is an important concept in thermochemistry. At absolute zero, there is no movement of molecules, and no energy available facilitate chemical reactions. 
6 0
3 years ago
Using standard heats of formation, calculate the standard enthalpy change for the following reaction. 2H2S(g) 3O2(g)2H2O(l) 2SO2
Marizza181 [45]

Answer:

\Delta _rH=-1124.14kJ/mol

Explanation:

Hello!

In this case, since the standard enthalpy change for a chemical reaction is stood for the enthalpy of reaction, for the given reaction:

2H_2S(g) +3O_2(g)\rightarrow 2H_2O(l) +2SO_2(g)

We set up the enthalpy of reaction considering the enthalpy of formation of each species in the reaction at the specified phase and the stoichiometric coefficient:

\Delta _rH=2\Delta _fH_{H_2O,liq}+2\Delta _fH_{SO_2,gas}-2\Delta _fH_{H_2S,gas}-3\Delta _fH_{O_2,gas}

In such a way, by using the NIST database, we find that:

\Delta _fH_{H_2O, liq}=-285.83kJ/mol\\\\\Delta _fH_{SO_2, gas}=-296.84kJ/mol\\\\\Delta _fH_{O_2,gas}=0kJ/mol\\\\\Delta _fH_{H_2S,gas}=-20.50kJ/mol

Thus, we plug in the enthalpies of formation to obtain:

\Delta _rH=2(-285.73kJ/mol)+2(-296.84kJ/mol)-2(-20.50kJ/mol)-3(0kJ/mol)\\\\\Delta _rH=-1124.14kJ/mol

Best regards!

8 0
3 years ago
identify the reagents you would use to convert each of the following compounds into pentanoic acid: (a) 1-pentene (b) 1-bromobut
Morgarella [4.7K]

a)BH3.THF is used to convert 1-pentane to pentanoic acid and b)NaCN is used to convert Bromobutane to pentanoic acid.

a) The conversion of 1-pentane to pentanoic acid using BH3, also known as hydroboration-oxidation, is a two-step reaction involving the reaction of 1-pentane with borane (BH3), followed by oxidation of the resulting 1-pentylborane with hydrogen peroxide or other oxidizing agents.

In the first step, 1-pentane reacts with borane (BH3) to form 1-pentylborane, through a process known as hydroboration. This reaction is catalyzed by a Lewis acid, such as aluminum chloride, and proceeds via a hydride transfer from the borane to the 1-pentane.

In the second step, the 1-pentylborane is oxidized to pentanoic acid using hydrogen peroxide (H₂O₂) or other suitable oxidizing agents. The oxidation is catalyzed by an acid, such as hydrochloric acid (HCl), and proceeds via a proton transfer from the 1-pentylborane to the hydrogen peroxide. The end result is the conversion of 1-pentane to pentanoic acid.

The overall chemical reaction for the conversion of 1-pentane to pentanoic acid using borane (BH₃) and hydrogen peroxide (H₂O₂) is as follows:

1-pentane + BH₃ + H₂O₂ → pentanoic acid + H₂O + BH₂

b)The conversion of 1-Bromo butane to pentanoic acid using sodium cyanide (NaCN) proceeds via a nucleophilic substitution reaction. The reaction mechanism involves the following steps:

1. Attack of the nucleophile, NaCN, on the carbon atom of 1-Bromo butane to form a tetrahedral intermediate.

2. Loss of a proton from the tetrahedral intermediate to form a carbanion.

3. Protonation of the carbanion by water (or another proton source) to form pentanoic acid.

The overall reaction can be represented as follows:

1-Bromo butane + NaCN → Pentanoic Acid + NaBr

To know more about reagents, click below:

brainly.com/question/26283409

#SPJ4

3 0
1 year ago
Help please?
Nadusha1986 [10]
If I am correct only 1
7 0
3 years ago
Read 2 more answers
2. The pressure of the oxygen gas inside a
Flauer [41]

Answer: 4.41 atm

Explanation:

Given that,

Original pressure of oxygen gas (P1) = 5.00 atm

Original temperature of oxygen gas (T1) = 25°C

[Convert 25°C to Kelvin by adding 273

25°C + 273 = 298K

New pressure of oxygen gas (P2) = ?

New temperature of oxygen gas (T2) = -10°C

[Convert -10°C to Kelvin by adding 273

-10°C + 273 = 263K

Since pressure and temperature are given while volume is held constant, apply the formula for Charle's law

P1/T1 = P2/T2

5.00 atm /298K = P2/263K

To get the value of P2, cross multiply

5.00 atm x 263K = 298K x V2

1315 atm•K = 298K•V2

V2 = 1315 atm•K / 298K

V2 = 4.41 atm

Thus, the new pressure inside the canister is 4.41 atmosphere

4 0
3 years ago
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