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ale4655 [162]
3 years ago
6

Which of the following flows is not driven by pressure differences?

Chemistry
1 answer:
AlekseyPX3 years ago
7 0

Answer:

I think it is option (b) river

You might be interested in
Which statement best describes properties of elements? A) All elements have a similar structure and hardness. B) All elements ar
Elis [28]

Answer:

D

Explanation:

That is the right answer

5 0
3 years ago
explain how energy changes from one form to another in a exothermic reaction. in an endothermic reaction.
Alexxx [7]

Answer:

Exothermic reaction: In exothermic reaction, energy is transferred to the surroundings, and the surrounding temperature increases, this is known as exothermic reaction. In other words energy exits in exothermic reaction. Some example of exothermic reactions are:

1) Neutralisation reaction.

2) Combustion reaction.

3) Some oxidation reaction.

Endothermic reaction: In endothermic reaction, energy is taken in from the surrounding, and the surrounding temperature decreases, this is known as endothermic reaction. In other words energy enters in endothermic reaction. Some example of exothermic reactions are:

1) Thermal decomposition.

2) Reaction between citric acid and sodium hydrogen carbonate.

6 0
3 years ago
6.02 x 1023<br> Significant figures
Bond [772]

Answer:

It depends on the number of significant figures you are changing to

Explanation:

6.02×1023=6158.46

1 sig fig = 6000

2 sig fig = 6200

3 sig fig = 6160

4 sig fig = 6158

5 sig fig = 6158.5

6 sig fig = 6158.46

When solving significant figures you have to consider the number after each number like in the case of changing to two sig fig the number following one is five and when the number is up to or greater than five you add a value of one to the number before it. But in a case where the number is less than five you just leave it like that like in the case of changing to one sig fig

3 0
3 years ago
Calculate ΔHo for the following reaction ussing the given bond dissociation energiesCH4(g) + 2O2(g) --&gt; CO2(g) + 2H2O(g)BOND
Mazyrski [523]

Answer:

The ΔH° for the following reaction is -794 kJ, hence exothermic reaction,

Explanation:

CH_4(g) + 2O_2(g)\rightarrow CO_2(g) + 2H_2O(g) ,ΔH° = ?

We are given with:

\Delta H_{O-O}=142 kJ/mol

\Delta H_{O=O}=498 kJ/mol

\Delta H_{H-O}=459 kJ/mol

\Delta H_{C-H}=411 kJ/mol

\Delta H_{C-O}=358 kJ/mol

\Delta H_{C=O}=799 kJ/mol

ΔH° =  

(Energies required to break bonds on reactant side) - (Energies released on formation of bonds on product side)

\Delta H^o=(1 mol\times 4\times \Delta H_{C-H}+2 mol\times 1\times \Delta H_{O=O})-(1 mol\times 2\times \Delta H_{C=O}+2 mol\times 2\times\Delta H_{H-O})

\Delta H^o=(1 mol\times 4\times 411 kJ/mol+2 mol\times 1\times 498 kJ/mol)-(1 mol\times 2\times 799 kJ/mol+2 mol\times 2\times 459 kJ/mol)

\Delta H^o=-794kJ

\Delta H^o>0 endothermic reaction

\Delta H^o exothermic reaction

The ΔH° for the following reaction is -794 kJ, hence exothermic reaction,

4 0
4 years ago
A 13.5 g sample of an unknown gas occupies 5.10 L at 149.83 kPa and 301 K. What is the molar mass of the gas ?
alisha [4.7K]

Answer:

The molar mass of the gas is 44.19 g/mol

Explanation:

Amount of sample of gas = m = 13.5 g

Volume occupied by the gas = V = 5.10 L

Pressure of the gas = P = 149.83 KPa

1 KPa = 0.00986 atm

P = 149.83 \textrm{ KPa} \times 0.00986 \textrm{ atm/KPa} = 1.48 \textrm{ atm}

Assuming M g/mol to be the molar mass of the gas

Assuming the gas is behaving as an ideal gas

\textrm{PV} =\textrm{nRT} \\\textrm{PV} = \displaystyle \frac{m}{M}\textrm{ RT } \\1.48 \textrm{ atm}\times 5.10 \textrm{ L} = \displaystyle \frac{13.5 \textrm{ g}}{M}\times 0.0821 \textrm{ L.atm.mol}^{-1}.K^{-1}\times 301\textrm{K} \\M = 44.19 \textrm{ g/mol}

The molar mass of gas is 44.19 g/mol

7 0
4 years ago
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