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2 years ago

Good evidence for sodium water

1 answer:

5 0

Answer: Understanding the timing and processes of synthesis and racemization of amino acids on asteroid parent bodies is crucial to showing how amino acids in living organisms on Earth grew to be predominantly left-handed. It has been postulated that racemization can occur rapidly depending on several factors, including the pH of the aqueous solution.Here, within the Tagish Lake carbonaceous chondrite, we perform a nanoscale geochemical analysis of a framboidal magnetite grain to show that the interlocking crystal structure formed within a sodium-rich, alkaline fluid environment. We study, in particular, on the discovery of Na-enriched subgrain boundaries and layers of Ca and Mg nanometer-scale surrounding individual framboids.

Explanation:

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Dalam ruang 1 liter, gas metana (ch4) direaksikan dengan 6 mol uap air sesuai reaksi:ch4(

kakasveta [241]

In the space of 1 liter, methane gas (ch4) is reacted with 6 moles of water vapor according to the reaction: if at equilibrium state is obtained 4 moles of hydrogen gas, how many moles of methane gas is needed for the equilibrium reaction?

the reaction is

CH4(g) + 2H2O(g) ----> CO2(g) + 4H2 (g)

Kc = 16 / 3

Kc = [CO2] [H2]^4 / [CH4] [H2O]^2

given :

equilibrium concentration

[H2] = 4 moles

so equilibrium concentration of CO2 must be 1 mole

equilibrium concentration of H2O = 6 - 2 = 4

putting values

16 /3 = [1] [4]^4 / [CH4] [4]^2

[CH4] = 0.333 moles

so moles of CH4 required = 1.33 moles

5 0

2 years ago

Land forms on earth change everyday. Some are worn away and even destroyed. Some are built up and even formed anew. Processes th

kherson [118]

Missing part of the question. Please add which are the process to be classified.

5 0

2 years ago

Read 2 more answers

Question 5

pantera1 [17]

Answer: The coefficient for the diatomic oxygen (O2) is 3.

Explanation:

To know the coefficient for the diatomic Oxygen, we need to balance the equation.

Fe + O2 -------> Fe2O3

LHS of the equation; Fe = 1 , O2 = 1

RHS of the equation; Fe = 2 , O = 3

∴ Multiply 'Fe' on the LHS of the equation by 4 and O2 by 3

Doing that will give the balance equation which is;

4 Fe + 3 O2 --------> 2 Fe2O3

The coefficient for the diatomic oxygen (O2) as seen from the equation is 3.

7 0

2 years ago

The heats of combustion of ethane (C2H6) and butane (C4H10) are 52 kJ/g and 49 kJ/g, respectively. We need to produce 1.000 x 10

LekaFEV [45]

Answer :

(1) The number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 19.23 g and 20.41 g respectively.

(2) The number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 0.641 moles and 0.352 moles respectively.

(3) The balanced chemical equation for the combustion of the fuels.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

(4) The number of moles of $CO_2$ produced by burning each fuel is 1.28 mole and 1.41 mole respectively.

The fuel that emitting least amount of $CO_2$ is $C_2H_6$

Explanation :

Part 1 :

First we have to calculate the number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

As, 52 kJ energy required amount of $C_2H_6$ = 1 g

So, 1000 kJ energy required amount of $C_2H_6$ = $\frac{1000}{52}=19.23g$

and,

As, 49 kJ energy required amount of $C_4H_{10}$ = 1 g

So, 1000 kJ energy required amount of $C_4H_{10}$ = $\frac{1000}{49}=20.41g$

Part 2 :

Now we have to calculate the number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

Molar mass of $C_2H_6$ = 30 g/mole

Molar mass of $C_4H_{10}$ = 58 g/mole

$\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=\frac{19.23g}{30g/mole}=0.641moles$

and,

$\text{ Moles of }C_4H_{10}=\frac{\text{ Mass of }C_4H_{10}}{\text{ Molar mass of }C_4H_{10}}=\frac{20.41g}{58g/mole}=0.352moles$

Part 3 :

Now we have to write down the balanced chemical equation for the combustion of the fuels.

The balanced chemical reaction for combustion of $C_2H_6$ is:

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

and,

The balanced chemical reaction for combustion of $C_4H_{10}$ is:

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

Part 4 :

Now we have to calculate the number of moles of $CO_2$ produced by burning each fuel to produce 1000 kJ.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

From this we conclude that,

As, 1 mole of $C_2H_6$ react to produce 2 moles of $CO_2$

As, 0.641 mole of $C_2H_6$ react to produce $0.641\times 2=1.28$ moles of $CO_2$

and,

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

From this we conclude that,

As, 1 mole of $C_4H_{10}$ react to produce 4 moles of $CO_2$

As, 0.352 mole of $C_4H_{10}$ react to produce $0.352\times 4=1.41$ moles of $CO_2$

So, the fuel that emitting least amount of $CO_2$ is $C_2H_6$

5 0

3 years ago

What is hydrogen ion concentration of an acid solution whose PH is 3.4?

meriva

It’s easy, if the PH of any acidic solution = -Log[H+], where [H+] is hydrogen ion concentration, multiply each term by (-1) then raise each term as a power to (10), so it will become like this:-
[H+] = 10^(-PH)

4 0

2 years ago