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Alinara [238K]
3 years ago
12

A white coloured solid salt “A” on heating gives out two colourless gases “B” and “C”.

Chemistry
1 answer:
QveST [7]3 years ago
6 0
I think it say go rggdggy and do something with one of those other things and get a work order and get the t off and u I know I have
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a 6.7 volume of air, initially at 23 degrees celsius and .98 atm, is compressed to 2.7 L while heated to 125 degrees celsius. Wh
Annette [7]
Data:

V1 = 6.7 liter
T1 = 23° = 23 + 273.15 K = 300.15 K
P1 = 0.98 atm

V2 = 2.7 liter
T2 = 125° = 125 + 273.15 K = 398.15 K
P2 = ?

Formula:

Combined law of ideal gases: P1 V1 / T1 = P2 V2 / T2

=> P2 = P1 V1 T2 / (T1 V2)

P2 = 0.98 atm * 6.7 liter * 398.15 K / (300.15K * 2.7 liter)

P2 = 3.22 atm
6 0
4 years ago
.) Calculate the mass in grams of each of the following samples.
Alik [6]
 12+12+6(1)+16 =46      46*.251 = 11.546 gram
4 0
3 years ago
Read 2 more answers
A chemical substance that increases the reaction rate and lowers the activation energy needed to start a reaction is
Elza [17]
A catalyst lowers the activation energy needed to start a reaction
4 0
3 years ago
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HELP PLS! I NEED TO GET THIS RIGHT! Curious Carl and his lab partner were conducting a variety of experiments to produce gases:
katen-ka-za [31]

Answer:

0.25 mole of H₂

Explanation:

The equation for the reaction is given below:

Mg + 2HCl —> MgCl₂ + H₂

From the question given above, we were told that the reaction produced 5.6 L of H₂.

Thus, we can obtain obtain the number of mole of H₂ that occupied 5.6 L at STP as follow:

1 mole of H₂ occupied 22.4 L at STP.

Therefore, Xmol of H₂ will occupy 5.6 L at STP i.e

Xmol of H₂ = 5.6 / 22.4

Xmol of H₂ = 0.25 mole

Therefore, 0.25 mole of H₂ were obtained from the reaction.

8 0
3 years ago
Calculate the standard enthalpy of formation of liquid methanol, CH3OH(l), using the following information: C(graphite) + O2 --&
Darya [45]

Answer : The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation :

Standard formation of reaction : It is a chemical reaction that forms one mole of a substance from its constituent elements in their standard states.

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of CH_3OH will be,

C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g)    \Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

(1) C(graphite)+O_2(g)\rightarrow CO_2(g)    \Delta H_2=-393.5kJ/mole

(2) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_3=-285.8kJ/mole

(3) CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l)     \Delta H_1=-726.4kJ/mole

Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, we get :

(1) C(graphite)+O_2(g)\rightarrow CO_2(g)    \Delta H_1=-393.5kJ/mole

(2) 2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)    \Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol

(3) CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g)     \Delta H_3=726.4kJ/mole

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)

\Delta H=-238.7kJ/mole

Therefore, the standard enthalpy of formation of methanol is, -238.7 kJ/mole

6 0
3 years ago
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