The equation is as follows;
2CH3OH(g) = 2CH4(g) + O2(g), ΔH= +252.8 kJ
From the equation; for the reaction to produce 2 moles of methane (32g) an energy of 252.8 kJ is released.
Therefore; for an energy of 82.3 kJ the number of moles that will be produced will be; = (2×82.3)/252.3
= 0.6524 moles
which is equivalent to 0.6524 × 16 = 10.438 g
There, the mass of CH4 produced will be 10.438 g
When 0.34 of HNO₃ is titrated to equivalence using 0.14 l of 0.1 m NaOH then the concentration of HNO₃ is 0.041 M
The reaction of neutralization of HNO₃ with NaOH is
HNO₃ + NaOH → H₂O + NaNo₃
When 1 mole of HNO₃ react with 1 mole of NaOH, based on chemical rection the moles of NaOH at equivalence point are equal to moles of HNO₃ present in solution: -
With the mole and volumes, we can find molarity as follows:
Moles of NaOH = moles HNO₃
⁼ 0.14 L X (0.1 mol NaOH/L) = 0.014 mole NaOH
=0.014 mol HNO₃
Molarity: -

= 0.041 M
Thus, from above solution we concluded that the concentration of HNO₃ solution is 0.041 M.
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It would be D. Cs
Because if you look at a periodic table the first column are the group 1 metals - I included a picture of group 1
Answer:
sry i don't know that language
Answer:
Single displacement and reduction
Explanation:
In a single-displacement reaction, one element exchanges partners with another.

This is a single-displacement reaction, because the element Ca exchanges partners with H.
This is also a reduction/oxidation (redox) reaction, because the optically active of Ca increases from 0 to +2 (oxidation), while the oxidation number of H decreases from +1 to 0 (reduction),
The most common types of reactions are:
- Combination
- Decomposition
- Single displacement
- Double displacement
- Reduction/oxidation