The process by which water falls to earth as rain,snow,hail,or sleet is called what

This type of chemical bond forms when electrons are either gained or lost by an atom.

Ket [755]

It's ionic because electrictrons are only lost or gained in ionic bonds

5 0

3 years ago

Using the equations 2 Sr(s) + O₂ (g) → 2 SrO (s) ∆H° = -1184 kJ/mol SrO (s) + CO₂ (g) → SrCO₃ (s) ∆H° = -234 kJ/mol CO₂ (g) → C(

kkurt [141]

Answer: The $\Delta H^o_{rxn}$ for the reaction is 72 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

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

The given chemical reaction follows:

$2SrCO_3(s)\rightarrow 2Sr(s)+2C(s)+3O_2(g)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $2Sr(s)+O_2(g)\rightarrow 2SrO(s)$ $\Delta H_1=-1184kJ$

(2) $SrO(s)+CO_2(g)\rightarrow SrCO_3(s)$ $\Delta H_2=-234kJ$ ( × 2)

(3) $CO_2(g)\rightarrow C(s)+O_2(g)$ $\Delta H_3=394kJ$ ( × 2)

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[2\times (-\Delta H_2)]+[2\times (\Delta H_3)]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-1184))+(2\times -(-234))+(2\times (394))]=72kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is 72 kJ.

4 0

3 years ago

The quantity of heat required to change the temperature of 1 g of a substance by 1 c is defined as

slega [8]

I think it's calorie
Hope this helps

6 0

3 years ago

What compound is formed when 2,2-dimethyloxirane (2-methyl-1,2-epoxypropane) is treated with ethanol containing sulfuric acid

vladimir1956 [14]

Answer:

2-ethoxy-2-methylpropan-1-ol

Explanation:

On this reaction, we have an "epoxide" (2-methyl-1,2-epoxypropane). Additionally, we have acid medium (due to the sulfuric acid $H_2SO_4$ ). The acid medium will produce the hydronium ion ( $H^+$ ). This ion would be attacked by the oxygen of the epoxide. Then a carbocation would be produced, in this case, the most stable carbocation is the tertiary one. Then an ethanol molecule acts as a nucleophile and will attack the carbocation. Finally, a deprotonation step takes place to produce 2-ethoxy-2-methylpropan-1-ol.

See figure 1

I hope it helps!

3 0

3 years ago

What is the molar mass of BaBr2? multiple choice on the picture

laiz [17]

B. 297.1 g/mol

Is your answer! ;)

3 0

3 years ago

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