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AleksandrR [38]
3 years ago
7

Uldilges of State

Chemistry
1 answer:
Harman [31]3 years ago
3 0

Answer:

Explanation:

Evaporation is a phase change from liquid to vapor. The right equation must show this change.

During evaporation, liquid acquires more heat energy and changes to gaseous form. In such processes, the molecules of the liquid break lose and becomes gases that are randomized.

            H₂O_{(l)}    →   H₂O_{(g)}

The expression above clearly shows water going from liquid to solid.

This should help to have a glimpse of how to solve this problem since the options are poorly written.    

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The chemical equation for the formation of ammonia is unbalanced. 3H2(g) + N2(g) -----> NH3(g) If three molecules of hydrogen
mestny [16]

Answer: 2 molecules of ammonia

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

The balanced chemical equation for the formation of ammonia is:

3H_2(g)+N_2(g)\rightarrow 2NH_3

According to stoichiometry,

3 molecules of hydrogen combines with 1 molecule of nitrogen to give 2 molecules of ammonia.

8 0
3 years ago
Read 2 more answers
What is the mass of a 1.68-l sample of a liquid that has a density of 0.921g/ml?
Gwar [14]
Hey there!

Volume in mL :

1.68 L  * 1000 => 1680 mL

Density = 0.921 g/mL

Therefore:

Mass = density * Volume

Mass = 0.921 * 1680

Mass = 1547.28 g 
7 0
3 years ago
Hydrogen iodide decomposes slowly to H2 and I2 at 600 K. The reaction is second order in HI, and the rate constant is 9.7×10−6M−
Lady bird [3.3K]

Answer : The molarity after a reaction time of 5.00 days is, 0.109 M

Explanation :

The integrated rate law equation for second order reaction follows:

k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)

where,

k = rate constant = 9.7\times 10^{-6}M^{-1}s^{-1}

t = time taken  = 5.00 days

[A] = concentration of substance after time 't' = ?

[A]_o = Initial concentration = 0.110 M

Now put all the given values in above equation, we get:

9.7\times 10^{-6}=\frac{1}{5.00}\left (\frac{1}{[A]}-\frac{1}{(0.110)}\right)

[A]=0.109M

Hence, the molarity after a reaction time of 5.00 days is, 0.109 M

8 0
3 years ago
Consider the intermediate chemical reactions. 2 equations. First: upper C a (s) plus upper C upper O subscript 2 (g) plus one ha
DochEvi [55]

<u>Answer:</u> When the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed.

<u>Explanation:</u>

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 overall chemical reaction follows:

CaO(s)+CO_2\rightarrow CaCO_3(s)     \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) Ca(s)+CO_2(g)+\frac{1}{2}O_2(g)\rightarrow CaCO_3(s)    \Delta H_1=-812.8kJ  

(2) 2Ca(s)+O_2(g)\rightarrow 2CaO(s)     \Delta H_2=-1269kJ

The expression for enthalpy of the reaction follows:

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

Hence, when the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed.

3 0
3 years ago
Write the empirical formula for at least four ionic compounds that could be formed from the following ions:
love history [14]

Answer:

FeSO_{4},FeCH_{3}CO,Fe_{2}(SO_{4})_{3},Fe_{2}(CH_{3}CO)_{3}

Explanation:

Empirical formula of ionic compound formed by two ions A^{x+} and B^{y-} is A_{y}B_{x} (for x\neq y) of AB (for x = y)

The above empirical formula is in accordance with charge neutrality principle

Here each cation (Fe^{3+} and Fe^{2+}) can form two ionic compounds by combining with two given anions (SO_{4}^{2-} and CH_{3}CO^{2-}).

So the four ionic compounds are: FeSO_{4},FeCH_{3}CO,Fe_{2}(SO_{4})_{3},Fe_{2}(CH_{3}CO)_{3}

4 0
4 years ago
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