What does the data in a science experiment show?

UNO [17]

Dc like a battery mate

3 0

3 years ago

Dissolving brass requires an oxidizing acid such as concentrated nitric acid. Nitrogen dioxide is produced as a byproduct in thi

polet [3.4K]

Answer:

Cu + 4 HNO₃ → Cu(NO₃)₂ + 2 NO₂ + 2 H₂O

Explanation:

Step 1: Write down the chemical formulas of given substances,

Copper Metal = Cu

Nitric Acid = HNO₃

Copper (II) Nitrate = Cu(NO₃)₂

Nitrogen Dioxide = NO₂

Water = H₂O

Step 2: Write down the unbalance Chemical equation,

Cu + HNO₃ → Cu(NO₃)₂ + NO₂ + H₂O

Step 3: Balance Cu atoms on both sides;

The number of Cu atoms on both sides are same. Hence, there number will remain the same.

Step 4: Balance N atoms on both sides;

As there is 1 N atom on left hand side and 3 N atoms on right hand side, so we will multiply HNO₃ by 3 to balance N on both sides, hence,

Cu + 3 HNO₃ → Cu(NO₃)₂ + NO₂ + H₂O

Step 5: Balance O atoms on both sides;

As there are 9 O atom on left hand side and 9 O atoms on right hand side, so they are balance.

Step 6: Balance H atoms on both sides;

As there are 3 H atom on left hand side and 2 H atoms on right hand side, so we will multiply H₂O by 2 as,

Cu + 3 HNO₃ → Cu(NO₃)₂ + NO₂ + 2 H₂O

By doing so the number of O atoms got imbalanced, so to balance O atoms again we will multiply HNO₃ by 4 as,

Cu + 4 HNO₃ → Cu(NO₃)₂ + NO₂ + 2 H₂O

Now, The Cu and H atoms are balanced, and the O atoms are greater on left hand side and the N atoms are greater on right hand side, therefore we will multiply NO₂ by 2 to balance both N and O as,

Cu + 4 HNO₃ → Cu(NO₃)₂ + 2 NO₂ + 2 H₂O

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

Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,

WITCHER [35]

Answer: Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

$2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]$

We are given:

$\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?$