Write the dissociation equation for NH4OH.

Deep sea divers can experience the bends due to an increased amount of nitrogen in the blood

garri49 [273]

b is the correct answer

6 0

4 years ago

In the Energy and Specific Heat lab, you measure the temperature change of water to study the specific heat of a metal. What sta

pshichka [43]

The question is incomplete, the complete question is;

In the Energy and Specific Heat lab, you measure the temperature change of water to study the specific heat of a metal. What statement explains the relationship between the water and the metal you are studying? Select one: O The heat lost by the metal plus the heat gained by the water equals 100. O The temperature change of the metal is equal to the temperature change of the water. O The heat lost by the metal is equal to the heat gained by the water. The initial temperature of the metal equals the initial temperature of the water

Answer:

The heat lost by the metal is equal to the heat gained by the water.

Explanation:

When the piece of metal is put into water, heat is lost by the metal and gained by the water.

Recall that energy is conserved hence heat lost by metal must be equal to heat gained by water.

Thus, the relationship between the metal under study and the water is that the metal looses heat to the water and heat lost by metal is equal to heat gained by water.

3 0

3 years ago

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

Use significant digits to express the weight of a 150.0 lb person in kilograms ( 1 lb = 453.6 g)

slamgirl [31]

Answer:

150.0 = 68.038856

Explanation:

kilograms

4 0

2 years ago

50cm3 of a mixture of methane and hydrogen were mixed with excess oxygen and exploded, the product after cooling to the original

marin [14]

Answer:the initial composition of the reactants is

40cm^3 of CH4

40cm^3of H2

100cm^3 of H2O

Explanation:

Balanced reaction is

CH4 +H2+5/2O2______

CO2 +3H2O

Excess KOH at room temperature absorbs CO2 whose volume is given by 40cm^3 i.e the volume by which the solution decreases

So using Gay lussac combining ratio which states that gases combine in volumes that are in simple ratio to each other if gases.

Since CO2 in the equation is 1 mole

Means 1mole represent 40cm^3

So CH4:H2:O2 are in ratio of 1:1:5/2=(40:40:100)cm^3 respectively.

8 0

3 years ago

Write the dissociation equation for NH4OH.​

Write the dissociation equation for NH4OH.