4. Why does liquid water take on the shape of a cup it is poured into, but solid ice cubes do

Consider the chemical equation. 2H2 + O2 mc022-1.jpg 2H2O What is the percent yield of H2O if 87.0 g of H2O is produced by combi

ch4aika [34]

Answer:

88.50 %

The balance chemical equation is as follow,

2 H₂ + O₂ → 2 H₂O

Step 1: Find the limiting reactant;

According to eq.

4.032 g (2 mole) H₂ reacts with = 32 g (1 mole) of O₂

So,

11 g of H₂ will react with = X g of O₂

Solving for X,

X = (11 g × 32 g) ÷ 4.032 g

X = 87.301 g of O₂

Therefore, H₂ is the limiting reactant as O₂ is present in excess.

Step 2: Calculating %age Yield;

According to eq.

4.032 g (2 mole) H₂ produces = 36.032 g (1 mole) of H₂O

So,

11 g of H₂ will react with = X g of H₂O

Solving for X,

X = (11 g × 36.032 g) ÷ 4.032 g

X = 98.301 g of H₂O

So,

Actual Yield = 87 g

Theoretical Yield = 98.301 g

Using formula = Actual Yield / Theoretical Yield × 100

= 87 g / 98.301 × 100

= 88.50 %

4 0

4 years ago

Read 2 more answers

How can one kg of iron melt more ice than 1 kg lead at 100 °C

Vanyuwa [196]

Answer:

Due to the specific heat capacity of iron, 0.444 J/(g·°C), is more than the specific heat capacity for lead, 0.160 J/(g·°C)

Explanation:

The given parameters are;

The metals provided to melt the ice and their temperature includes;

One kg (1000 g) of iron;

Specific heat capacity = 0.444 J/(g·°C)

Temperature = 100°C

1 kg (1000 g) of lead

Specific heat capacity = 0.160 J/(g·°C)

Temperature = 100°C

Therefore, the heat provided to the ice of mass m, and latent heat of 334 J/g at 0°C by the metals are as follows;

For iron, we have;

ΔQ = Mass × Specific heat capacity × Temperature change

ΔQ $_{iron}$ = Heat obtained from the iron by the ice

ΔQ $_{iron}$ = 0.444 m × 1000 × (100 - 0) = 44400 J

Heat absorbed by the ice for melting, H $_l$ = Heat obtained from the iron

∴ Heat absorbed by the ice for melting, H $_l$ = Mass of ice × Latent heat of ice

H $_l$ = Mass of ice × 334 J/g = 44400 J

Mass of ice melted by the iron = 44400 J/334 (J/g) ≈ 132.9 g

Mass of ice melted by the iron ≈ 132.9 g

For lead, we have;

ΔQ = Mass × Specific heat capacity × Temperature change

ΔQ $_{lead}$ = Heat obtained from the iron by the ice

ΔQ $_{lead}$ = 0.160 m × 1000 × (100 - 0) = 16000 J

Heat absorbed by the ice for melting, H $_l$ = Heat obtained from the iron

∴ Heat absorbed by the ice for melting, H $_l$ = Mass of ice × Latent heat of ice

H $_l$ = Mass of ice × 334 J/g = 16000 J

Mass of ice melted by the lead = 16000 J/334 (J/g) ≈ 47.9 g

Mass of ice melted by the lead ≈ 47.9 g

Therefore, mass of ice melted by the iron, approximately 132.9 g, is more than mass of ice melted by the lead, approximately 47.9 g.

3 0

4 years ago

Which of the following is the correct Lewis dot diagram for Br found in LiBr.

Bas_tet [7]

The correct option is D.

Hope this helps!! (:

4 0

3 years ago

Read 2 more answers

If the reaction N2 (g) + 3 H2 (g) --> 2 NH3 (g) has the concentrations 1.1 M for nitrogen, 0.75 M for hydrogen and 0.25 M fo

Luba_88 [7]

Answer: The value of $K_c$ is 0.136 and is reactant favored.

Explanation:

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{c}$

For the chemical reaction between carbon monoxide and hydrogen follows the equation:

$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)$

The expression for the $K_{c}$ is given as:

$K_{c}=\frac{[NH_3]^2}{[N_2][H_2]^3}$

We are given:

$[NH_3]=0.25M$

$[H_2]=0.75M$

$[N_2]=1.1M$

Putting values in above equation, we get:

$K_c=\frac{(0.25)^2}{1.1\times (0.75)^3}$

$K_c=0.135$

There are 3 conditions:

When $K_{c}>1$ ; the reaction is product favored.
When $K_{c}$ ; the reaction is reactant favored.
When $K_{c}=1$ ; the reaction is in equilibrium.

For the given reaction, the value of $K_c$ is less than 1. Thus, the reaction is reactant favored.

Hence, the value of $K_c$ is 0.136 and is reactant favored.

4 0

3 years ago

What factors could have affected your results?

AleksandrR [38]

Answer:

See below

Step-by-step explanation:

heat gained by metal + heat lost by water = 0

m₁C₁ΔT₁ + m₂C₂ΔT₂ = 0

C₁ = -(m₂C₂ΔT₂)/(m₁ΔT₁)

The factors determining C₁ are

mass of water
temperature change of water (T_f - Ti)
mass of metal
temperature change of metal (T_f - Ti)

Any factor that makes the numerator higher or the denominator lower than what you thought, will give a calculated C₁ that is too high (and vice versa).

The major sources of uncertainty are probably in determining the temperatures, especially the initial and final temperatures of the metal. However, you will have to decide what the principal factors were in your experiment.

For example, did the metal have a chance to cool during the transfer to the calorimeter? How easy was it to determine the equilibrium temperature, etc?

Factors Affecting the Calculation of Specific Heat Capacity

Too Low Too high

Water Water

Mass less than thought Mass more than thought

Ti lower Ti higher

T_f higher T_f lower

Metal Metal

Mass more than thought Mass less than thought

Ti higher Ti lower

8 0

3 years ago