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Aleonysh [2.5K]

3 years ago

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Isopropyl alcohol is a clear liquid with a density of 0.78 g/cm3. If the same small sphere (with a density of 0.88 g/cm3) is dro

pped in a beaker of isopropyl alcohol, predict whether the sphere will float or sink?

1 answer:

Burka [1]3 years ago

6 0

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Find the standard enthalpy of formation of ethylene, C2H4(g), given the following data: C2H4(g) + 3 O2(g) --> 2CO2(g) + 2 H2O

Brums [2.3K]

Answer : The standard enthalpy of formation of ethylene is, 52.4 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

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

The formation reaction of $C_2H_4$ will be,

$2C(s)+2H_2(g)\rightarrow C_2H_4(g)$ $\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

(1) $C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)$ $\Delta H_1=-1411kJ$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.5kJ$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_3=-285.8kJ$

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equation, we get :

(1) $2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)$ $\Delta H_1=+1411kJ$

(2) $2C(s)+2O_2(g)\rightarrow 2CO_2(g)$ $\Delta H_2=2\times (-393.5kJ)=-787kJ$

(3) $2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_3=2\times (-285.8kJ)=-571.6kJ$

The expression for enthalpy of formation of $C_2H_4$ will be,

$\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(+1411kJ)+(-787kJ)+(-571.6kJ)$

$\Delta H=52.4kJ$

Therefore, the standard enthalpy of formation of ethylene is, 52.4 kJ

7 0

3 years ago

If you rip a dollar bill in half is it qualified as 50 cents

omeli [17]

Yes, 50 pennies plus 50 pennies equals 100 pennies minus 50pennies equals 50 pennies.

7 0

3 years ago

Ammonium nitrate is an ingredient in cold packs used by sports trainers for injured athletes. Calculate the change in temperatur

Ronch [10]

Answer:

See explanation

Explanation:

First, we need to use the correct expression:

Q = m*Cp*ΔT (1)

Q = n*ΔH (2)

These are the 2 expressions to calculate heat or energy.

Now, we want to know the change of temperature of the nitrate in water after being added, so with the innitial data of nitrate, we can calculate heat using the second expression. First, we need to calculate moles with the molecular mass:

n = m/MM

n = 42/80.1 = 0.52 moles

With these moles, we can calculate heat with the ΔH of this reaction:

Q = 0.52 * 25.7 = 13.364 kJ or 13,364 J

However, this heat as is being absorbed, the value would be negative.

Now that we have heat, we can use expression (1) and plug these values to solve for ΔT, but before, we need to know the total mass of the solution (water + nitrate)

m = 250 + 42 = 292 g

now, solving for ΔT:

-13,364 = 292 * 4.18 * ΔT

ΔT = -13,363 / (292 * 4.18)

ΔT = -10.95 °C

3 0

3 years ago

For the reaction 2A(g) â B(g), the equilibrium constant is Kp = 0.76. A reaction mixture initially contains 4.0 atm of gas (PA =

timofeeve [1]

Answer: (a) The reaction mixture will proceed toward products.

Explanation:

Equilibrium constant is defined as the ratio of pressure of products to the pressure of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_p$

K is the constant of a certain reaction when it is in equilibrium, while Q is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

For the given chemical reaction:

$2A(g)\rightleftharpoons B(g)$

The expression for $Q_p$ is written as:

$Q_p=\frac{p_B}{(p_A)^2}$

$Q_p=\frac{(2.0)}{(2.0)^2}$

$Q_p=0.5$

$K_p=0.76$

Thus as $K_p>Q_p$ , the reaction will shift towards the right i.e. towards the product side.

7 0

3 years ago

74.5 g of KCl was dissolved in 1000. mL of water. What is the

Degger [83]

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3 0

3 years ago

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