Como se ordena la palabra ecrfinaine

A 360. g iron rod is placed into 750.0 g of water at 22.5°C. The water temperature rises to 46.7°C. What was the initial tempera

Grace [21]

Answer: The initial temperature of the iron was $515^0C$

Explanation:

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of iron = 360 g

$m_2$ = mass of water = 750 g

$T_{final}$ = final temperature = $46.7^0C$

$T_1$ = temperature of iron = ?

$T_2$ = temperature of water = $22.5^oC$

$c_1$ = specific heat of iron = $0.450J/g^0C$

$c_2$ = specific heat of water= $4.184J/g^0C$

Now put all the given values in equation (1), we get

$-360\times 0.450\times (46.7-x)=[750\times 4.184\times (46.7-22.5)]$

$T_i=515^0C$

Therefore, the initial temperature of the iron was $515^0C$

4 0

3 years ago

Oxygen gas is collected over water. The atmospheric pressure is 748 mmHg, and the temperature of the oxygen is 29 ºC. The pressu

KonstantinChe [14]

The pressure of the oxygen gas collected : 718 mmHg

<h3>Further explanation</h3>

Given

P tot = 748 mmHg

P water vapour = 30 mmHg

Required

P Oxygen

Solution

Dalton's law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases

Can be formulated:

P tot = P1 + P2 + P3 ....

The partial pressure is the pressure of each gas in a mixture

P tot = P H₂O + P Oxygen

P Oxygen = 748 mmHg - 30 mmHg

P Oxygen = 718 mmHg

6 0

3 years ago

A key step in the extraction of iron from its ore isFeO(s) + CO(g) ⇄ Fe(s) + CO₂(g) Kp = 0.403 at 1000°CThis step occurs in the

Kisachek [45]

A key step in the extraction of iron from its ore isFeO(s) + CO(g) ⇄ Fe(s) + CO₂(g) Kp = 0.403 at 1000°CThis step occurs in the 700°C to 1200°C zone within a blast furnace. the partial pressure of CO2 at equilibrium is 1.58 atm.

The complete reaction of the problem, for better illustration, is

FeO(s) + CO(g) <--> Fe(s) + CO2(g)

The double-tailed arrow signifies that the reaction is in a dynamic chemical equilibrium. When the system is in equilibrium, the forward and the backward reaction rates have an equal ratio of Kp = 0.403 at 1000°C. The formula for Kp is

Kp = [partial pressure of products]/[partial pressure of reactants]

So, first, let's find the partial pressure of the compounds in the reaction.

FeO(s) + CO(g) <--> Fe(s) + CO2(g)

Initial x 1.58 0 0

Change -1.58 -1.58 +1.58 +1.58

------------------------------------------------------------------

Equilbrium x-1.58 0 1.58 1.58

Kp = [(1.58)(1.58)]/[(x-1.58)] = 0.403

x = 7.77 atm (this is the amount of excess FeO)

Therefore, the partial pressure of CO2 at equilibrium is 1.58 atm. There is no more CO because it has been consumed due to excess FeO.

To know more about partial pressure of CO2 at equilibrium

brainly.com/question/16022760

#SPJ4

6 0

2 years ago

At 298 K, the rate constant for a reaction is 0.0346 s-1. What is the rate constant at 350K if the Ea = 50.2kJ/mol

frutty [35]

Answer:

0.702 /s

Explanation:

Rate constant at $[298 \mathrm{~K}, \mathrm{~K}_{1}=3.46 \times 10^{-2} \mathrm{~s}^{-1}$

Rate constant at $350 \mathrm{~K}, \mathrm{~K}_{2}=?$

$T_{1}=298 \mathrm{~K}$

$T_{2}=350 \mathrm{~K}$

Activation energy, $\mathrm{Ea}=50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}$

Use the following equation to calculate $K_{2}$ at $350 \mathrm{~K}$

$\ln \frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}=\frac{\mathrm{Ea}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right]$

Therefore,

$\ln \left(\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}}\right) &=\frac{50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}}{8.314 \mathrm{JK}^{-1} \mathrm{~mole}^{-1}}\left[\frac{1}{298 \mathrm{~K}}-\frac{1}{350 \mathrm{~K}}\right]$

$\ln \left(\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}}\right) &=\frac{50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}}{8.314 \mathrm{JK}^{-1} \mathrm{~mole}^{-1}}\left[\frac{52 \mathrm{~K}}{298 \mathrm{~K} \times 350 \mathrm{~K}}\right]$