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3 years ago

What is the SI unit for distance ? A) cm B) km C) m D) mm

Chemistry

2 answers:

Nuetrik [128]3 years ago

7 0

<span>The SI base unit for distance is the meter</span>

Dafna1 [17]3 years ago

4 0

The answer is C) Meter.

Hope this helps!!

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Enter your answer in the provided box. Oxidation of gaseous ClF by F2 yields liquid ClF3, an important fluorinating agent. Use t

MakcuM [25]

Answer : The value of $\Delta H_{rxn}$ for the reaction is, -135.2 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 of $ClF_3$ will be,

$ClF(g)+F_2(g)\rightarrow ClF_3(l)$ $\Delta H_{rxn}=?$

The intermediate balanced chemical reaction will be,

(1) $2ClF(g)+O_2(g)\rightarrow Cl_2O(g)+OF_2(g)$ $\Delta H_1=167.5kJ$

(2) $2F_2(g)+O_2(g)\rightarrow 2OF_2(g)$ $\Delta H_2=-43.5kJ$

(3) $2ClF_3(l)+2O_2(g)\rightarrow Cl_2O(g)+3OF_2(g)$ $\Delta H_3=394.4kJ$

We are dividing the reaction 1, 2 and 3 and reversing reaction 3 and then adding all the equations, we get :

(1) $ClF(g)+\frac{1}{2}O_2(g)\rightarrow \frac{1}{2}Cl_2O(g)+\frac{1}{2}OF_2(g)$ $\Delta H_1=\frac{167.5kJ}{2}=83.75kJ$

(2) $F_2(g)+\frac{1}{2}O_2(g)\rightarrow OF_2(g)$ $\Delta H_2=\frac{-43.5kJ}{2}=-21.75kJ$

(3) $\frac{1}{2}Cl_2O(g)+\frac{3}{2}OF_2(g)\rightarrow ClF_3(l)+O_2(g)$ $\Delta H_3=\frac{-394.4kJ}{2}=-197.2kJ$

The expression for enthalpy of formation of $ClF_3$ will be,

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

$\Delta H_{rxn}=(83.75kJ)+(-21.75kJ)+(-197.2kJ)$

$\Delta H_{rxn}=-135.2kJ$

Therefore, the value of $\Delta H_{rxn}$ for the reaction is, -135.2 kJ

5 0

4 years ago

Some items, such as rubber, float on the water, while some items, such as gold, sink in

elena-14-01-66 [18.8K]

Answer:

If an item is less dense, it floats on the water.

Explanation:

We know this since the more dense something it usually is heavier than if it was less dense. Which ways it down resulting to it sinking.

8 0

3 years ago

A solution contains two isomers, n-propyl alcohol and isopropyl alcohol, at 25°C. The total vapor pressure is 38.6 torr. What ar

Leona [35]

Answer:

Mole fraction of alcohols in liquid phase $x_1=0.2727\& x_2=0.7273$ .

Mole fraction of alcohols in vapor phase $y_1=0.1468\& y_2=0.8516$ .

Explanation:

The total vapor pressure of the solution = p =38.6 Torr

Partial vapor pressure of the n-propyl alcohol = $p^{o}_1=21.0 Torr$

Partial vapor pressure of the isopropyl alcohol = $p^{o}_2=45.2 Torr$

$p=x_1\times p^{o}_1+x_2\times p^{o}_2$ (Raoult's Law)

$p=x_1\times p^{o}_1+(1-x_1)\times p^{o}_2$

$38.6 Torr=x_1\times 21.0 Torr+(1-x_1)\times 45.2 Torr$

$x_1=0.2727$

$x_2=1-0.2727=0.7273$

$x_1\& x_2$ is mole fraction in liquid phase.

Mole fraction of components in vapor phase $y_1\& y_2$

$p_1=y_1\times p$ (Dalton's law of partial pressure)

$y_1=\frac{p_1}{38.6 Torr}=\frac{p^{o}_1\times x_1}{38.6 Torr}$

$y_1=\frac{21.0 Torr\times 0.2727}{38.6 Torr}=0.1468$

$y_1=\frac{p_2}{38.6 Torr}=\frac{p^{o}_2\times x_2}{38.6 Torr}$

$y_2=\frac{45.2 Torr\times 0.7273}{38.6 Torr}=0.8516$

Mole fraction of alcohols in vapor phase $y_1=0.1468\& y_2=0.8516$

5 0

3 years ago

Salt water can be desalinated by distillation. How much energy is needed to convert 175 g of salt water at 28.0 °C to water vapo

Vinil7 [7]

Answer:

$\Delta H=4.63x10^{5}J=463kJ$

Explanation:

Hello,

In this case, we find the following states:

a. Liquid salt water at 28.0 °C.

b. Liquid salt water at 102.5 °C.

c. Vapor salt water at 102.5 °C.

The first process (1) is to heat the liquid water from 28.0 °C to 102.5 °C and the second one (2) to vaporize the liquid salt water. In such a way, each process has an amount of energy that when added, yields the total energy for the process as shown below:

$\Delta H=\Delta H_1+\Delta H_2\\\Delta H=mCp\Delta T+m\Delta _VH\\\Delta H=175g*5.19\frac{J}{g*K}*(375.65K-301.15K)+175g*2.26\frac{kJ}{g} *\frac{1000J}{1kJ} \\\Delta H=4.63x10^{5}J=463kJ$

Best regards.

4 0

3 years ago

Here is a more complex redox reaction involving the permanganate ion in acidic solution: 5fe2+ + 8h+ + mno4− → 5fe3+ +mn2+ + 4h2

dezoksy [38]

<span>Answer: Here is a more complex redox reaction involving the permanganate ion in acidic solution: 5Fe2+ + 8H+ + MnO4 --> 5Fe3+ +Mn2+ + 4H2O Classify each reactant as the reducing agent, oxidizing agent, or neither.</span>

4 0

3 years ago