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3 years ago

238 U 92 is an isotope notation. what is the atomic number, mass number, number of protons, and number of neutrons?

Chemistry

1 answer:

sergejj [24]3 years ago

5 0

Atomic Number: 92

Mass Number: 238

Number of Protons: 92

Number of Neutrons: 146 (you calculate this by subtracting 238 from 92)

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Given these reactions, where X represents a generic metal or metalloid 1 ) H 2 ( g ) + 1 2 O 2 ( g ) ⟶ H 2 O ( g ) Δ H 1 = − 241

anygoal [31]

Answer : The enthalpy of the given reaction will be, -1048.6 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.

The main reaction is:

$XCl_4(s)+2H_2O(l)\rightarrow XO_2(s)+4HCl(g)$ $\Delta H=?$

The intermediate balanced chemical reactions are:

(1) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(g)$ $\Delta H_1=-241.8kJ$

(2) $X(s)+2Cl_2(g)\rightarrow XCl_4(s)$ $\Delta H_2=+461.9kJ$

(3) $\frac{1}{2}H_2(g)+\frac{1}{2}Cl_2(g)\rightarrow HCl(g)$ $\Delta H_3=-92.3kJ$

(4) $X(s)+O_2(g)\rightarrow XO_2(s)$ $\Delta H_4=-789.1kJ$

(5) $H_2O(g)\rightarrow H_2O(l)$ $\Delta H_5=-44.0kJ$

Now reversing reaction 2, multiplying reaction 3 by 4, reversing reaction 1 and multiplying by 2, reversing reaction 5 and multiplying by 2 and then adding all the equations, we get :

(1) $2H_2O(g)\rightarrow 2H_2(g)+O_2(g)$ $\Delta H_1=2\times 241.8kJ=483.6kJ$

(2) $XCl_4(s)\rightarrow X(s)+2Cl_2(g)$ $\Delta H_2=-461.9kJ$

(3) $2H_2(g)+2Cl_2(g)\rightarrow 4HCl(g)$ $\Delta H_3=4\times -92.3kJ=-369.2kJ$

(4) $X(s)+O_2(g)\rightarrow XO_2(s)$ $\Delta H_4=-789.1kJ$

(5) $2H_2O(l)\rightarrow 2H_2O(g)$ $\Delta H_5=2\times 44.0kJ=88.0kJ$

The expression for enthalpy of main reaction will be:

$\Delta H=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4+\Delta H_5$

$\Delta H=(483.6)+(-461.9)+(-369.2)+(-789.1)+(88.0)$

$\Delta H=-1048.6kJ$

Therefore, the enthalpy of the given reaction will be, -1048.6 kJ

4 0

3 years ago

What is he difference between ligaments and tendons

inn [45]

Tendons may also attach muscles to structures such as the eyeball. A tendon serves to move the bone or structure. A ligament is a fibrous connective tissue which attaches bone to bone, and usually serves to hold structures together and keep them stable.

7 0

3 years ago

Read 2 more answers

The second-order rate constant for the following gas-phase reaction is 0.041 1/MLaTeX: \cdotâs. We start with 0.438 mol C2F4 in

pantera1 [17]

Answer:

134.8 seconds is the half-life (in seconds) of the reaction for the initial $C_2F_4$ concentration

Explanation:

Half life for second order kinetics is given by:

$t_{\frac{1}{2}=\frac{1}{k\times a_0}$

Integrated rate law for second order kinetics is given by:

$\frac{1}{a}=kt+\frac{1}{a_0}$

$t_{\frac{1}{2}$ = half life

k = rate constant

$a_0$ = initial concentration

a = Final concentration of reactant after time t

We have :

$C_2F_4 \longrightarrow \frac{1}{2} C_4F_8$

Initial concentration of $C_2F_4=[a_o]=\frac{0.438 mol}{2.42 L}=0.1810 mol/L$

Rate constant = k = $0.041 M^{-1} s^{-1}$

$t_{\frac{1}{2}=\frac{1}{k\times a_0}$

$=\frac{1}{0.041 M^{-1} s^{-1}\times 0.1810 mol/L}$

$t_{1/2}=134.8 s$

134.8 seconds is the half-life (in seconds) of the reaction for the initial $C_2F_4$ concentration

3 0

3 years ago

Aluminum metal reacts with bromine, a red-brown liquid with a noxious odor. The reaction is vigorous and produces aluminum bromi

GuDViN [60]

Answer: The mass of bromine reacted is 160.6 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of aluminium = 18.1 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium}=\frac{18.1g}{27g/mol}=0.670mol$

The chemical equation for the reaction of aluminium and bromide follows:

$2Al+3Br_2\rightarrow 2AlBr_3$

By Stoichiometry of the reaction:

2 moles of aluminium reacts with 3 moles of bromine gas

So, 0.670 moles of aluminium will react with = $\frac{3}{2}\times 0.670=1.005mol$ of bromine gas.

Now, calculating the mass of bromine gas, we use equation 1:

Moles of bromine gas = 1.005 moles

Molar mass of bromine gas = 159.81 g/mol

Putting values in equation 1, we get:

$1.005mol=\frac{\text{Mass of bromine}}{159.81g/mol}\\\\\text{Mass of bromine}=(1.005mol\times 159.81g/mol)=160.6g$

Hence, the mass of bromine reacted is 160.6 grams.

5 0

3 years ago

When people smoke, carbon monoxide is released into the air. In a room of volume 70 m3, air containing 5% carbon monoxide is int

KonstantinChe [14]

Answer:

$\frac {dQ}{dt}=0.0001-\frac {Q}{70}\times 0.002$

Explanation:

Given that the volume is $70\ m^3$

The air flows (F) at the rate of $0.002\ m^3/min$ . The concentration of carbon monoxide in the air is 5%.

Let Q be the amount of the carbon monoxide present in the room at time t. then,

$\frac {dQ}{dt}=Rate\ of\ Q_{in}-Rate\ of\ Q_{out}$

Also,

$Rate\ of\ Q_{in}=5\%\ of\ 0.002=0.0001$

$Rate\ of\ Q_{out}=\frac {Q}{70}\times 0.002$

Thus,

$\frac {dQ}{dt}=0.0001-\frac {Q}{70}\times 0.002$

6 0

3 years ago