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Korvikt [17]
3 years ago
9

examples of chemical reactions except __________. A. flowers wilting and turning from white to brown B. the heating element on a

stove turning from black to orange C. bleach spilled on a shirt turning it from red to pink
Chemistry
2 answers:
sergey [27]3 years ago
4 0

ANSWER: B) The heating element on a stove turning from black to orange.

The reason being is that, its a physical reaction, the color changes but the chemical structure of the element stays the same. A's changing and breaking down on a molecular level, that's a chemical reaction. Bleach bonded with the clothing making it from red to pink.

azamat3 years ago
4 0

The answer would be A. It wouldn't be B because the gasoline is a chemical which lights when caught with air. Bleach is a chemical that can clean and change the color of things. I don't exactly know how bleach does those things , but I do know the answer would be A.

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Coal gasification is a multistep process to convert coal into cleaner-burning fuels. In one step, a coal sample reacts with supe
ddd [48]

Answer :

The enthalpy of reaction is, -187.6 kJ/mol

The total heat will be, -2251 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.

(a) The formation of CH_4 will be,

2C(coal)+2H_2O(g)\rightarrow CH_4(g)+CO_2(g)    \Delta H_{rxn}=?

The intermediate balanced chemical reaction will be,

(1) C(coal)+H_2O(g)\rightarrow CO(g)+H_2(g)     \Delta H_1=29.7kJ

(2) CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)    \Delta H_2=-41kJ

(3) CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)    \Delta H_3=-206kJ

We are multiplying equation 1 by 2 and then adding all the equations, we get :

(b) The expression for enthalpy of reaction will be,

\Delta H_{rxn}=2\times \Delta H_1+\Delta H_2+\Delta H_3

\Delta H_{rxn}=(2\times 29.7)+(-41)+(-206)

\Delta H_{rxn}=-187.6kJ/mol

Therefore, the enthalpy of reaction is, -187.6 kJ/mol

(c) Now we have to calculate the total heat.

\Delta H=\frac{q}{n}

or,

q=\Delta H\times n

where,

\Delta H = enthalpy change = -187.6 kJ/mol

q = heat = ?

n = number of moles of coal = \frac{1.00\times 1000g}{12.00g/mol}=83.33mol

Now put all the given values in the above formula, we get:

q=(-187.6kJ/mol)\times (83.33mol)=-2.251kJ

Thus, the total heat will be, -2251 kJ

4 0
4 years ago
HURRY PLEASE! True or false: Making an observation is the first step of the scientific method.
lions [1.4K]

Answer:

True

Explanation:

You must first observe your data then form a hypothesis.

8 0
4 years ago
Given the chemical name, choose which answer provides the correct chemical formula. Please use the periodic table that has been
Andrews [41]

Answer:CuNO3

Explanation:

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7 0
3 years ago
Calculate the initial rate for the formation of c at 25 ∘c, if [a]=0.50m and [b]=0.075m
Nataly_w [17]
The rate of formation of a product depends on the the concentrations of the reactants in a variable way.

If two products, call them A and B react together to form product C, a general equation for the formation of C has the form:

rate = k*[A]^m * [B]^n

Where the symbol [ ] is the concentration of each compound.

Then, plus the concentrations of compounds A and B you need k, m and n.

Normally you run controled trials in lab which permit to calculate k, m and n .

Here the data obtained in the lab are:

<span>Trial      [A]      [B]         Rate </span><span>
<span>            (M)     (M)          (M/s) </span>
<span>1         0.50    0.010      3.0×10−3 </span>
<span>2         0.50    0.020       6.0×10−3 </span>
<span>3         1.00 0  .010       1.2×10−2</span></span>


Given that for trials 1 and 2 [A] is the same you can use those values to find n, in this way

rate 1 = 3.0 * 10^ -3 = k [A1]^m * [B1]^n

rate 2 = 6.0*10^-3 = k [A2]^m * [B2]^n

divide rate / rate 1 => 2 = [B1]^n / [B2]^n

[B1] = 0.010 and [B2] = 0.020 =>

6.0 / 3.0  =( 0.020 / 0.010)^n =>

2 = 2^n => n = 1

 
Given that for data 1 and 3 [B] is the same, you use those data to find m

rate 3 / rate 1 = 12 / 3.0   = (1.0)^m / (0.5)^m =>

4 = 2^m => m = 2

Now use any of the data to find k

With the first trial: rate = 3*10^-3 m/s = k (0.5)^2 * (0.1) =>

k = 3.0*10^-3 m/s / 0.025 m^3 = 0.12 m^-2 s^-1

Now that you have k, m and n you can use the formula of the rate with the concentrations given

rate = k[A]^2*[B] = 0.12 m^-2 s^-1 * (0.50m)^2 * (0.075m) = 0.0045 m/s = 4.5*10^=3 m/s

Answer: 4.5 * 10^-3 m/s
8 0
3 years ago
Read 2 more answers
Express in scientific notation. Make sure your answer has the same number of significant figures as the starting value.
ZanzabumX [31]
It has  5 digits so the exponent of 10 will be 4:- 
61,103 =  
6.1103 x 10^4

B
6 0
4 years ago
Read 2 more answers
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