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frozen [14]
3 years ago
15

Metallic materials are becoming more and more widely used. They have

Chemistry
2 answers:
weqwewe [10]3 years ago
7 0

Answer:

Its B

Explanation:

hope its help by the way iam dream

Tema [17]3 years ago
6 0
It’s B hehehehhdhdhfgfgfhrhthhrhdhr
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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
Convert the composition of the following alloy from atom percent to weight percent (a) 44.9 at% of silver, (b) 46.3 at% of gold,
vesna_86 [32]

Answer:

Mass percentage of gold : 33.35%

Mass percentage of silver: 62.80%

Mass percentage of copper : 3.85%

Explanation:

N=n\times N_A

Where : N = Number of atoms

n = moles

N_A=6.022\times 10^{23} = Avogadro number

Let the total atoms present in the alloy be 100.

Atom percentage of silver = 44.9%

Atoms of of silver = 44.9% of 100 atoms = 44.9

Atomic weight of silver = 107.87 g/mol

Mass of 44.9 silver atoms = 107.87 g/mol ×  44.9 = 4,843.363 g/mol

Atom percentage of gold= 46.3%

Atoms of of gold = 46.3% of 100 atoms = 46.3

Atomic weight of gold = 196.97 g/mol

Mass of 44.9 gold atoms = 196.97 g/mol × 46.3 = 9,119.711 g/mol

Atom percentage of copper = 8.8 %

Atoms of of copper = 8.8% of 100 atoms = 8.8

Atomic weight of copper = 63.55 g/mol

Mass of 44.9 copper atoms = 63.55 g/mol ×  8.8 = 559.24 g/mol

Total mass of an alloy :4,843.363 g/mol + 9,119.711 g/mol +  559.24 g/mol

Mass percentage of gold :

\frac{4,843.363 g/mol}{4,843.363 g/mol + 9,119.711 g/mol +  559.24 g/mol}\times 100=33.35 \%

Mass percentage of silver:

\frac{9,119.711 g/mol}{4,843.363 g/mol + 9,119.711 g/mol +  559.24 g/mol}\times 100=62.80\%

Mass percentage of copper :

\frac{559.24 g/mol}{4,843.363 g/mol + 9,119.711 g/mol +  559.24 g/mol}\times 100=3.85\%

7 0
3 years ago
Help please this is also science idk why brainly doesn't have that option
solniwko [45]

Answer:

fffffffffffdddddddddddddddddd

Explanation:

fdddddddddddddddddddddddddd

6 0
2 years ago
Read 2 more answers
The oceanic crust is mostly associated with the —————.
Lilit [14]
I think it would the hydrosphere
4 0
2 years ago
Equation 3x-5=-2x+ 10?<br> O<br> 0<br> x= 5<br> -5 = x<br> -15 = -5x<br> -5x = 15
Harman [31]

Answer:

are those the answer choice

7 0
3 years ago
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