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ArbitrLikvidat [17]
2 years ago
7

Susan and her friend found a piece of metal which they want to

Chemistry
1 answer:
joja [24]2 years ago
4 0

Answer:

The given  metal is beryllium.

Explanation:

Given data:

Mass of metal = 74 g

Volume of metal = 40 cm³

Which metal is this = ?

Solution:

First of all we will calculate the density.

it is mass divided by volume.

d = m/v

d = 74 g/40 cm³

d = 1.85 g/cm³

In literature it is given that the density of beryllium is 1.85 g/cm³. Thus given  metal is beryllium.

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Liula [17]

Answer:

each continent shifts 2-5 inches per year so once it was joined then it is what it is now.In a million years again it will comes closer and wjo knows what will happen.moutains will be formed as the tectonic plates will collide.

5 0
3 years ago
Suppose you could watch radioactive atoms decay. It would probably get quite boring as time went by. Why?
kari74 [83]

Answer: the answer should and most definitely be D.

Explanation: I mean think about it after a while only a few radioactive nuclei are left which means it will dye down after a while which also makes it very boring hope this helps :)

6 0
3 years ago
Read 2 more answers
A)
Andrews [41]

sorry what us this i don't get it

7 0
2 years ago
The following data were obtained in a kinetics study of the hypothetical reaction A + B + C → products. [A]0 (M) [B]0 (M) [C]0 (
Vladimir [108]

Answer:

B. First order, Order with respect to C = 1

Explanation:

The given kinetic data is as follows:

A + B + C → Products

     [A]₀     [B]₀    [C]₀       Initial Rate (10⁻³ M/s)

1.   0.4      0.4     0.2       160

2.  0.2      0.4      0.4       80

3.   0.6     0.1       0.2       15

4.   0.2     0.1       0.2        5

5.   0.2     0.2      0.4       20

The rate of the above reaction is given as:

Rate = k[A]^{x}[B]^{y}[C]^{z}

where x, y and z are the order with respect to A, B and C respectively.

k = rate constant

[A], [B], [C] are the concentrations

In the method of initial rates, the given reaction is run multiple times. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.

Order w.r.t A : Use trials 3 and 4

\frac{Rate3}{Rate4}= [\frac{[A(3)]}{[A(4)]}]^{x}

\frac{15}{5}= [\frac{[0.6]}{[0.2]}]^{x}

3 = 3^{x} \\\\x =1

Order w.r.t B : Use trials 2 and 5

\frac{Rate2}{Rate5}= [\frac{[B(2)]}{[B(5)]}]^{y}

\frac{80}{20}= [\frac{[0.4]}{[0.2]}]^{y}

4 = 2^{y} \\\\y =2

Order w.r.t C : Use trials 1 and 2

\frac{Rate1}{Rate2}= [\frac{[A(1)]}{[A(2)]}]^{x}[\frac{[B(1)]}{[B(2)]}]^{y}[\frac{[C(1)]}{[C(2)]}]^{z}

we know that x = 1 and y = 2, substituting the appropriate values in the above equation gives:

\frac{160}{80}= [\frac{[0.4]}{[0.2]}]^{1}[\frac{[0.4]}{[0.4]}]^{2}[\frac{[0.2]}{[0.4]}]^{z}

1 = (0.5)^{z}

z = 1

Therefore, order w.r.t C = 1

8 0
3 years ago
True or false atoms with fewer then 4 outer elections lend elections?
Mamont248 [21]
The answer is false ....

3 0
3 years ago
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