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Nutka1998 [239]
4 years ago
15

01:29:45

Chemistry
1 answer:
marusya05 [52]4 years ago
8 0
Charle’s Law.

It states that, for a given mass of an ideal gas at constant pressure, the volume is directly proportional to its absolute temperature, assuming in a closed system.
You might be interested in
Can anybody help??
kvv77 [185]

The amount of heat energy added to silver to heat it from 25°C to 100°C is :

8737.5 J

<u>Given data: </u>

mass of silver ( m ) = 500 g

T1 = 25°C

T2 = 100°C

s ( specific heat of silver ) = 0.233 J/g.c

<h3 /><h3>Determine the amount of heat required </h3>

Applying the formula below for heat ( Q )

Q = ms * ΔT

   = 500 * 0.233 * ( 100 - 25 )

   = 8737.5 J

Hence we can conclude that the The amount of heat energy added to silver to heat it from 25°C to 100°C is :  8737.5 J.

Learn more about heat energy : brainly.com/question/13439286

5 0
3 years ago
What is a molecule???
ololo11 [35]
A molecule consists of group of atoms that are bonded together by chemical bond. It is the smallest particle of an element or a compound that has the properties of the element or the compound embedded in it. The atoms present in one element easily forms a bonding with the atoms of another element to form molecules. The size and complex nature of molecules varies hugely. Some molecules can have a single atom as is the case with helium, while in several other elements the number of atoms in the molecule can be more than one.
6 0
3 years ago
The element antimony has an atomic weight of 122 and consists of two stable isotopes antimony-121 and antimony-123. the isotope
grin007 [14]

Answer:-  123 amu.

Solution:- The formula to calculate the average atomic mass of an atom is:

Average atomic mass = mass of first isotope(abundance of first isotope) + mass of second isotope(abundance of second isotope)

Note: The percent abundance is converted to decimals.

mass of Sb-121 is 121 amu and it's percent abundance is 57.3% and in decimal it is 0.573. Percent abundance of Sb-123 is 42.8% and in decimal it is 0.428. We are asked to calculate it's mass. The average atomic mass of Sb is given as 122.

Let's say the mass of Sb-123 is M and  plug in the values in the formula and do calculations:

122 = 121(0.573) + M(0.428)

122 = 69.333 + M(0.428)

On rearrangement:-

M(0.428) = 122 - 69.333

M(0.428) = 52.667

M=\frac{52.667}{0.428}

M = 123

So, the mass of Sb-123 is 123 amu.

6 0
4 years ago
Can some one help me in this I hope this will be easy for you :)
kramer

Answer:

c) i)  Mg²⁺   ii) O²⁻

Explanation:

I don't know the answer to Q7. because you don't show the diagram

5 0
3 years ago
What is the rate law for the reaction 2A + 2B + 2C --&gt; products
-Dominant- [34]

Answer:

R = 47.19 [A]*([B]^2)*[C]

Explanation:

The rate law for the reaction 2A + 2B + 2C --> products

Is being sought.

The reaction rate R could be expressed as  

R = k ([A]^m)*([B]^n)*([C]^p)                      (1)

where m, n, and p are the reaction orders with respect to (w.r.t.) components A, B and C respectively. This could be reduced to

R = ka ([A]^m)                   (2)

Where ka=(k[B]^n)*([C]^p);    

R = kb ([B]^n)                    (3)

Where kb=(k[A]^m)*([C]^p); and  

R = kc ([C]^p)                     (4)

Where kc=(k[A]^m)*([B]^n).

Equations (2), (3) and (4) are obtained for cases when the concentrations of two components are kept constant, while only one component’s concentration is varied. We can determine the reaction wrt each component by employing these equations.  

The readability is very much enhanced when the given data is presented in the following manner:

Initial [A]  0.273   0.819   0.273   0.273

Initial [B]  0.763   0.763   1.526   0.763

Initial [C]  0.400   0.400    0.400   0.800

Rate           3.0       9.0       12.0       6.0

Run#  1  2 3  4

Additional row is added to indicate the run # for each experiment for easy reference.

First, we use the initial rate method to evaluate the reaction order w.r.t. each component [A], [B] and [C] based on the equations (2), (3) and (4) above.

Let us start with the order wrt [A]. From the given data, for experimental runs 1 and 2, the concentrations of reactants B and C were kept constant.

Increasing [A] from 0.273 to 0.819 lead to the change of R from 3.0 to 9.0, hence we can apply the relation based on equation (2) between the final rate R2, the initial rate R1 and the final concentration [A2] and the initial concentration [A1] as follows:

R2/R1=ka[A2]^m/ka[A1]^m=([A2]/[A1])^m

9.0/3.0 = (0.819/0.273)^m

3 = (3)^m = 3^1  -> m = 1

Similarly, applying experimental runs 1 and 3 could be applied for the determination of n, by employing equation (3):  

R3/R1=kb[B3]^n/kb[B1]^n=([B3]/[B1])^n

12/3= (1.526/0.763)^n

4= 2^n, -> n = 2

And finally for the determination of p we have using runs 4 and 1:

R4/R1=kc[C4]^p/kc[C1]^p=([C4]/[C1])^p

6/3= (0.8/0.4)^p

2= 2^p , -> p = 1

Therefore, plugging in the values of m, n and p into equation (1), the rate law for the reaction will be:

R = k [A]*([B]^2)*[C]

The value of the rate constant k could be estimated by making it the subject of the formula, and inserting the given values, say in run 1:

k = R /( [A]*([B]^2)*[C]) = 3/0.273*(0.763^2)*0.4 =

47.19

Finally, the rate law is

R = 47.19 [A]*([B]^2)*[C]

7 0
4 years ago
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