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kondor19780726 [428]
3 years ago
5

25.0 g sample of compound A is mixed with a 40.0 g sample of compound B. a chemical reaction occurs. Once the reaction is comple

te, the final mixture has a mass of 55.0g. What happens to the “missing” mass? How much mass is missing?
Chemistry
1 answer:
artcher [175]3 years ago
4 0

Answer:

Explanation:

Mass of compound A  = 25g

Mass of compound B = 40g

Mass of final mixture = 55g

What happens to the missing mass?

According to the law of conservation of mass, in chemical reaction, matter is transformed from one form to another but cannot be created nor destroyed.

We expect the final mass of the mixture and that of the reacting compounds to be the same but the opposite is the case.

There is a mass loss which typifies most chemical reaction.

The reason for this is that some of the masses must have been lost by the production of gaseous species which are unaccounted for.

The missing mass:

  Total mass expected  = mass of A + mass of B = 25 + 40 = 65g

Missing mass = expected mass - mass of final mixture = 65 - 55 = 10g

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Why would a scientist want to replicate another scientist's experiment?
choli [55]

Answer:

See the answer below

Explanation:

<u>A scientist B might want to replicate the experiment of another scientist A in order to assess the claims made by scientist A.</u> In other words, replication of the experiment of another scientist is done to see if a similar outcome would be arrived at or there would be variations.

<em>The claims made by a scientist while reporting the outcome of a particular experiment must be reproducible by another scientist under similar conditions. Otherwise, the claims are said to be false.</em>

7 0
3 years ago
A chemist wishing to do an experiment requiring 47-Ca2+ (half-life = 4.5 days) needs 5.0 μg of the nuclide. What mass of 47-CaCO
NARA [144]

Answer:

5.8μg

Explanation:

According to the rate or decay law:

N/N₀ = exp(-λt)------------------------------- (1)

Where N = Current quantity,  μg

            N₀ = Original quantity, μg

             λ= Decay constant day⁻¹

              t =  time in days

Since the half life is 4.5 days, we can calculate the  λ from (1) by  substituting N/N₀ = 0.5

0.5 = exp (-4.5λ)

ln 0.5  = -4.5λ

-0.6931 = -4.5λ

λ =   -0.6931 /-4.5

  =0.1540 day⁻¹

Substituting into (1)  we have :

N/N₀ = exp(-0.154t)----------------------------- (2)

To receive 5.0 μg of the nuclide with a delivery time of 24 hours or 1 day:

N = 5.0 μg

N₀ = Unknown

t = 1 day

Substituting into (2) we have

[5/N₀]   = exp (-0.154 x 1)

    5/N₀        = 0.8572

N₀  =  5/0.8572

     =    5.8329μg

    ≈     5.8μg

The Chemist must order 5.8μg  of 47-CaCO3

6 0
3 years ago
How many moles of water are in 1.23x10^18 water molecules
Harlamova29_29 [7]

By 1.23 x 1024 you mean 10 to the power of 24 molecules? If so all you need to do is divide the number of molecules you have by Avagadros number, 6.022 x 10^23. This will give you the mols of water, or the mols of anything, since there is always 6.022 x 10^23 molecules in 1 mol of substance.


1.23x10^24 atoms/6.022x10^23 atom/mol = 2.04 mol H20


6 0
3 years ago
Read 2 more answers
A 52.0 g of Copper (specific heat=0.0923cal/gC) at 25.0C is warmed by the addition of 299 calories of energy. find the final tem
Leto [7]

Answer : The final temperature of the copper is, 87.29^oC

Solution :

Formula used :

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

where,

Q = heat gained  = 299 cal

m = mass of copper = 52 g

c = specific heat of copper = 0.0923cal/g^oC      

\Delta T=\text{Change in temperature} 

T_{final} = final temperature = ?

T_{initial} = initial temperature = 25^oC

Now put all the given values in the above formula, we get the final temperature of copper.

299cal=52g\times 0.0923cal/g^oC\times (T_{final}-25^oC)

T_{final}=87.29^oC

Therefore, the final temperature of the copper is, 87.29^oC

4 0
3 years ago
Whats the atomic mass
Semmy [17]
The mass of a chemical substance is known as the atomic number or atomic mass
8 0
3 years ago
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