What is the limiting reagent when 9.517 g of Fe is allowed to react with 19.34 g of water according to

What is a DNA fingerprint and why is it considered individual evidence?

Answer:

DNA fingerprinting is a laboratory technique used to establish a link between biological evidence and a suspect in a criminal investigation. 

A DNA sample taken from a crime scene is compared with a DNA sample from a suspect. If the two DNA profiles are a match, then the evidence came from that suspect.

5 0

3 years ago

The area of a telescope lens is 6.668 × 103 mm2. What is the area in square feet (ft2)? Enter your answer in scientific notation

ollegr [7]

Answer:

$7.13\times 10^{-2}\ ft^2$

Explanation:

The area of a telescope lens is 6.668 × 10³ mm²

We need to convert the area in ft².

We know that,

$1\ mm^2=1.07\times 10^{-5}\ ft^2$

It means,

$6.668 \times 10^3\ mm^2=6.668 \times 10^3\times 1.07\times 10^{-5}\ ft^2\\\\=0.0713\ ft^2$

or

$0.0713\ ft^2=7.13\times 10^{-2}\ ft^2$

So, the required answer is $7.13\times 10^{-2}\ ft^2$ .

3 0

3 years ago

Which statement describes an Arrhenius acid

liq [111]

Answer:

According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs. Hence, the correct statement is arrhenius acid produces hydrogen ions in solution.

3 0

3 years ago

Be sure to answer all parts. Acetone is one of the most important solvents in organic chemistry. It is used to dissolve everythi

zimovet [89]

Answer:

a) 79.66 seconds is the half-life of the reaction.

b) It will take $4.776\times 10^1$ seconds for 34% of a sample of an acetone to decompose.

c) It will take $2.537\times 10^2$ seconds for 89% of a sample of an acetone to decompose.

Explanation:

The decomposition of acetone follows first order kinetics

The rate constant of the reaction = k = $8.7\times 10^{-3} s^{-1}$

a)

Half life of the reaction = $t_{1/2}$

For the first order kinetic half life is related to k by :

$t_{1/2}=\frac{0.693}{k}$

$t_{1/2}=\frac{0.693}{8.7\times 10^{-3} s^{-1}}=79.66 s=7.966\times 10^1 s$

79.66 seconds is the half-life of the reaction.

b)

Let the initial concentration of acetone be = $[A_o]$

Final concentration of acetone left after t time = [A]

$A=(100\%-34\%)[A_o]=66\%[A_o]=0.66[A_o]$

For the first order kinetic :

$[A]=[A_o]\times e^{-kt}$

$0.66[A_o]=[A_o]\times e^{-8.7\times 10^{-3} s^{-1}\times t}$

Solving for t;

$t=47.76 s$

It will take 47.76 seconds for 34% of a sample of an acetone to decompose.

c)

Let the initial concentration of acetone be = $[A_o]$

Final concentration of acetone left after t time = [A]

$A=(100\%-89\%)[A-o]=11\%[A_o]=0.11[A_o]$

For the first order kinetic :

$[A]=[A_o]\times e^{-kt}$

$0.11[A_o]=[A_o]\times e^{-8.7\times 10^{-3} s^{-1}\times t}$

Solving for t;

$t=2.537\times 10^2 s$

It will take $2.537\times 10^2$ seconds for 89% of a sample of an acetone to decompose.

3 0

3 years ago

What is different about most of the elements of the 7th period?

11Alexandr11 [23.1K]

Answer:

All elements of period 7 are radioactive. This period contains the actinides, which includes plutonium, the naturally occurring element with the heaviest nucleus; subsequent elements must be created artificially.

Explanation:

hope u understand :)

3 0

3 years ago

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