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3 years ago

3 sources of background radiation and one source of radiation that is not condisderd background radiation

1 answer:

5 0

Natural sources. Natural sources of background radiation include: Cosmic rays - radiation that reaches the Earth from space. Rocks and soil - some rocks are radioactive and give offradioactive radon gas.

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hydrolysis of decapeptide P with the enzyme trypsin affords the following fragments: Glu-Gly-Lys, Gln-Val-Ile, Ala-Ser-Phe-Lys.

ehidna [41]

Answer:

The sequence of an amino acid P is:

Glu-Gly-Lys-Ala-Ser-Phe-Lys-Gln-Val-Ile

Explanation:

Fragments obtained on hydrolysis of decapeptide P by the action of an enzyme named trypsin:

Glu-Gly-Lys,
Gln-Val-Ile
Ala-Ser-Phe-Lys

Fragments obtained on hydrolysis of decapeptide P by the action of an enzyme named chymotrypsin:

Lys-Gln-Val-Ile,
Glu-Gly-Lys-Ala-Ser-Phe

In order to determine the sequence of protein P , we will arrange fragments in such a way so that common fragments or the common parts of fragments should come under each other.

On arranging these fragments :

Glu-Gly-Lys-Ala-Ser-Phe

Glu-Gly-Lys

Ala-Ser-Phe-Lys

Lys-Gln-Val-Ile

Gln-Val-Ile

The sequence of an amino acid P is:

Glu-Gly-Lys-Ala-Ser-Phe-Lys-Gln-Val-Ile

3 0

3 years ago

How do scientists do background research on topics they have question about

Aliun [14]

They use information from other things that relate to that topic and then use that to do their research

3 0

3 years ago

What is the total energy change for the following reaction:CO+H2O-CO2+H2

Alekssandra [29.7K]

Answer:

$\large \boxed{\text{-41.2 kJ/mol}}$

Explanation:

Balanced equation: CO(g) + H₂O(g) ⟶ CO₂(g) + H₂(g)

We can calculate the enthalpy change of a reaction by using the enthalpies of formation of reactants and products

$\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)$

(a) Enthalpies of formation of reactants and products

$\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{CO(g)} & -110.5 \\\text{H$_{2}$O} & -241.8\\\text{CO$_{2}$(g)} & -393.5 \\\text{H$_{2}$(g)} & 0 \\\end{array}$

(b) Total enthalpies of reactants and products

$\begin{array}{ccr}\textbf{Substance} & \textbf{Contribution)/(kJ/mol})&\textbf{Sum} \\\text{CO(g)} & -110.5& -110.5 \\\text{H$_{2}$O(g)} &-241.8& -241.8\\\textbf{Total}&\textbf{for reactants} &\mathbf{ -352.3}\\&&\\\text{CO}_{2}(g) & -393.5&-393.5 \\\text{H}_{2} & 0 & 0\\\textbf{Total}&\textbf{for products} & \mathbf{-393.5}\end{array}$

(c) Enthalpy of reaction $\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)= \text{-393.5 kJ/mol - (-352.3 kJ/mol}\\= \text{-393.5 kJ/mol + 352.3 kJ/mol} = \textbf{-41.2 kJ/mol}\\ \text{The total enthalpy change is $\large \boxed{\textbf{-41.2 kJ/mol}}$}$

4 0

3 years ago

If the half-life of a radioactive element is 4 days, how long will it take for three- fourths of a sample of the element to deca

Jlenok [28]

Answer:

$\boxed{\text{8 da}}$

Explanation:

The question will be easier to solve if we interpret it as, " How long will it take until one-fourth of a sample of the element remains,?"

The half-life of the element is the time it takes for half of it to decay.

After one half-life, half (50 %) of the original amount will remain.

After a second half-life, half of that amount (25 %) will remain, and so on.

We can construct a table as follows:

$\begin{array}{cccl}\textbf{No. of} & & \textbf{Fraction} & \\\textbf{half-lives} & \textbf{t/da} & \textbf{remaining} & \\1 & 4 & \dfrac{1}{2} & \\\\2 & 8 & \dfrac{1}{4}& \\\\3 & 12 & \dfrac{1}{8}& \\\end{array}$

$\text{We see that 8 da is two half-lives, and the fraction of the element remaining is $\frac{1}{4}$.}\\\text{It takes $\boxed{\textbf{8 da}}$ for three-fourths of the element to decay}$

3 0

3 years ago

Which two kinds of information describe weather and not climate?

Fynjy0 [20]

Answer: b & c

Explanation: A-P-E-X

5 0

3 years ago

Read 2 more answers