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STatiana [176]
3 years ago
7

A mineral that helps in clotting of blood________.

Chemistry
2 answers:
sashaice [31]3 years ago
7 0

Answer:

Vitamin K

Explanation:

this is the answer

Rufina [12.5K]3 years ago
6 0

Answer:

<h2> VITAMIN K</h2><h2>VITAMIN K IS A FAT-SOLUBLE</h2><h2>VITAMIN FOUND IN GREEN LEAFY VEGETABLES, ESSENTIAL FOR BLOOD CLOTTING.</h2>
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A chemist needs to find the concentration of a solution of barium hydroxide.
Brums [2.3K]

Method 1: gravimetry

advantages: Impurities in the sample can be identified

disadvantages: The process is long, because it goes through several stages

Method 2: titration

advantages: the process is fast, because the titrate and titrant react immediately

disadvantages: Sometimes the determination of the end point of the titration is carried out too fast or too slowly so that the calculations carried out are inaccurate

3 0
2 years ago
Describe each highlighted bond in terms of the overlap of atomic orbitals. (If the highlighted bond is not a pi bond, select the
Anestetic [448]

The image of the bonds are missing, so i have attached it.

Answer:

A) - Sigma bond

-Sp³ and Sp³

- None

B) - Sigma and pi bond

- Sp² of C and p of O

- p of C and P of O

Explanation:

A) For compound 1;

- the molecular orbital type is sigma bond due to the end-to-end overlapping.

- Atomic orbitals in the sigma bond will be Sp³ and Sp³

- Atomic orbitals in the pi bond would be nil because there is no pi bond.

B) For compound 2;

- the molecular orbital type is sigma and pi bond

-Atomic orbitals in the sigma bond would be Sp² of C and p of O

- The Atomic orbitals in the pi bond will be; p of C and p of O

6 0
3 years ago
As an athlete exercises, sweat is produced and evaporated to help maintain a proper body temperature. On average, an athlete los
dmitriy555 [2]

<u>Answer:</u> For the given amount of sweat lost, the amount of energy required will be 692,899 Joules.

<u>Explanation:</u>

We are given:

Heat of vaporization for water = 2257 J/g

Amount of sweat lost = 307 grams

Applying unitary method:

For 1 g of sweat lost, the energy required is 2257 Joules

So, for 307 grams of sweat lost, the energy required will be = \frac{2257J}{1g}\times 307g=692,899J

Hence, for the given amount of sweat lost, the amount of energy required will be 692,899 Joules.

7 0
3 years ago
Which of the following statements about C3 carbon fixation is true? a. C3 carbon fixation is an adaptation for plants exposed to
saul85 [17]

Answer: The answer is B

Explanation:

RiP BoZo. shout out to faze gabi staright up bopped potato girl.

8 0
3 years ago
Read 2 more answers
Calculate the frequency of the n=2 line in the lyman series of hydrogen
Alona [7]

Answer:

Approximately 2.47\times 10^{15}\; \rm Hz.

Explanation:

The Lyman Series of a hydrogen atom are due to electron transitions from energy levels n \ge 2 to the ground state where n = 1. In this case, the electron responsible for the line started at n = 2 and transitioned to

A hydrogen atom contains only one electron. As a result, Bohr Model provides a good estimate of that electron's energy at different levels.

In Bohr's Model, the equation for an electron at energy level n (

\displaystyle - \frac{k\, Z^2}{n^2} (note the negative sign in front of the fraction,)

where

  • k = 2.179 \times 10^{-18}\; \rm J is a constant.
  • Z is the atomic number of that atom. Z = 1 for hydrogen.
  • n is the energy level of that electron.

The electron that produced the n = 2 line was initially at the

\begin{aligned} &E_{n = 2} \cr &= -\frac{k\, Z^2}{n^2} \cr &= -\frac{2.179 \times 10^{-18} \times 1}{2^2} \cr & \approx -5.4475\times 10^{-19}\; \rm J\end{aligned}.

The electron would then transit to energy level n = 1. Its energy would become:

\begin{aligned} &E_{n = 1} \cr &= -\frac{k\, Z^2}{n^2} \cr &= -\frac{2.179 \times 10^{-18} \times 1}{1^2} \cr & \approx -2.179 \times 10^{-18} \; \rm J\end{aligned}.

The energy change would be equal to

\begin{aligned}&\text{Initial Energy} - \text{Final Energy} \cr &= E_{n = 2} - E_{n = 1} \cr &= -5.4475 \times 10^{-19} - \left(-2.179 \times 10^{-18}\right) \cr & \approx 1.63425\times 10^{-18}\; \rm J \end{aligned}.

That would be the energy of a photon in that n = 2 spectrum line. Planck constant h relates the frequency of a photon to its energy:

E = h \cdot f, where

  • E is the energy of the photon.
  • h \approx 6.62607015\times 10^{-34}\; \rm J \cdot s is the Planck constant.
  • f is the frequency of that photon.

In this case, E \approx 1.63425 \times 10^{-18}\; \rm J. Hence,

\begin{aligned} f &= \frac{E}{h} \cr &\approx \frac{1.63425\times 10^{-18}}{6.62607015\times 10^{-34}} \cr & \approx 2.47 \times 10^{15}\; \rm s^{-1}\end{aligned}.

Note that 1 \; \rm Hz = 1 \; \rm s^{-1}.

6 0
3 years ago
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