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3 years ago

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