Ask question

Ask question

All categories

skelet666 [1.2K]

2 years ago

14

Greater than ___________ mcg/dl is considered a high level of morning cortisol.

1 answer:

Natalija [7]2 years ago

5 0

Greater than 23mcg/dl is considered a high level of morning cortisol.

Normally, cortisol levels rise during the early morning hours and are highest about 7AM. They drop very low in the evening and during the early phase of sleep. If you do not have this daily change (diurnal rhythm) in cortisol levels, you may have overactive adrenal glands. This condition is called Cushing's syndrome.

You might be interested in

All cell activity is controlled by the A. nucleus

tangare [24]

Answer:

A

Explanation:

it also stores all the cells dna

3 0

2 years ago

Read 2 more answers

5.943x10^24 molecules of H3PO4 will need how many grams of Mg(OH)2 in the reaction below? 3 Mg(OH)2 + 2 H3PO4 -------> 1 Mg3(

professor190 [17]

Answer:

Mass of Mg(OH)₂ required for the reaction = 863.13 g

Explanation:

3Mg(OH)₂ + 2H₃PO₄ -------> Mg₃(PO₄)₂ + 6H₂O

(5.943 x 10²⁴) molecules of H₃PO₄ is available fore reaction. Mass of Mg(OH)₂ required for reaction.

According to Avogadro's theory, 1 mole of all substances contain (6.022 × 10²³) molecules.

This can allow us find the number of moles that (5.943 x 10²⁴) molecules of H₃PO₄ represents.

1 mole = (6.022 × 10²³) molecules.

x mole = (5.943 x 10²⁴) molecules

x = (5.943 x 10²⁴) ÷ (6.022 × 10²³)

x = 9.87 moles

From the stoichiometric balance of the reaction,

2 moles of H₃PO₄ reacts with 3 moles of Mg(OH)₂

9.87 moles of H₃PO₄ will react with y moles of Mg(OH)₂

y = (3×9.87)/2 = 14.80 moles

So, 14.8 moles of Mg(OH)₂ is required for this reaction. We them convert this to mass

Mass = (number of moles) × Molar mass

Molar mass of Mg(OH)₂ = 58.3197 g/mol

Mass of Mg(OH)₂ required for the reaction

= 14.8 × 58.3197 = 863.13 g

Hope this Helps!!!

5 0

3 years ago

Read 2 more answers

Help ASAAPPPPPPPPPP!!!!

Verdich [7]

Third choice is correct.

6 0

2 years ago

7. Write the equation for the positron emission of barium-127.

ziro4ka [17]

The reaction is given by

$\\ \rm\Rrightarrow {}^{127}_{56}Ba\longrightarrow {}^{0}_{+1}\beta+{}^{127}_{55}Cs$

Barium goes underneath beta decay to form Ceaseum

Cs is very mellable element
It can melt on your hand

8 0

1 year ago

A chemist determined by measurements that 0.030 moles of barium participated in a chemical reaction. Calculate the mass of bariu

schepotkina [342]

Answer: 4.1 g of barium precipitated.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar Mass}}$

Given : moles of barium = 0.030

Molar mass of barium = 137 g/mol

$0.030=\frac{x}{137}$

x= 4.1 g

Thus there are 4.1 g of barium that precipitated.

3 0

2 years ago