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slamgirl [31]
3 years ago
9

Blood test can be used to check a person's blood glucose and hormone levels. The tabular column given below shows the result of

two blood tests carried out on three people to check their blood glucose levels. Person 1 is healthy.
A:- Compare the glucose levels of person 1 with the glucose levels of person 2 after fasting for 12 hours.

B:- Compare the glucose levels of person 3 with the glucose levels of person 1, 2 hours after drinking 75g glucose.

C:- Person 3 cannot be produce the hormone that control blood glucose levels. State the hormone that person 3 cannot produce.

( THE ORDER OF THE PHOTO IS:- person 1, and person 2, and person 3)​


Pls it is urgent
Chemistry
1 answer:
Anastasy [175]3 years ago
3 0

Answer:

Explanation:

1) person 2 had a slightly higher blood glucose level than person 1 after fasting 12 hrs (0.2mmols)

2) person 3 has a much higher blood glucose level than person 1 after taking glucose (5.6mmols)

3) insulin

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How many electrons occupy a filled 7s sublevel
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Explanation:

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3 years ago
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How many milliliters of 0.150 M NaOH are required to neutralize 85.0 mL of 0.300 M H2SO4 ? The balanced neutralization reaction
nekit [7.7K]

Answer : The volume of NaOH required to neutralize is, 340 mL

Explanation :

To calculate the volume of base (NaOH), we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=0.300M\\V_1=85.0mL\\n_2=1\\M_2=0.150M\\V_2=?

Putting values in above equation, we get:

2\times 0.300M\times 85.0mL=1\times 0.150M\times V_2\\\\V_2=340mL

Hence, the volume of NaOH required to neutralize is, 340 mL

4 0
4 years ago
Write a ground state electron configuration for each neutral atom
Gre4nikov [31]

Answer:

Pb[lead] [Xe]4f^145d^106s^26p^2

U[uranium] 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4e^14 5d^10 6p^6

7s^2 5f^4

This notation can be written in core notation or noble gas notation by replacing the

1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4e^14 5d^10 6p^6

7s^2 5f^4

with the noble gas [Rn].

[Rn]7s25f4

N[nitrogen] The full electron configuration for nitrogen is 1s^2 2s^2 2p^3.

Ti[titanium] Ti2+:[Ar]3d^2

Ti:1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 4s^2

1s^2 2s^2 2p^6 3s^2 3p^5 = 17 electrons

(1) electron gain will result to a

negative charge (−), and

(2) electron loss will result to a positive charge (+),

1s^2 2s^2 2p^6 3s^2 3p^6 = 18 electrons

Hg[mercury] You should then find its atomic number is 80. It has a Xe core, so in shorthand notation, you can include [Xe]instead of

1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6,

for 54 electrons. For the 6th row of the periodic table, we introduce the 4f orbitals, and proceed to atoms having occupied 5d orbitals. We, as usual, have the ns orbitals, and n=? for the 6th period?

Mercury has a regular electron configuration. It becomes:

[Xe]4f145d106s2

Explanation:

socratic.org helped me! I'm really sorry if this is wrong!

6 0
3 years ago
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