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2 years ago

10

Mr. Smith is hyperglycemic with a blood glucose level of 300mg/ml of blood. Explain how homeostasis would regulate his glucose l

evels through a negative feedback system.

1 answer:

aleksley [76]2 years ago

6 0

Answer:

When the animal has eaten food and the blood glucose level in the body increases. The pancreas cells in the body detects the increase in the blood glucose which leads to increase the insulin hormone.

This decreases the blood glucose level in the level. This is how the negative feedback works in the body if the level of glucose increases.

Negative feedback is the way by which the body maintains homeostasis and maintains equilibrium in the body.

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When an athlete gets hurt they apply sometimes an instant cold pack to the injury. The instant cold pack works by mixing two che

NeX [460]

Answer:

The evidence showing that there is a chemical reaction taking place is the instantaneous temperature drop once the cold pack is shaken.

Explanation:

When an athlete applies a cold pack to the injury, they shake it before, mixing the water and <em>ammonium-nitrate fertilizer</em> inside the cold pack. This mixing is an endothermic reaction, which means it absorbs heat. In turn, the temperature falls to 35 F for around 10 minutes.

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2 years ago

In a covalently bonded molecule, the number of electrons that an atom shares with others is usually equal to the number of elect

3241004551 [841]

In the last shell (valence shell)

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2 years ago

A 0.380 kg sample of aluminum (with a specific heat of 910.0 J/(kg x K)) is heated to 378 K and then placed in 2.40 kg of water

lbvjy [14]

Answer:

The equilibrium temperature of the system is 276.494 Kelvin.

Explanation:

Let consider the system formed by the sample of aluminium and water as a control mass, in which the sample is cooled and water is heated until thermal equilibrium is reached. The energy process is represented by First Law of Thermodynamics:

$Q_{water} -Q_{sample} = 0$

$Q_{water} = Q_{sample}$

Where:

$Q_{water}$ - Heat received by water, measured in joules.

$Q_{sample}$ - Heat released by the sample of aluminium, measured in joules.

Given that no mass is evaporated, the previous expression is expanded to:

$m_{w}\cdot c_{p,w}\cdot (T-T_{w}) = m_{s}\cdot c_{p,s}\cdot (T_{s}-T)$

Where:

$m_{s}$ , $m_{w}$ - Mass of water and the sample of aluminium, measured in kilograms.

$c_{p,s}$ , $c_{p,w}$ - Specific heats of the sample of aluminium and water, measured in joules per kilogram-Kelvin.

$T_{s}$ , $T_{w}$ - Initial temperatures of the sample of aluminium and water, measured in Kelvin.

$T$ - Temperature which system reaches thermal equilibrium, measured in Kelvin.

The final temperature is now cleared:

$(m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s})\cdot T = m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}$

$T = \frac{m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}}{m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s}}$

Given that $m_{s} = 0.380\,kg$ , $m_{w} = 2.40\,kg$ , $c_{p,s} = 910\,\frac{J}{kg\cdot K}$ , $c_{p,w} = 4186\,\frac{J}{kg\cdot K}$ , $T_{s} = 378\,K$ and $T_{w} = 273\,K$ , the final temperature of the system is:

$T = \frac{(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)\cdot (378\,K)+(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)\cdot (273\,K)}{(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)+(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)}$

$T = 276.494\,K$

The equilibrium temperature of the system is 276.494 Kelvin.

7 0

2 years ago

A magnesium ion, Mg2+, with a charge of 3.2×10−19C and an oxide ion, O2−, with a charge of −3.2×10−19C, are separated by a dista

baherus [9]

Explanation:

Formula for work done is as follows.

W = $-k \frac{q_{1}q_{2}}{d}$

where, k = proportionality constant = $8.99 \times 10^{9} Jm/C^{2}$

$q_{1} = charge of Mg^{2+} = 3.2 \times 10^{-19} C$

$q_{2} = charge of O_{2-} = -3.2 \times 10^{-19} C$

d = separation distance = 0.45 nm = $0.45 \times 10^{-9} m$

Now, we will put the given values into the above formula and calculate work done as follows.

W = $-k \frac{q_{1}q_{2}}{d}$

= $\frac{-[8.99 \times 10^{9} Jm/C^{2} \times 3.2 \times 10^{-19} C \times -3.2 \times 10^{-19} C]}{0.25 \times 10^{-9} m}$

= $3.68 \times 10^{-18} J$

Thus, we can conclude that work required to increase the separation of the two ions to an infinite distance is $3.68 \times 10^{-18} J$ .

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2 years ago

A compound with molecular formula C4H6O4 produces a broad signal between 2500 and 3600 cm-1 in its IR spectrum and produces two

tiny-mole [99]

Answer:

Succinic Acid

Explanation:

We have to start using the info given by the IR spectrum. In this case, we have a <u>broad signal in 3600</u>. This indicates the presence of OH in the structure. Therefore we can have an <u>alcohol or an acid</u> in the structure.

Now, the NMR info tells us that we only have 2 signals, this indicates that we have a <u>very symmetric structure</u>. Also, we have a signal in 12 ppm therefore we can affirm that we have an O-H bond with <u>high polarity</u> (a downfield signal) this behavior is given in the <u>acid groups</u>.

The structure that fulfill these requirements it the succinic acid. In which we only have 2 signals in the 1H- NMR. We have an acid group and we have a formula of $C_4H_6O_4$ .

<u>For further explanations see the attached images.</u>

3 0

2 years ago