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gayaneshka [121]
3 years ago
5

Please help this is due tonight

Chemistry
1 answer:
Elenna [48]3 years ago
5 0
4. the nervous system uses electrical impulses to send messages through neurons while endocrine glands use hormones to send messages

5. Hormones are released from the endocrine glands

6. a. neck
b. Chest

7. help regulate important functions, such as growth, blood pressure and reproduction; it is small and small and bean shaped.

8. produces hormones that regulate the body's metabolic rate controlling heart, muscle and digestive function, brain development.

9. regulate your metabolism, immune system, blood pressure, respond to stress etc.

10. “enzymes” - break down sugars, fats, and starches

11. cells are unable to use glucose (can’t break down sugar to use as energy)
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How would the concentration of silver ions compare in a 1.0 x 10-16 L saturated solution to a 1.5 x 10-16 L saturated solution?
VMariaS [17]
The second one is more concentrated as they both times with the same thing but the second one (1.5) is bigger
3 0
3 years ago
A 2.00 kg piece of lead at 40.0°C is placed in a very large quantity of water at 10.0°C,and thermal equilibrium is eventually re
sveticcg [70]

Answer:

Δ S = 26.2 J/K

Explanation:

The change in entropy can be calculated from the formula  -

Δ S = m Cp ln ( T₂ / T₁ )

Where ,

Δ S = change in entropy

m = mass  = 2.00 kg

Cp =specific heat of lead is 130 J / (kg ∙ K) .

T₂ = final temperature  10.0°C + 273 = 283 K

T₁ = initial temperature ,  40.0°C + 273 = 313 K

Applying the above formula ,

The change in entropy is calculated as ,

ΔS = m Cp ln ( T₂ / T₁ )  = (2.00 )( 130 ) ln( 283 K / 313 K )

ΔS = 26.2 J/K

6 0
3 years ago
How to balance <br> Mg + O2 &gt; MgO
katovenus [111]
<span>Mg + O2 > MgO. In reactant side, 2 O atoms and 1 Mg are present. In product side, 1 Mg and O atoms are present. Put 2 in product side to balance O atoms and 2 at Mg in reactant side to balance Mg atoms. Therefore the balanced equation becomes, 2Mg + O2 ----> 2MgO. Hope it helps.</span>
5 0
3 years ago
Consider a sample of 3.5 mol of N2(g) at T1 = 350 K, that undergoes a reversible and adiabatic change in pressure from p1 = 1.50
devlian [24]

Answer:

Part A is just T2 = 58.3 K

Part B ∆U = 10967.6 x C_{V} You can work out C_{V}

Part C

Part D

Part E

Part F

Explanation:

P = n (RT/V)

V = (nR/P) T

P1V1 = P2V2

P1/T1 = P2/T2

V1/T1 = V2/T2

P = Pressure(atm)

n = Moles

T = Temperature(K)

V = Volume(L)

R = 8.314 Joule or 0.08206 L·atm·mol−1·K−1.

bar = 0.986923 atm

N = 14g/mol

N2 Molar Mass 28g

n = 3.5 mol N2

T1 = 350K

P1 = 1.5 bar = 1.4803845 atm

P2 = 0.25 bar = 0.24673075 atm

Heat Capacity at Constant Volume

Q = nCVΔT

Polyatomic gas: CV = 3R

P = n (RT/V)

0.986923 atm x 1.5 = 3.5 mol x ((0.08206 L atm mol -1 K-1 x 350 K) / V))

V = (nR/P) T

V = ((3.5 mol x 0.08206 L atm mol -1 K-1)/(1.5 x 0.986923 atm) )x 350K

V = (0.28721/1.4803845) x 350

V = 0.194 x 350

V = 67.9036 L

So V1 = 67.9036 L

P1V1 = P2V2

1.4803845 atm x 67.9036 L = 0.24673075 x V2

100.52343693 = 0.24673075 x V2

V2 = P1V1/P2

V2 = 100.52343693/0.24673075

V2 = 407.4216 L

P1/T1 = P2/T2

1.4803845 atm / 350 K = 0.24673075 atm / T2

0.00422967 = 0.24673075 /T2

T2 = 0.24673075/0.00422967

T2 = 58.3 K

∆U= nC_{V} ∆T

Polyatomic gas: C_{V} = 3R

∆U= nC_{V} ∆T

∆U= 28g x C_{V} x (350K - 58.3K)

∆U = 28C_{V} x 291.7

∆U = 10967.6 x C_{V}

5 0
3 years ago
How would a flood be a limitation for synthetic polymers that rely on natural rubber in its production?
Zepler [3.9K]
A flood, if it hits the environment of the natural rubbers, would destroy how the rubber is being produced. to have a large amount of limitation, the flood would destroy a large percentage of rubber trees. This natural rubber is needed to make synthetic polymers. Without the rubber (because of damages to it's ecosystem through the flood), there would be a limited supply, and a substancial drop on synthetic polymers.


hope this helps
3 0
3 years ago
Read 2 more answers
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