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Gala2k [10]
3 years ago
5

List 4 functions of proteins.1.2.3.4.Help !!​

Chemistry
1 answer:
Jet001 [13]3 years ago
8 0

Answer:

List 4 functions of proteins.

1.Growth and Maintenance.

2.Balances Fluids.

3.Bolsters Immune Health.

4.Transports and Stores Nutrients.

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Substances have more kinetic energy in the _______ state than in the _______ state.
lions [1.4K]

Answer:

Substances have more kinetic energy in the gas state than in the Solid state.

Gas have more kinetic energy as it flows from here and there.It flows really fast as it have less space between the molecules

Solids have less Kinetic energy as it is hard and can't flow so easily like gas . It has tightly packed molecules

5 0
3 years ago
What is the vapor pressure (in mm Hg) of a solution of 17.5 g of glucose (C6H12O6) in 82.0 g of methanol (CH3OH) at 27∘C? The va
vichka [17]

Answer:

134.8 mmHg is the vapor pressure for solution

Explanation:

We must apply the colligative property of lowering vapor pressure, which formula is: P° - P' = P° . Xm

P° → Vapor pressure of pure solvent

P' → Vapor pressure of solution

Xm → Mole fraction for solute

Let's determine the moles of solute and solvent

17.5 g . 1 mol/180 g = 0.0972 moles

82 g . 1mol / 32 g = 2.56 moles

Total moles → moles of solute + moles of solvent → 2.56 + 0.0972 = 2.6572 moles

Xm → moles of solute / total moles = 0.0972 / 2.6572 = 0.0365

We replace the data in the formula

140 mmHg - P' = 140 mmHg . 0.0365

P' = - (140 mmHg . 0.0365 - 140mmHg)

P' = 134.8 mmHg

5 0
3 years ago
Pressure gauge at the top of a vertical oil well registers 140 bars. The oil well is 6000 m deep and filled with natural gas dow
andreyandreev [35.5K]

Explanation:

(a)  The given data is as follows.

              Pressure on top (P_{o}) = 140 bar = 1.4 \times 10^{7} Pa       (as 1 bar = 10^{5})

              Temperature = 15^{o}C = (15 + 273) K = 288 K

         Density of gas = \frac{PM}{ZRT}

                \frac{dP}{dZ} = \rho \times g

               \frac{dP}{dZ} = \int \frac{PM}{ZRT}

                \int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ

           ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}

                              = 0.4548

                     P_{1} = P_{o} \times e^{0.4548}

                                 = 1.4 \times 10^{7} Pa \times 1.5797

                                 = 2.206 \times 10^{7} Pa

Hence, pressure at the natural gas-oil interface is 2.206 \times 10^{7} Pa.

(b)   At the bottom of the tank,

                 P_{2} = P_{1}  + \rho \times g \times h

                             = 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]

                             = 309.8 \times 10^{5} Pa

                             = 309.8 bar

Hence, at the bottom of the well at 15^{o}C pressure is 309.8 bar.

6 0
4 years ago
If there are 7 moles of copper dissolved in 15 moles of zinc to make 22 moles of a brass solution. What is the mole fraction of
iragen [17]
Mole fraction of copper is 7/22
6 0
3 years ago
Read the passage. Look at the underlined section marked number 9. There may be a mistake in the way the sentence is written. If
Alexus [3.1K]

The answer is D just did it

5 0
3 years ago
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