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3 years ago

I need help with this one please it’s science

Chemistry

2 answers:

amid [387]3 years ago

6 0

Answer:

Explanation:

Taya2010 [7]3 years ago

5 0

Answer: A change in appearance. (Please say thanks!)

Explanation:

The biggest sign of a chemical change is change in appearance.

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Provide three specific examples of how this cloning is used.

lara [203]

Answer:

Cloning is used to duplicate different organisms

5 0

3 years ago

on the periodic table, elements are arranged by which of the following. A. mass numbers. B. increasing atomic number. C. alphabe

Sever21 [200]

Its B. Increasing Atomic Number.

3 0

3 years ago

When NEM is added to a purified solution of creatine kinase, Cys 278 is alkylated, but no other Cys residues in the protein are

USPshnik [31]

Answer:

The answer is given below

Explanation:

Cys 278 residue is the only available cysteine which is alkylated by the addition of N-Ethylmaleimide or NEM (alkylating agent). It works by only alkylating the sulfhydryls. In this case, Cys 278 residue is the only one which has exposed cysteine residue.

While the other residues have their sulfhydryls group either involved in the synthesis of disulfide bonds of proteins or their Cys residues are intrinsically placed in the proteins and cannot be alkylated with NEM.

NEM cannot alkylate if its protein is not available in the free form or it is in bounded form. For NEM to alkylate Cys 278, it should be free and should have sulfhydryls available for alkylation.

Alkylation: it is the transfer of alkyl groups. Alkyl groups contain Hydrogen and Carbon in their structure.

5 0

4 years ago

A researcher dispenses distilled deionized water from a 20-ml volumetric pipet into an empty 8.4376 g weighting bottle. If the t

bonufazy [111]

As given that some volume of water has been dispensed say "x mL"

The initial weight of bottle =8.4376 g

The final weight of bottle + water =28.5845 g

So weight of water transferred = 28.5845 g - 8.4376 g = 20.1469 g

Now there is a relation between density, mass and volume

density = mass / volume

Therefore

Volume = mass / density

So volume of water dispensed = mass dispensed / density =20.1469 g / 0.9967867 g/ml.

Volume of water = 20.2118467 mL

7 0

4 years ago

Calculate the volume of carbon dioxide at 20.0°C and 0.941 atm produced from the complete combustion of 4.00 kg of methane. Comp

tankabanditka [31]

Answer:

The volume of carbon dioxide at 20.0°C and 0.941 atm produced was 6390.89 Liters.

More volume of carbon dioxide gas was released on combustion of 4.00 kg of propane in comparison to the volume of carbon dioxide released on combustion of 4.00 kg of methane.

Explanation:

Methane

$CH_4+2O_2\rightarrow CO_2+2H_2O$

Mass of methane = 4.00 kg = 4000 g (1 kg = 1000 g)

Moles of methane = $\frac{4000 g}{16 g/mol}=250 mol$

According to reaction, 1 mole of methane gives 1 mole of carbon dioxide gas,then 250 moles of methane will give :

$\frac{1}{1}\times 250 mol=250 mol$ of carbon dioxide gas

Moles of carbon dioxide gas = n = 250 mol

Pressure of carbon dioxide gas = P = 0.941 atm

Temperature of carbon dioxide gas = T = 20.0°C = 20.0+273 K = 293 K

Volume of carbon dioxide gas = V

$PV=nRT$ (Ideal gas equation)

$V=\frac{nRT}{P}=\frac{250 mol\times 0.0821 atm L/mol K\times 293 K}{0.941 atm}=6390.89 L$

The volume of carbon dioxide at 20.0°C and 0.941 atm produced was 6390.89 Liters.

Propane

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

Mass of propane = 4.00 kg = 4000 g (1 kg = 1000 g)

Moles of propane = $\frac{4000 g}{44 g/mol}=90.91 mol$

According to reaction, 1 mole of propane gives 3 mole of carbon dioxide gas,then 90.91 moles of propane will give :

$\frac{3}{1}\times 90.91 mol=272.73 mol$ of carbon dioxide gas

Moles of carbon dioxide gas = n = 272.73 mol

Pressure of carbon dioxide gas = P = 0.941 atm

Temperature of carbon dioxide gas = T = 20.0°C = 20.0+273 K = 293 K

Volume of carbon dioxide gas = V

$PV=nRT$ (Ideal gas equation)

$V=\frac{nRT}{P}=\frac{272.73 mol\times 0.0821 atm L/mol K\times 293 K}{0.941 atm}=6,971.95 L$

The volume of carbon dioxide at 20.0°C and 0.941 atm produced was 6,971.95 Liters.

Volume of carbon dioxide given by combustion of 4.00 kg of methane = 6390.89 Liters.

Volume of carbon dioxide given by combustion of 4.00 kg of propane = 6,971.95 Liters.

6390.89 Liters < 6,971.95 Liters

More volume of carbon dioxide gas was released on combustion of 4.00 kg of propane in comparison to the volume of carbon dioxide released on combustion of 4.00 kg of methane.

7 0

3 years ago