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3 years ago

Help please!!!!!!!!!!!

Chemistry

1 answer:

EleoNora [17]3 years ago

4 0

water acts as the solvent for this experiment

we use to calculate R f values and compare with know to determine the compound is made of example in forensic science to find DNA, amino acids

most soluble one will travel the most

c to b to a (A most further travelled)

Explanation:

This quest in requires general knowledge on chromatography

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Please help ASAP 100 points points help I will mark you as brainlister

Vlada [557]

this is lithium ion with a +1 charge

7 0

4 years ago

Read 2 more answers

If kerosene has a specific gravity of 0.820, what force will be exerted on the circular bottom of a cylindrical kerosene tank th

Svet_ta [14]

Answer:

$F = 774146.534\,N$

Explanation:

The pressure at the bottom of the tank is:

$P_{bottom} = (0.820)\cdot (1000\,\frac{kg}{m^{3}})\cdot (9.807\,\frac{m}{s^{2}})\cdot (30\,ft)\cdot (\frac{0.305\,m}{1\,ft} )$

$P_{bottom} = 73581.921\,Pa$

The force exerted on the circular bottom is:

$F=(73581.921\,Pa)\cdot (\frac{\pi}{4} )\cdot [(12\,ft)\cdot (\frac{0.305\,m}{1\,ft} )]^{2}$

$F = 774146.534\,N$

4 0

3 years ago

Fernando places 50 mL of HCl in a 100 mL beaker and 50 mL of NaOH in a 100 mL beaker. he places them on a scale together and mea

olchik [2.2K]

Answer:

The correct options are:

B. The HCl is the solute.

D. The H₂O is the solvent.

E. This mixture conducts electricity.

Explanation:

For the preparation of the HCl solution, 10 mL of HCl and 40 mL of water were mixed.

The compound that is present in the solution in a smaller ratio is the solute and the other compount -present in higher ratio- is the solvent.

So, the solute is HCl and the solvent is water (H₂O). Furthermore, water is considered as an universal solvent (for most of the polar and ionic compounds). According to this, the following options are correct:

B. The HCl is the solute.

D. The H₂O is the solvent.

Then, a condition for a solution that is able to conduct electricity is the presence of ions in the solution. When HCl is dissolved in water, it dissociates in ions according to the following equilibrium:

HCl ⇔ H⁺ + Cl⁻

The ions H⁺ and Cl⁻ are electrically charged so the solution can conduct electricity. According to this, the following option is also correct:

E. This mixture conducts electricity.

7 0

3 years ago

1. Determine the number of atoms in 4.64 g gold.

liubo4ka [24]

Answer:

The answer to your question is below

Explanation:

1.-

a) Look for the mass number of Gold in the periodic table

Mass number = 197 g

Use proportions to answer this question

197 g of gold ------------- 6.023 x 10 ²³ atoms

4.64 g of gold ---------- x

x = (4.64 x 6.023 x 10²³) / 197

x = 1.42 x 10²² atoms

2.- mass of 4.12 x 10²⁴ molecules of glucose

Molecular mass = (6 x 12) + (12 X 1) + (6 x 16)

= 72 + 12 + 96

= 180 g

180 g of glucose ------------------- 6.023 x 10 ²³ molecules

x ------------------- 4.12 x 10 ²⁴ molecules

x = (4.12 x 10²⁴ x 180) / 6.023 x 10²³

x = 1231.3 g of glucose

4 0

3 years ago

Ethanol (C2H5OH) is produced from the fermentation of sucrose in the presence of enzymes.

Klio2033 [76]

Explanation:

to calculate theoratical yield...first we must calculate how much Sucrose moles we have..in order to to do so...we must calculate molar mass of sucrose.

molar mass of sucrose is 342.3 g/mol.

now we can calculate how many sucrose moles we have by dividing the mass with molar mass of sucrose

735g/(342.3g/mol)=2.147mol

theoratically...according to stoichiometry..every 1 mole of sucrose yields 4 moles of ethanol...so...

2.147mole yield 2.147*4mol=8.588mol

now we must calculate weight of that much ethanol..molar mass of ethanol is 46.07 g/mol...

so we can multiple moles by molar mass to obtain the weight 8.588mol*46.07g/mol=395.649g

but we only obtained 310.5g...so percentage we have is

$\frac{310.5}{395.649} \times 100$

78.47%

if there is trouble with molar mases I used....use what you calculated...

If my explanation is good enough...please mark it as brainliest.thanks

5 0

3 years ago