CHEMISTRY COLLEGE

Whats the difference between china and Europe

Answers

Answer 1

Answer:

Answer:

the difference is they arnt close

Explanation:

Answer 2

Answer:

Answer:

China is communist, Europe isn't.

Explanation:

Organic chemistry is the chemistry of carbon based compounds while biochemistry is the study of directly biologically relevant chemistry..

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What is the empirical formula of a compound that is 64.3 % c, 7.2 % h, and 28.5 % o by mass?

Answers

Asnwer : Empirical formula of a compound is : $C_(3)H_(4)O$

Given information : C = 64.3 % , H = 7.2 % , O = 28.5 %

Step 1 : Convert the given percentage (%) to grams.

Explanation : Let the total mass of the compound be 100 grams.

Mass of C = 64.3 g

$(100g)* ((64.3percent))/((100percent)) = 64.3g$

Mass of H = 7.2 g

$(100g)* ((7.2percent))/((100percent)) = 7.2g$

Mass of O = 28.5 g

$(100g)* ((28.5percent))/((100percent)) = 28.5g$

Step 2 : Convert the grams of each compound to moles.

$Moles = (Grams)/(Molar mass)$

Molar mass of C = 12.0g/mol

Molar mass of H = 1.0 g/mol

Molar mass of O = 16.0g/mol

$Moles of C = (64.3g)/(12.0(g)/(mol))$

Moles of C = 5.36 mol

$Moles of H = (7.2g)/(1.0(g)/(mol))$

Moles of H = 7.2 mol

$Moles of O = (28.5g)/(16.0(g)/(mol))$

Moles of O = 1.78 mol

Step 3 : Find the mole ratio of C , H and O

Mole ratio is calculated by dividing the mole values by the smallest value.

Mole of C = 5.36 mol , Mole of H = 7.2 mol , Mol of O = 1.78 mol

Out of the three mole values , mole value of O that is 1.78 mol is less , so we divide all the mole values by 1.78 mol.

$Mole of C = (5.36mol)/(1.78mol) = 3.0$

$Mole of H = (7.2mol)/(1.78mol) = 4.0$

$Mole of O = (1.78mol)/(1.78mol) = 1.0$

C : H : O = 3 : 4 : 1

So empirical formula of the compound is $C_(3)H_(4)O_(1)$ or $C_(3)H_(4)O$

A student puts 0.020 mol of methyl methanoate into an empty and rigid 1.0 L vessel at 450 K. The pressure is measured to be 0.74 atm. When the experiment is repeated using 0.020 mol of ethanoic acid instead of methyl methanoate, the measured pressure is lower than 0.74 atm. The lower pressure for ethanoic acid is due to the following reversible reaction.CH3COOH(g)+CH3COOH(g) ⇋ (CH3COOH)2(g)+Assume that when equilibrium has been reached, 50 percent of the ethanoic acid molecules have reacted.i. Calculate the total pressure in the vessel at equilibrium at 450 K.ii. Calculate the value of the equilibrium constant, Kp, for the reaction at 450 K

Answers

Explanation:

Starting moles of ethanol acid = 0.020 mol

At the equilibrium 50 % of the ethanol acid molecules reacted

∴ Moles of ethanol acid reacted = 0.020 mol * 50 %/100 %

= 0.010 mol

Moles of ethanol acid remain = 0.020 mol + 0.010 mol = 0.010 mol

Moles of the product $(CH3COOH)^(2)$ gas formed are calculated as

0.010 mol CH3COOH * 1 mol $(CH3COOH)^(2)$ / 2 mol CH3COOH

= 0.005 mol $(CH3COOH)^(2)$

Therefore at the equilibrium total moles of gas present in the vessel are 0.010 mol CH3COOH and 0.005 mol $(CH3COOH)^(2)$

That is total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol

Now Calculate the pressure :

0.020 mol gas has pressure of 0.74 atm therefore at the same condition what will be the pressure exerted by 0.015 mol gas

P1/n1 = P2/n2

P2 = P1*n2 / n1

= 0.74 atm * 0.015 mol / 0.020 mol

= 0.555 atm

You have 2.2 mol Xe and 2.0 mol F2, but when you carry out thereaction you end up with only 0.25 mol XeF4. What is the percent yield

of this experiment?

Xe(g) + 2 F2 (g) - XeF. (g)

Answers

The percentage yield of XeF from the concentration of the given reactants will be 25%.

The reaction states that 1 mole of Xe will give 1 mole of XeF, and 2 moles of fluorine, will gives 1 mole of XeF.

The limiting reactant can be calculated:

1 mole Xe = 2 moles of $\rm F_2$

2.2 moles of Xe = 2.2 $*$ 2 moles of $\rm F_2$

2.2 moles of Xe = 4.4 moles of $\rm F_2$

Since the amount of available $\rm F_2$ has been in the limiting, thus $\rm F_2$ has been the limiting reactant.

So, the yield of XeF in terms of $\rm F_2$ will be:

2 moles $\rm F_2$ = 1 mole XeF

Thus the theoretical yield of XeF is 1 mole.

The yield of XeF we get = 0.25 moles.

Thus the percentage yield = $\rm (Obtained)/(Theoretical\;yield)\;*\;100$

Percentage yield = $\rm (0.25)/(1)\;*\;100$

Percentage yield = 25%

Thus the percentage yield of XeF from the concentration of the given reactants will be 25%.

For more information about the percent yield, refer to the link:

brainly.com/question/12809634

Does dissolved potassium chloride affect the surface tension between water molecules?

Answers

Answer:

Inorganic impurities present in the bulk of a liquid such as KCl tend to increase the surface tension of water.

Explanation:

As potassium chloride (KCl) dissolves in water, the ions are hydrated. ... When ionic compounds dissolve in water, the ions in the solid separate and disperse uniformly throughout the solution because water molecules surround and solvate the ions, reducing the strong electrostatic forces between them.

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For the Bradford assay, the instructor will make a Bradford reagent dye by mixing 50 ml of 95% v/v ethanol with 100 mg of Coomassie Blue followed by the addition of 50 ml of 85% v/v phosphoric acid. This entire mixture is then diluted to 1 liter with water. What is the final concentration of phosphoric acid?

Answers

Answer:

4,25% v/v H3PO4

Explanation:

The concentration of phosphoric acid (H3PO4) is expressed as a volume / volume percentage, which means:

%v/v H3PO4 = (mL of pure H3PO4/mL of solution)*100%

In other words, we are only interested in the final volume of the solution to which the phosphoric acid was diluted, regardless of its composition. Which in this case is 1 L (1000 mL).

We can then apply the following equation, commonly used to calculate the initial or final concentration (or volume) of a substance when it is diluted:

Ci*Vi=Cf*Vf

Where:

Ci, is the initial concentration of the substance.

Vi, the initial volume of the substance

Cf, the final concentration reached after dilution

Vf, the final volume of the solution at which the substance was diluted

In this case, the incognite would be the final concentration of H3PO4 reached after dilution, that is, Cf. Therefore, we proceed to clear Cf from the previous equation and replace our data:

Cf = (Ci*Vi)/Vf = (85% v/v * 50 mL)/1000 mL = 4,25 % v/v

Note that being up and down in the division, the mL unit is canceled to result in% v / v.

Other Questions

A compound contains only carbon, hydrogen, and oxygen. Combustion of 139.1 g of the compound yields 208.6 g of CO2 and 56.93 g of H2O. The molar mass of the compound is 176.1 g/mol. *Each part of this problem should be submitted separately to avoid losing your work* 1. Calculate the grams of carbon (C) in 139.1 g of the compound: grams 2. Calculate the grams of hydrogen (H) in 139.1 g of the compound. grams 3. Calculate the grams of oxygen (O) in 139.1 g of the compound. grams

Sodium phosphate and calcium chloride react to form sodium chloride and calcium phosphate. If you have 379.4 grams of calcium chloride and an excess of sodium phosphate, how much calcium phosphate can you make?

6 CO2 + 6 H2OHow many atoms of each element are inthe equation?

Aqueous solutions of barium chloride and silver nitrate are mixed to form solid silver chloride and aqueous barium nitrate. The complete ionic equation contains which of the following species (when balanced in standard form)? A. NO (aq) B. 2Ba (aq) C. 2Ag (aq) D. CI(aq)