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2 years ago

In 1.00 mol of potassium zirconium sulfate trihydrate, K4Zr(SO4)4.3 H20, there are

Chemistry

1 answer:

LenKa [72]2 years ago

8 0

Answer:

C. 4 × 6.02 × 10²³ potassium atoms.

Explanation:

In 1.00 mol of potassium zirconium sulfate trihydrate, K₄Zr(SO₄)₄.3 H₂O, there are

A . 3 × 6.02 × 10²³ hydrogen atoms. NO. There are 3 × 2 moles of hydrogen atoms = 6 moles of hydrogen atoms = 6 × 6.02 x 10²³ hydrogen atoms

B. 6.02 × 10²³ sulfur atoms. NO. There are 4 moles of sulfur atoms = 4 × 6.02 × 10²³ sulfur atoms

C. 4 × 6.02 × 10²³ potassium atoms. YES. There are 4 moles of potassium atoms = 4 × 6.02 × 10²³ potassium atoms

D . 4 moles of oxygen atoms. NO. There are 4 × 4 moles of oxygen atoms = 16 moles of oxygen atoms

E . 4 moles of zirconium atoms NO. There is 1 mole of zirconium atoms

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Dr. Franck cuts a bar of pure gold into smaller and smaller pieces. Will this action change the element that makes up the bar? E

Anvisha [2.4K]

Answer:

i believe this is a chemical or physical question? well your answer to that is no the element does not change because the gold is still gold it is still physical because you have just cut it into piece it is still gold

Explanation:

lmk if it was helpful :/

7 0

2 years ago

12. The vapor pressure of water at 90°C is 0.692 atm. What is the vapor pressure (in atm) of a solution made by dissolving 3.68

luda_lava [24]

Answer : The vapor pressure (in atm) of a solution is, 0.679 atm

Explanation : Given,

Mass of $H_2O$ = 1.00 kg = 1000 g

Moles of $CsF$ = 3.68 mole

Molar mass of $H_2O$ = 18 g/mole

Vapor pressure of water = 0.692 atm

First we have to calculate the moles of $H_2O$ .

$\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{1000g}{18g/mole}=55.55mole$

Now we have to calculate the mole fraction of $H_2O$

$\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CsF}=\frac{55.55}{55.55+3.68}=0.938$

Now we have to partial pressure of solution.

According to the Raoult's law,

$P_{Solution}=X_{H_2O}\times P^o_{H_2O}$

where,

$P_{Solution}$ = vapor pressure of solution

$P^o_{H_2O}$ = vapor pressure of water = 0.692 atm

$X_{H_2O}$ = mole fraction of water = 0.938

$P_{Solution}=X_{H_2O}\times P^o_{H_2O}$

$P_{Solution}=0.938\times 0.692atm$

$P_{Solution}=0.649atm$

Therefore, the vapor pressure (in atm) of a solution is, 0.679 atm

5 0

2 years ago

In this experiment, 0.170 g of caffeine is dissolved in 10.0 ml of water. the caffeine is extracted from the aqueous solution th

zmey [24]

solution:

Weight of caffeine is W = 0.170 gm.

Volume of water is V= 10 ml

Volume of methylene chloride which extracted caffeine is v= 5ml

No of portions n=3

Distribution co-efficient= 4.6

Total amount of caffeine that can be unextracted is given by

$w_{n}=w\times[\frac{k_{Dx}v}{k_{Dx}v+v}]^n\\w_{3}=0.170[\frac{4.6\times10}{(4.6\times10+5)}]^3\\=0.170[\frac{46}{46+5}]^3\\=0.170[\frac{46}{51}]^3\\=0.170[\frac{97336}{132651}]\\=0.170\times0.734=0.125gms$

amount of caffeine un extracted is 0.125gms

amount of caffeine extracted=0.170-0.125

=0.045 gms

6 0

2 years ago

Give the number of significant figures in this number: 40.00

Tatiana [17]

A significant figure is every symbol that made the number itself.

In this case, the number 40.00 has four figures but only two of them are significant 40, this is because you haven't got any more decimals than the first zero.

If you have a case with zeros in front, you take to the first non zero digits.

For example, 0.071004 you wold express as 0.071 and those 7, and 1 are the significant ones.

4 0

2 years ago

Read 2 more answers

Determine the PH at the point in the titration of 40.0ml of 0.200M HC4H7o2 with 0.100 M Sr(OH)2 after 100ml of the strong base h

Dmitriy789 [7]

Answer:

Check the explanation

Explanation:

Mols HC4H7O2 = (volume in L)*(molarity) = (40.0 mL)*(0.200 M)

= (40.0 mL)*(1 L)/(1000 mL)*(0.200 M)

= 8.00*10-3 mol.

Mols Sr(OH)2 corresponding to 10.0 mL of 0.100 M solution =

(volume in L)*(molarity)

= (10.0 mL)*(0.100 M)

= (10.0 mL)*(1 L)/(1000 mL)*(0.100 M)

= 1.00*10-3 mol.

Consider the ionization of Sr(OH)2 as below.

Sr(OH)2 (aq) ----------> Sr2+ (aq) + 2 OH- (aq)

As per the stoichiometric equation,

1 mol Sr(OH)2 = 2 mols OH-.

Therefore,

0.0010 mol Sr(OH)2 = [0.0010 mol Sr(OH)2]*(2 mols OH-)/[1 mole Sr(OH)2]

= 0.0020 mol

= 2.00*10-3 mol

Set up the ICE charts as below.

HC4H7O2 (aq) + OH- (aq) ------------> H2O (l) + C4H7O2- (aq)

Before (mol) 8.00*10-3 2.00*10-3 - -

Change (mol) -2.00*10-3 -2.00*10-3 - +2.00*10-3

After (mol) 6.00*10-3 0 - 2.00*10-3

The change in a pure substance, e.g., H2O is not considered in an acid-base reaction.

Volume of the solution = (40.0 + 10.0) mL = 50.0 mL = (50.0 mL)*(1 L)/(1000 mL) = 0.05 L.

The initial concentrations are obtained by dividing the numbers of moles by the volume, 0.05 L.

Set up the ICE charts as below.

HC4H7O2 (aq) + OH- (aq) ------------> H2O (l) + C4H7O2- (aq)

Initial (M) 0.160 0.0400 - -

Change (M) -0.0400 -0.0400 - +0.0400

Equilibrium (M) 0.120 0 - 0.0400

The acid-ionization constant is written as

Ka = [H3O+][C4H7O2-]/[HC4H7O2] = 1.5*10-5

Plug in the known values and get

Ka = [H3O+]*(0.0400)/(0.120) = 1.5*10-5

======> [H3O+] = (1.5*10-5)*(0.120)/(0.0400) (ignore units)

======> [H3O+] = 4.5*10-5

The proton concentration of the solution is 4.5*10-5 M.

pH = -log (4.5*10-5 M)

= 4.346

≈ 4.35 (ans).

8 0

2 years ago