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jenyasd209 [6]
3 years ago
13

What volume of a 1.5 M KOH solution is needed to provide 3.0 moles of KOH?

Chemistry
1 answer:
Brums [2.3K]3 years ago
8 0

Answer:

  • Volume = <u>2.0 liter</u> of 1.5 M solution of KOH

Explanation:

<u>1) Data:</u>

a) Solution: KOH

b) M = 1.5 M

c) n = 3.0 mol

d) V = ?

<u>2) Formula:</u>

Molarity is a unit of concentration, defined as number of moles of solute per liter of solution:

  • M = n / V in liter

<u>3) Calculations:</u>

  • Solve for n: M = n / V ⇒ V = n / M

  • Substitute values: V = 3.0 mol / 1.5 M = 2.0 liter

You must use 2 significant figures in your answer: <u>2.0 liter.</u>

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What is the chemical equation of ccl4, please explain thank you! :-D
Gekata [30.6K]
It could be
C + 2Cl2 ➡️ CCl4
I decided this because I believe this is the only way you could make this a balanced equation. To be a balanced equation Both sides of the equation has the same number of atoms for each element. Carbon has 1 atom and chlorine has 4.
3 0
3 years ago
Calculate the volume of 5.0 grams of NO gas at STP.
Sveta_85 [38]

Answer:

The volume of  5.0 g CO  2  is  2.6 L CO  2  at STP

Explanation:

STP

STP is currently  

0

∘

C

or  

273.15 K

, which are equal, though the Kelvin temperature scale is used for gas laws; and pressure is  

10

5

.

Pascals (Pa)

, but most people use  

100 kPa

, which is equal to  

10

5

.

Pa

.

You will use the ideal gas law to answer this question. Its formula is:

P

V

=

n

R

T

,

where  

P

is pressure,  

V

is volume,  

n

is moles,  

R

is a gas constant, and  

T

is temperature in Kelvins.

Determine moles

You may have noticed that the equation requires moles  

(

n

)

, but you have been given the mass of  

CO

2

. To determine moles, you multiply the given mass by the inverse of the molar mass of  

CO

2

, which is  

44.009 g/mol

.

5.0

g CO

2

×

1

mol CO

2

44.009

g CO

2

=

0.1136 mol CO

2

Organize your data

.

Given/Known

P

=

100 kPa

n

=

0.1136 mol

R

=

8.3145 L kPa K

−

1

mol

−

1

https://en.wikipedia.org/wiki/Gas_constant

T

=

273.15 K

Unknown:  

V

Solve for volume using the ideal gas law.

Rearrange the formula to isolate  

V

. Insert your data into the equation and solve.

V

=

n

R

T

P

V

=

0.1136

mol

×

8.3145

.

L

kPa

K

−

1

mol

−

1

×

273.15

K

100

kPa

=

2.6 L CO

2

rounded to two significant figures due to  

5.0 g

Answer link

Doc048

May 18, 2017

I got 2.55 Liters

Explanation:

1 mole of any gas at STP = 22.4 Liters

5

g

C

O

2

(

g

)

=

5

g

44

(

g

mole

)

=

0.114

mole

C

O

2

(

g

)

Volume of 0.114 mole  

C

O

2

(

g

)

= (0.114 mole)(22.4 L/mole) = 2.55 Liters  

C

O

2

(g) at STP

6 0
3 years ago
ASAP!IF YOU PUT LINKS YOUR REPORTED!<br><br> how does sand sifting collect plastic
ANEK [815]

Answer:

Dirty sand is piled on a sheet of fine mesh stretched between two long poles, the mesh collects the mircoplastic and other materials while allowing the sand to fall through.

5 0
3 years ago
8. Which of the following elements reacts like calcium (Ca) but with less atomic
Salsk061 [2.6K]

Answer:

Magnesium (Mg)

Explanation:

An element with a lower atomic mass than Calcium that has similar chemical property is the element Magnesium (Mg).

This because, elements in the same group or family on the periodic table has similar chemical properties.

Magnesium belongs to the same group as Calcium, other elements are:

     Be  Mg  Ca    Sr  Ba  Ra

The group is made of the alkaline earth metals. They have two valence electrons in their outer shell.

7 0
3 years ago
Glucose (C6H12O6) is a key nutrient for generating chemical potential energy in biological systems. We were provided 16.55 g of
harina [27]

Answer:

a) 40 %

b) 4.04~g~CO_2

c) 5.53x10^2^3~molecules~of~O_2

Explanation:

For a) we will have to calculate the <u>molar mass</u> of C_6H_1_2O_6, so the first step is to find the <u>atomic mass</u> of each atom and multiply by the <u>amount of atoms</u> in the molecule.

C => 12*(6) = 72

H => 1*(12) = 12

O => 6*(16) = 96

Molar mass = 180 g/mol

Then we can calculate the percentage by mass:

Percentage~=~\frac{72}{180}*100=40

For b) we have to start with the <u>reaction of glucose</u>:

C_6H_1_2O_6~+~6O_2~->~6CO_2~+~6H_2O

Then we have to convert the grams of glucose to moles, the moles of glucose to moles of carbon dioxide and finally the moles of carbon dioxide to grams. To do this we have to take into account the<u> following conversion ratios</u>:

-) 180 g of glucose = 1 mol glucose

-) 1 mol glucose = 6 mol carbon dioxide

-) 1 mol carbon dioxide = 44 g carbon dioxide

16.55~g~C_6H_1_2O_6\frac{1~mol~C_6H_1_2O_6}{180~g~C_6H_1_2O_6}\frac{6~mol~CO_2}{1~mol~C_6H_1_2O_6}\frac{44~g~CO_2}{1~mol~CO_2}=4.04~g~CO_2

For C, we have to start with the conversion from grams of glucose to moles, the moles of glucose to moles of oxygen and finally the moles of oxygen to molecules. To do this we have to take into account the <u>following conversion ratios</u>:

-) 180 g of glucose = 1 mol glucose

-) 1 mol glucose = 6 mol oxygen

-) 1 mol oxygen = 6.023x10^23 molecules of O2

16.55~g~C_6H_1_2O_6\frac{1~mol~C_6H_1_2O_6}{180~g~C_6H_1_2O_6}\frac{6~mol~O_2}{1~mol~C_6H_1_2O_6}\frac{6.023x10^2^3~molecules~O_2}{1~mol~O_2}=~5.53x10^2^3~molecules~of~O_2

4 0
3 years ago
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