Jane is sliding down a slide. What kind of motion is she demonstrating?

How many liters of water do we need to add to 5.00 moles to get a 0.648 M solution?

Tanya [424]

Answer: This is a typical acid/base equilibrium problem, that involves the use of logarithms.

Explanation:We assume that both nitric acid and hydrochloric acid dissociate to give stoichiometric

H

3

O

+

.

Moles of nitric acid:

26.0

×

10

−

3

⋅

L

×

8.00

⋅

m

o

l

⋅

L

−

1

=

0.208

⋅

m

o

l

H

N

O

3

(

a

q

)

.

And, moles of hydrochloric acid:

88.0

×

10

−

3

⋅

L

×

5.00

⋅

m

o

l

⋅

L

−

1

=

0.440

⋅

m

o

l

H

C

l

(

a

q

)

.

This molar quantity is diluted to

1.00

L

. Concentration in moles/Litre =

(

0.208

+

0.440

)

⋅

m

o

l

1

L

=

0.648

⋅

m

o

l

⋅

L

−

1

.

Now we know that water undergoes autoprotolysis:

H

2

O

(

l

)

⇌

H

+

O

H

−

. This is another equilibrium reaction, and the ion product

[

H

+

]

[

O

H

−

]

=

K

w

. This constant,

K

w

=

10

−

14

at

298

K

.

So

[

H

+

]

=

0.648

⋅

m

o

l

⋅

L

−

1

;

[

O

H

−

]

=

K

w

[

H

+

]

=

10

−

14

0.648

=

?

p

H

=

−

log

10

[

H

+

]

=

−

log

10

(

0.648

)

=

?

Alternatively, we know further that

p

H

+

p

O

H

=

14

. Once you have

p

H

,

p

O

H

is easy to find. Take the antilogarithm of this to get

[

O

H

−

]

.

Answer link

4 0

2 years ago

For the following balanced equation: 3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l) a) How many moles of HNO3 will r

s344n2d4d5 [400]

Answer:

a) 26.67 moles HNO3

b) 0.33 moles NO

c) 0.40 moles NO is produced

d).157 moles Cu

e) 0.105 moles NO

f) 26.4 grams HNO3

g) Cu is in excess

h) 2.41 grams Cu remain

i) 2.37 grams NO

Explanation:

Step 1: Data given

Molar mass of Cu = 63.55 g/mol

Molar mass of HNO3 = 63.01 g/mol

Molar mass of Cu(NO3)2 = 187.56 g/mol

Molar mass of NO = 30.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: The balanced equation

3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l)

a) How many moles of HNO3 will react with 10 moles of Cu?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 10 moles Cu we need 8/3 *10 = 26.67 moles HNO3

b) How many moles of NO will form if 0.50 moles of Cu reacts?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.50 moles Cu we'll have 2/3 *0.50 = 0.33 moles NO

c) If 0.80 moles of H2O forms, how much NO must also form?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

If 0.80 moles H2O is produced, 0.80/2 = 0.40 moles NO is produced

d) How many moles of Cu are in 10.0 grams of Cu?

Moles Cu = 10.0 grams / 63.55 g/mol = 0.157 moles

In 10.0 grams Cu we have 0.157 moles Cu

e) If 10.0 g of Cu reacts, how many moles of NO will form?

10.0 grams Cu = 0.157 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.157 moles Cu we'll have 2/3 * 0.157 = 0.105 moles NO

f) If 10.0 g of Cu reacts, how many grams of HNO3 are required?

10.0 grams Cu = 0.157 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.157 moles Cu we'll need 0.419 moles HNO3

This is 0.419 moles * 63.01 g/mol = 26.4 grams HNO3

g) If 10.0 g of Cu and 20.0 g of HNO3 are put together in a reaction vessel, which one will be in excess?

Moles Cu = 0.157 moles

Moles HNO3 = 20.0 grams / 63.01 g/mo = 0.317 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

The limiting reactant is HNO3. It will completely be consumed (0.317 moles). Cu is in excess. There will react 3/8 * 0.317 = 0.119 moles Cu

There will remain 0.157 - 0.119 = 0.038 moles

h) How many grams of the excess substance will be left over?

There will react 3/8 * 0.317 = 0.119 moles Cu

There will remain 0.157 - 0.119 = 0.038 moles

This is 0.038 moles * 63.55 g/mol = 2.41 grams

i) How many grams of NO will form in the reaction described in part g?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.317 moles HNO3 we'll have 0.317/4 = 0.0793 moles NO

This is 0.079 mol * 30.01 g/mol = 2.37 grams NO

3 0

3 years ago

If the car still produces 8500N, but now it accelerates at only 4 m/s2 what is the MASS

Mashutka [201]

Answer:

You can do that yourself, but there's a example question below. And, if for example, I just answer your question and you don't even try to answer. it dosent matter.

Explanation:Force=Mass x Acceleration -or- F=ma

where F is the force, m is the mass, and a is the acceleration. The units are Newtons (N) for force, kilograms (kg) for mass, and meters per second squared (m/s2) for acceleration. The other forms of the equation can be used to solve for mass or acceleration.

m=F/a and a=F/m Example:

Engineers at the Johnson Space Center must determine the net force needed for a rocket to achieve an acceleration of 70 m/s2. If the mass of the rocket is 45,000 kg, how much net force must the rocket develop?

Using Newton's second law, F=ma

F=(45,000 kg)(70 m/s2) = 3,150,000 kg m/s2 F=3,150,000 N Note that the units kg m/s2 and newtons are equivalent; that is, 1 kg m/s2

6 0

3 years ago

Taken all together the reactions of the human body are

jasenka [17]

The reactions of the human body are exothermic

8 0

3 years ago

How many moles are found in 10.5 g of Pb?

deff fn [24]

Answer:

Number of moles = 0.05 mol

Explanation:

Given data:

Mass of Pb = 10.5 g

Number of moles of Pb = ?

Solution:

Formula:

Number of moles = mass/molar mass

Molar mass of Pb is 207.2 g/mol.

Now we will put the values in formula.

Number of moles = 10.5 g/207.2 g/mol

Number of moles = 0.05 mol

6 0

3 years ago