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Anni [7]
4 years ago
12

To neutralized 1.65g LiOH, how much .150M HCl would be needed?

Chemistry
1 answer:
katovenus [111]4 years ago
8 0
0.1 M x 0.5L = 0.05 mols HCl.
Adding 25 mL 2M NaOH is
2M x 0.025 L = 0.05 mols NaOH.
What you want to do is to back off very slightly with the NaOH (you might try something like 24.95 mL which I calculate to give 0.00019 M or a pH of 3.7. Inching closer, 24.99 mL would leave H^+ of 4E-5 for pH 4.4. The problem here is two-fold.
hope it helps
 
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2.50 L of a gas at standard temperature and pressure is compressed to 575 mL. What is the new pressure of the gas
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Answer:

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Explanation:

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5 0
3 years ago
Two children on roller skates stand facing each other. The child on the right puts her arms out and pushes away from her partner
shutvik [7]

Answer:

They both go backward because of force.

Explanation:

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8 0
2 years ago
A 32.5 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 63.0 ∘C, in an insulated container. The f
Allisa [31]

Answer:

The mass of water m_{w} = 39.18 gm

Explanation:

Mass of iron m_{iron} = 32.5 gm

Initial temperature of iron T_{1} = 22.4°c = 295.4 K

Specific heat of iron  C_{iron} = 0.448 \frac{KJ}{kg K}

Mass of water = m_{w}

Specific heat of water  C_{w} = 4.2 \frac{KJ}{kg  K}

Initial temperature of water T_{2} = 336 K  

Final temperature after equilibrium T_{f} = 59.7°c = 332.7 K

When iron rod is submerged into water then

Heat lost by water  = Heat gain by iron rod

m_{w} C_{w} (T_{2} - T_{f} ) =  m_{iron} C_{iron} ( T_{f} - T_{1} )

Put all the values in above formula we get

m_{w} × 4.2 × ( 336 - 332.7 ) = 32.5 × 0.448 × ( 332.7 - 295.4 )

m_{w} = 39.18 gm

Therefore the mass of water m_{w} = 39.18 gm

8 0
3 years ago
Name two compounds in unpolluted air?
goblinko [34]
Nitrogen and oxygen are in unpolluted air

6 0
4 years ago
Earth has approximately 600,000,000 meters of coastline. If we assume this entire length of coastline has sandy beaches 60 meter
VashaNatasha [74]

Answer:

7 and 11

Explanation:

The amount of sand on the beaches can be found using this formula:

volume (m3) = length (m) × width (m) × depth (m)

(6 × 108 m) × 60 m × 20 m = 7 × 1011 m3

Therefore, there would be a total of 7 × 1011 cubic meters of sand on the beaches.

4 0
4 years ago
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