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leonid [27]
2 years ago
6

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

Chemistry
1 answer:
CaHeK987 [17]2 years ago
4 0
The reaction between LiOH and HCl is;   
LiOH + HCl → LiCl + H₂O

The stoichiometric ratio between LiOH and HCl is 1 : 1
moles of LiOH added = moles of HCl needed to neutralize.

Molar mass of LiOH = 24 g/mol

moles = mass / molar mass
LIOH moles  = 1.65 / 24 = 0.06875 mol
Hence needed HCl moles = 0.06875 mol

Molarity = moles (mol) / Volume (L)
Hence needed HCl volume = moles / molarity                                           
                                            = 0.06875 mol / 0.150 mol/L                                           
                                            = 0.458 L = 458 mL

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At 10°c one volume of water dissolves 3.10 volumes of chlorine gas at 1.00 atm pressure. what is the henry's law constant of cl2
s344n2d4d5 [400]
Answer is:  0,133 mol/ l· atm.
T(chlorine) = 10°C = 283K.
p(chlorine) = 1 atm.
V(chlorine) = 3,10 l.
R - gas constant, R = 0.0821 atm·l/mol·K. 
Ideal gas law: p·V = n·R·T
n(chlorine) = p·V ÷ R·T.
n(chlorine) = 1atm · 3,10l ÷ 0,0821 atm·l/mol·K · 283K = 0,133mol.
Henry's law: c = p·k.
k - <span>Henry's law constant.
</span>c - solubility of a gas at a fixed temperature in a particular solvent.
c = 0,133 mol/l.
k = 0,133 mol/l ÷ 1 atm = 0,133 mol/ l· atm.

4 0
2 years ago
What is the percent CdSO4 by mass in a 1.00 m aqueous CdSO4 solution?
Svet_ta [14]
I don't know but I'm wasting 5 seconds of your time you can't take back sorry
3 0
2 years ago
A 20.0 -\,L volume of an ideal gas in a cylinder with a piston is at a pressure of 3.2 atm. Enough weight is suddenly removed fr
zzz [600]

Answer:

1. ΔE = 0 J

2. ΔH = 0 J

3. q = 3.2 × 10³ J

4. w = -3.2 × 10³ J

Explanation:

The change in the internal energy (ΔE) and the change in the enthalpy (ΔH) are functions of the temperature. If the temperature is constant, ΔE = 0 and ΔH = 0.

The gas initially occupies a volume V₁ = 20.0 L at P₁ = 3.2 atm. When the pressure changes to P₂ = 1.6 atm, we can find the volume V₂ using Boyle's law.

P₁ × V₁ = P₂ × V₂

3.2 atm × 20.0 L = 1.6 atm × V₂

V₂ = 40 L

The work (w) can be calculated using the following expression.

w = - P . ΔV

where,

P is the external pressure for which the process happened

ΔV is the change in the volume

w = -1.6 atm × (40L - 20.0L) = -32 atm.L × (101.325 J/1atm.L) = -3.2 × 10³ J

The change in the internal energy is:

ΔE = q + w

0 = q + w

q = - w = 3.2 × 10³ J

6 0
2 years ago
How much heat is released as the temperature of 25.2 grams of iron is decreased from 72.1°C to 9.8°C? The specific heat of iron
prisoha [69]

Answer:

Q=-697.06\ J

Negative sign says that release of heat.

Explanation:

The expression for the calculation of the heat released or absorbed of a process is shown below as:-

Q=m\times C\times \Delta T

Where,  

\Delta H  is the heat released or absorbed

m is the mass

C is the specific heat capacity

\Delta T  is the temperature change

Thus, given that:-

Mass = 25.2 g

Specific heat = 0.444 J/g°C

\Delta T=9.8-72.1\ ^0C=-62.3\ ^0C

So,  

Q=25.2\times 0.444\times -62.3\ J=-697.06\ J

Negative sign says that release of heat.

8 0
2 years ago
A 3.5 kg iron shovel is left outside through the winter. The shovel, now orange with rust, is rediscovered in the spring. Its ma
Anton [14]

Answer:

a

Explanation:

8 0
2 years ago
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