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2 years ago

HELP!!!!!! PLEASE!!!

Chemistry

1 answer:

mixas84 [53]2 years ago

7 0

Answer:

Explanation:

sorry if im wrong!!

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The pH of a solution is 3.81. What is the OH concentration in the solution?

disa [49]

Answer:

6.46 × 10⁻¹¹ M

Explanation:

Step 1: Given data

pH of the solution: 3.81

Step 2: Calculate the pOH of the solution

We will use the following expression.

pH + pOH = 14.00

pOH = 14.00 - pH = 14.00 - 3.81 = 10.19

Step 3: Calculate the concentration of OH⁻ ions

We will use the definition of pOH.

pOH = -log [OH⁻]

[OH⁻] = antilog -pOH = antilog -10.19 = 6.46 × 10⁻¹¹ M

3 0

2 years ago

Rain washing away soil from a hillside

Aneli [31]

Answer:

deposition

Explanation:

8 0

2 years ago

A 50.0 mL sample of a 1.00 M solution of CuSO4 is mixed with 50.0 mL of 2.00 M KOH in a calorimeter. The temperature of both sol

Reika [66]

Answer : The enthalpy change for the process is 52.5 kJ/mole.

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the solution

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]$

where,

q = heat released by the reaction

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the solution

$c_1$ = specific heat of calorimeter = $12.1J/^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_2$ = mass of water or solution = $Density\times Volume=1/mL\times 100.0mL=100.0g$

$\Delta T$ = change in temperature = $T_2-T_1=(26.3-20.2)^oC=6.1^oC$

Now put all the given values in the above formula, we get:

$q=[(12.1J/^oC\times 6.1^oC)+(100.0g\times 4.18J/g^oC\times 6.1^oC)]$

$q=2623.61J$

Now we have to calculate the enthalpy change for the process.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat released = 2626.61 J

n = number of moles of copper sulfate used = $Concentration\times Volume=1M\times 0.050L=0.050mole$

$\Delta H=\frac{2623.61J}{0.050mole}=52472.2J/mole=52.5kJ/mole$

Therefore, the enthalpy change for the process is 52.5 kJ/mole.

8 0

3 years ago

Help I really help on this

nikdorinn [45]

I think it’s C but not certainly positive

8 0

2 years ago

Calculate the mass of silver bromide produced from 22.5 g of silver nitrate.

Orlov [11]

Equation is as follow,

 2 AgNO₃ + MgBr₂ → 2 AgBr + Mg(NO₃)₂

According to eq.

339.74 g (2 moles) AgNO₃ produces = 375.54 g (2 moles) of AgBr
So,
22.5 g AgNO₃ will produce = X g of AgBr

Solving for X,
X = (22.5 g × 375.54 g) ÷ 339.74 g

X = 24.87 g of AgBr

6 0

2 years ago