What is the scientific method describes as nonlinear

A solution has a poh of 7. 1 at 10∘c. what is the ph of the solution given that kw=2. 93×10−15 at this temperature? remember to

Gnom [1K]

A solution has a pOH of 7. 1 at 10∘c. Then the pH of the solution given that kw=2. 93×10−15 at this temperature is 7.4 .

It is given that,

pOH of solution = 7.1

Kw =2.93×10^(-15)

Firstly, we will calculate the value of pKw

The expression which we used to calculate the pKw is,

pKw=-log [Kw]

Now by putting the value of Kw in this expression,

pKw =−log{2.93×10^(-15)}

pKw =15log(2.93)

pKw=14.5

Now we have to calculate the pH of the solution.

As we know that,

pH+pOH=pKw

Now put all the given values in this formula,

pH+7.1=14.5

pH=7.4

Therefore, we find the value of pH of the solution is, 7.4.

learn more about pH value:

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7 0

11 months ago

1. A 10 g gold coin is heated from 25°C to 50°C (CAu is 0.13 J/g-°C). What is the H?

Oksanka [162]

The correct answer is + 32.5 J because heat is absorbed here by copper not evolved also after calculation we will find the H value is positive, lets calculate it:
H = m C Δt where:
m = 10 g
C = 0.13 J/g.°C
Δ t = final temperature - initial temperature = 50 - 25 = + 25 °C
so H = 10 x 0.13 x (+25) = + 32.5 J

4 0

3 years ago

A calorimeter contains 22.0 mL of water at 14.0 ∘C . When 2.50 g of X (a substance with a molar mass of 82.0 g/mol ) is added, i

Alik [6]

Answer:

The enthalpy change in the the reaction is -47.014 kJ/mol.

Explanation:

$X(s)+H_2O(l)\rightarrow X(aq)$

Volume of water in calorimeter = 22.0 mL

Density of water = 1.00 g/mL

Mass of the water in calorimeter = m

$m=1.00 g/mL\times 22.0 mL=22 g$

Mass of substance X = 2.50 g

Mass of the solution = M = 2.50 g + 22 g = 24.50 g

Heat released during the reaction be Q

Change in temperature =ΔT = 28.0°C - 14.0°C = 14.0°C

Specific heat of the solution is equal to that of water :

c = 4.18J/(g°C)

$Q=Mc\times \Delta T$

$Q=24.50 g\times 4.18 J/g ^oC\times 14.0^oC=1,433.74 J=1.433 kJ$

Heat released during the reaction is equal to the heat absorbed by the water or solution.

Heat released during the reaction =-1.433 kJ

Moles of substance X= $\frac{2.50 g}{82.0 g/mol}=0.03048 mol$

The enthalpy change, ΔH, for this reaction per mole of X:

$\Delta H=\frac{-1.433 kJ}{0.03048 mol}=-47.014 kJ/mol$

5 0

2 years ago

True or false A(n) rift valley forms where plates diverge on land.

motikmotik

Answer:

True

Explanation:

4 0

2 years ago

Read 2 more answers

For the following reaction, 4.21 grams of hydrogen gas are allowed to react with 10.6 grams of ethylene (C2H4) . hydrogen(g) + e

sergiy2304 [10]

Answer:a) 11.34 g of ethane $(C_2H_6)$ can be formed

b) $C_2H_4$ is the limiting reagent

c) 3.44 g of the excess reagent remains after the reaction is complete

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}$

1. $\text{Moles of} H_2=\frac{4.21}{2}=2.10moles$

2. $\text{Moles of} C_2H_4=\frac{10.6}{28}=0.378moles$

$H_2(g)+C_2H_4(g)\rightarrow C_2H_6(g)$

According to stoichiometry :

1 mole of $C_2H_4$ require 1 mole of $H_2$

Thus 0.378 moles of $C_2H_4$ will require= $\frac{1}{1}\times 0.378=0.378moles$ of $H_2$

Thus $C_2H_4$ is the limiting reagent as it limits the formation of product and $H_2$ is the excess reagent.

moles of $H_2$ left = (2.10-0.378) = 1.72 moles

mass of $H_2$ left= $moles\times {\text {Molar mass}}=1.72moles\times 2g/mol=3.44g$

According to stoichiometry :

As 1 mole of $C_2H_4$ give = 1 mole of $C_2H_6$

Thus 0.378 moles of $C_2H_4$ give = $\frac{1}{1}\times 0.378=0.378moles$ of $C_2H_6$

Mass of $C_2H_6=moles\times {\text {Molar mass}}=0.378moles\times 30g/mol=11.34g$

Thus 11.34 g of ethane is formed.

4 0

3 years ago