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3 years ago

A solution of pH 3 was mixed with a solution of pH 9 what The final pH could be?

Chemistry

1 answer:

Gekata [30.6K]3 years ago

5 0

Answer:

Explanation:

can't be higher or lower than what you started with

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A perfect cube of aluminum metal was found to weigh 20.00 g. The density of aluminum is 2.7 g/ml. What are the dimensions of the

denis23 [38]

Answer:

Height = 1.9493 cm

Width = 1.9493 cm

Depth = 1.9493 cm

Solution:

Data Given:

Mass = 20 g

Density = 2.7 g/mL

Step 1: Calculate the Volume,

As,

Density = Mass ÷ Volume

Or,

Volume = Mass ÷ Density

Putting values,

Volume = 20 g ÷ 2.7 g/mL

Volume = 7.407 mL or 7.407 cm³

Step 2: Calculate Dimensions of the Cube:

As we know,

Volume = length × width × depth

So, we will take the cube root of 7.407 cm³ which is 1.9493 cm.

Hence,

Volume = 1.9493 cm × 1.9493 cm × 1.9493 cm

Volume = 7.407 cm³

4 0

3 years ago

Number of electrons for all of these even the ones with a number by them ? Will give brainlist!!!!

Bumek [7]

Explanation:

s=2

p=6

d=10

f=14

pls mark me brainlist

7 0

2 years ago

Which of the following is a chemical change?

frutty [35]

C. rusting is the correct answer

3 0

3 years ago

The meaning of atoms

REY [17]

Answer:

They make up all matter.

4 0

3 years ago

Read 2 more answers

Five kilograms of liquid carbon tetrachloride undergo a mechanically reversible, isobaric change of state at 1 bar during which

Drupady [299]

Answer:

Explanation:

From the information given:

Mass of carbon tetrachloride = 5 kg

Pressure = 1 bar

The given density for carbon tetrachloride = 1590 kg/m³

The specific heat of carbon tetrachloride = 0.84 kJ/kg K

From the composition, the initial volume of carbon tetrachloride will be: $= \dfrac{5 \ kg }{1590 \ kg/m^3}$

= 0.0031 m³

Suppose $\beta$ is independent of temperature while pressure is constant;

Then:

The change in volume can be expressed as:

$\int ^{V_2}_{V_1} \dfrac{dV}{V} =\int ^{T_2}_{T_1} \beta dT$

$In ( \dfrac{V_2}{V_1}) = \beta (T_2-T_1)$

$V_2 = V_1 \times exp (\beta (T_2-T_1))$

$V_2 = 0.0031 \ m^3 \times exp (1.2 \times 10^{-3} \times 20)$

$V_2 = 0.003175 \ m^3$

However; the workdone = -PdV

$W = -1.01 \times 10^5 \ Pa \times ( 0.003175 m^3 - 0.0031 \ m^3)$

W = - 7.6 J

The heat energy Q = Δ h

$Q = mC_p(T_2-T_1)$

$Q = 5 kg \times 0.84 \ kJ/kg^0 C \times 20$

Q = 84 kJ

The internal energy is calculated by using the 1st law of thermodynamics; which can be expressed as;

ΔU = ΔQ + W

ΔU = 84 kJ + ( -7.6 × 10⁻³ kJ)

ΔU = 83.992 kJ

3 0

3 years ago