Ask question

Ask question

All categories

3 years ago

5

Two students were in Chemistry Class completing an assigned lab experiment. The two students were to mix 5.34g CaCl2 with CuSO4

to make a 6.3L solution
. Complete and balance the reaction with the correct Products:
CaCl2 + CuSO4  ______ + ______

1 answer:

skelet666 [1.2K]3 years ago

3 0

Answer:

CaCl₂ + CuSO₄ → CaSO₄ + CuCl₂.

Explanation:

It is a replacement reaction, that cations and anions are replaced with each other.

So, it gives CaSO₄ and CuCl₂.

To balance the reaction, we apply the law of conversation of mass, that the no. of atoms in the reactants side is equal to that in the products side.

So, the balanced reaction is:

<em>CaCl₂ + CuSO₄ → CaSO₄ + CuCl₂.</em>

You might be interested in

∆G° for the process benzene (l) benzene (g) is 3.7 Kj/mol at 60 °C , calculate the vapor pressure of benzene at 60 °C [R=0.0821

Alex787 [66]

Answer:

4) 0.26 atm

Explanation:

In the process:

Benzene(l) → Benzene(g)

ΔG° for this process is:

ΔG° = -RT ln Q

<em>Where Q = P(Benzene(g)) / P°benzene(l) P° = 1atm</em>

ΔG° = 3700J/mol = -8.314J/molK * (60°C + 273.15) ln P(benzene) / 1atm

1.336 = ln P(benzene) / 1atm

0.26atm = P(benzene)

Right answer is:

<h3>4) 0.26 atm </h3><h3 />

6 0

3 years ago

Of the reactions involved in the photodecomposition of ozone (shown below), which are photochemical? 1. o2 (g) + hν → o (g) + o

Taya2010 [7]

Answer:

The answers are...

Explanation:

2 and 4.

8 0

3 years ago

Calculate the percent of each component in the mixture. Show your calculations. Circle final answers.

Colt1911 [192]

Answer:

See Explanation

Explanation:

The question is incomplete; as the mixtures are not given.

However, I'll give a general explanation on how to go about it and I'll also give an example.

The percentage of a component in a mixture is calculated as:

$\%C_E = \frac{E}{T} * 100\%$

Where

E = Amount of element/component

T = Amount of all elements/components

Take for instance:

In $(Ca(OH)_2)$

The amount of all elements is: (i.e formula mass of $(Ca(OH)_2)$ )

$T = 1 * Ca + 2 * H + 2 * O$

$T = 1 * 40 + 2 * 1 + 2 * 16$

$T = 74$

The amount of calcium is: (i.e formula mass of calcium)

$E = 1 * Ca$

$E = 1 * 40$

$E = 40$

So, the percentage component of calcium is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{40}{74} * 100\%$

$\%C_E = \frac{4000}{74}\%$

$\%C_E = 54.05\%$

The amount of hydrogen is:

$E = 2 * H$

$E = 2 * 1$

$E = 2$

So, the percentage component of hydrogen is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{2}{74} * 100\%$

$\%C_E = \frac{200}{74}\%$

$\%C_E = 2.70\%$

Similarly, for oxygen:

The amount of oxygen is:

$E = 2 * O$

$E = 2 * 16$

$E = 32$

So, the percentage component of oxygen is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{32}{74} * 100\%$

$\%C_E = \frac{3200}{74}\%$

$\%C_E = 43.24\%$

5 0

3 years ago

what is the predicted order of fist ionization energies from highest to lowest for beryllium, calcium, magnesium, and stronium?

MrRa [10]

Lowest to highest:
Ca
Mg
Be
S

4 0

3 years ago

Read 2 more answers

The head loss in a turbulent flow in a pipe varies Approximant as square of velocity • Direct as the velocity • Invers as square

Eddi Din [679]

Answer:

Head loss in turbulent flow is varying as square of velocity.

Explanation:

As we know that head loss in turbulent flow given as

$h_F=\dfrac{FLV^2}{2gD}$

Where

F is the friction factor.

L is the length of pipe

V is the flow velocity

D is the diameter of pipe.

So from above equation we can say that

$h_F\alpha V^2$

It means that head loss in turbulent flow is varying as square of velocity.

We know that loss in flow are of two types

1.Major loss :Due to surface property of pipe

2.Minor loss :Due to change in momentum of fluid.

5 0

3 years ago

Read 2 more answers