Balance this equation and determine what the coefficients should be.

Ron and Hermione begin with 1.50 g of the hydrate copper(II)sulfate·x-hydrate (CuSO4·xH2O), where x is an integer. Part of their

iris [78.8K]

Answer:

5

Explanation:

We can obtain the value of x by doing the following:

Mass of hydrated salt (CuSO4.xH2O) = 1.50g

Mass of anhydrous salt (CuSO4) = 0.96g

Mass of water molecule(xH2O) = 1.50 — 0.96 = 0.54g

Molar Mass of CuSO4.xH2O = 63.5 + 32 + (16x4) + x(2 +16) = 63.5 + 32 + 64 + 18x = 159.5 + 18x

Mass of water(xH2O) molecules in the hydrate salt is given by:

xH2O/CuSO4.xH2O = 0.54/1.5

18x/(159.5 + 18x) = 0.36

Cross multiply to express in linear form

18x = 0.36 (159.5 + 18x)

18x = 57.42 + 6.48x

Collect like terms

18x — 6.48x = 57.42

11.52x = 57.42

Divide both side by 11.52

x = 57.42/11.52

x = 5

Therefore, the unknown integer x is 5 and the formula for the hydrated salt is CuSO4.5H2O

8 0

4 years ago

Consider the reaction to produce methanolCO(g) + 2H2 (g) <-----> CH3OHAn equilibrium mixture in a 2.00-L vessel is found t

MariettaO [177]

Answer : The value of $K_c$ of the reaction is 10.5 and the reaction is product favored.

Explanation : Given,

Moles of $CH_3OH$ at equilibrium = 0.0406 mole

Moles of $CO$ at equilibrium = 0.170 mole

Moles of $H_2$ at equilibrium = 0.302 mole

Volume of solution = 2.00 L

First we have to calculate the concentration of $CH_3OH,CO\text{ and }H_2$ at equilibrium.

$\text{Concentration of }CH_3OH=\frac{\text{Moles of }CH_3OH}{\text{Volume of solution}}=\frac{0.0406mole}{2.00L}=0.0203M$

$\text{Concentration of }CO=\frac{\text{Moles of }CO}{\text{Volume of solution}}=\frac{0.170mole}{2.00L}=0.085M$

$\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{0.302mole}{2.00L}=0.151M$

Now we have to calculate the value of equilibrium constant.

The balanced equilibrium reaction is,

$CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)$

The expression of equilibrium constant $K_c$ for the reaction will be:

$K_c=\frac{[CH_3OH]}{[CO][H_2]^2}$

Now put all the values in this expression, we get :

$K_c=\frac{(0.0203)}{(0.085)\times (0.151)^2}$

$K_c=10.5$

Therefore, the value of $K_c$ of the reaction is, 10.5

There are 3 conditions:

When $K_{c}>1$ ; the reaction is product favored.

When $K_{c}$ ; the reaction is reactant favored.

When $K_{c}=1$ ; the reaction is in equilibrium.

As the value of $K_{c}>1$ . So, the reaction is product favored.

7 0

3 years ago

Convert 2.8 pounds to kg

Pani-rosa [81]

Answer:

1.27006 kg

Explanation:

1 lb is equal to 0.907185 kg

5 0

4 years ago

Read 2 more answers

What is the molarity of a solution that has a volume of 5.90 L and contains 0.21 moles of calcium chloride?

allsm [11]

Answer:

.036 M

Explanation:

Molarity = number of moles/volume.

Molarity = .21/5.90 = .036

- Hope this helps! If youstill do not understand or need a more in-depth explanation please let me know.

4 0

3 years ago

What is the specific activity (in Bq/g) if 8.58 mg of an isotope emits 7.4x10⁴ α particles per minute?

Monica [59]

Specific activity is the activity per unit mass of a radionuclide and is a physical property of that radionuclide. Activity is a quantity (for which the SI unit is the becquerel) related to radioactivity, and is defined as the number of radioactive transformations per second that occur in a particular radionuclide.

<h3>How do you find the specific activity of an isotopes?</h3>

Activity = λN.
= (0.693/8 days) x (1/86,400 sec/day) x (3 x 1017 atoms)
= 3 x 1011 atoms/sec I-131.
= 3 x 1011 dps I-131.

<h3>What is the specific activity of Co 60?</h3>

The average specific activity of 60Co in continental aquatic sediments is in the region of 1 Bq. kg-1 dry weight (Eyrolle et al., 2008). Cobalt distribution in the abiotic compartments of continental aquatic systems is characterised by the strong affinity of the element for sediment.

Learn more about specific activity of isotopes here:

<h3>brainly.com/question/364529</h3><h3 /><h3>#SPJ4</h3>

6 0

2 years ago