All categories

3 years ago

Having difficulty understanding formulas involved. Determine the quantity (g) of pure CaCl2 in 7.5 g of CaCl2•9H2O.

1 answer:

7 0

To determine the quantity of pure CaCl2 from the given amount of CaCl2.9H2O, we use the ratio of CaCl2 to CaCl2.9H2O. From the formula the ratio would 1 is to 1 We calculate as follows:

7.5 g CaCl2.9H2O (1 mol / 273.1215 g) ( 1 mol CaCl2 / 1 mol CaCl2.9H2O ) ( <span>110.98 g / 1 mol ) = 3.05 g CaCl2

Hope this answers the question. Have a nice day. Feel free to ask more questions.</span>

You might be interested in

What pressure (in atm) is required to contain 0.034 moles of oxygen gas in a 5.1 L container at a temperature of 25.00C

kherson [118]

Answer:

0.16 atm

Explanation:

Let's look at the formula PV=nRT. First of all, T should be in kelvin, so let's add 273. Now solve for P so we get P = nRT/V.

Now let's plug these in...

P = (0.034 mol)(0.08206 L atm mol^-1 K^-1)(298 K)/5.1 L

After solving and rounding to sig figs we get 0.16 atm

5 0

2 years ago

How many milliliters of a 2.00 M KBr are needed to prepare 250.0 mL of 0.60 M KBr?

PIT_PIT [208]

Answer:

75 ml

Explanation:

I did it on a Solution Calculator online.

Heres the link to the calculator to do it yourself if you want:

- https://www.sigmaaldrich.com/chemistry/stockroom-reagents/learning-center/technical-library/solution-dilution-calculator.html

7 0

3 years ago

Read 2 more answers

2 H2O2(aq)-> 2H20(l) + O2(g) ΔΗ -196 kJ/molrun The decompostion of H202(aq) is represented by the equation above. A student m

Aleks [24]

Answer:

the approximate pressure in the flask after 400. s = 1.2atm

Explanation:

The detailed explanation is shown below

5 0

3 years ago

Which best describes a scientist who solves a new problem using something already learned?

adelina 88 [10]

Answer:

The answer should be Skeptical

6 0

3 years ago

What is the standard potential, e∘celle∘cell, for this galvanic cell? use the given standard reduction potentials in your calcul

Olin [163]

The standard potential for the given galvanic cell is 0.477 V

<h3>What is electrode potential?</h3>

The electrode potential is the electromotive force of a galvanic cell built using a standard reference electrode and another electrode whose potential is to be found.

There are two types of electrode potential

Oxidation potential - The potential associated with oxidation reaction is known as oxidation potential

Reduction potential - The potential associated with reduction reaction is known as reduction potential

At the anode, oxidation occurs

$Sn(s)\rightarrow Sn^{2+}(aq)+2e^-$

At the cathode, reduction occurs

$Cu^{2+}(aq)+2e^-\rightarrow Cu(s)$

$E^o_{cell} =E^o_{cathode} -E^o_{anode}$

= 0.337 - (-0.140)

= 0.477 V

Thus, The standard potential for the given galvanic cell is 0.477 V

Learn more about electrode potential:

brainly.com/question/17362810

#SPJ4

Disclaimer: The question was given incomplete on the portal. Here is the complete question

Question: What is the standard potential, E∘cell, for this galvanic cell? Use the given standard reduction potentials in your calculation as appropriate.

$Sn^{2+}(aq)+2e^-\rightarrow Sn(s)$ , E°red=−0.140 V

$Cu^{2+}(aq)+2e^-\rightarrow Cu(s)$ , E°red=+0.337 V

4 0

2 years ago

Read 2 more answers