All categories

2 years ago

Calculate the mass of 2.50 mol of CH,OH(1). Show your work. Use the appropriate

Chemistry

1 answer:

Sophie [7]2 years ago

7 0

Answer:

80.1 grams

Explanation:

Find the molar mass of CH3OH first by using the periodic table values.

12.011 g/mol C + (1.008*3 g/mol H) + 15.999g/mol O + 1.008 g/mol H

=32.042 so that is the molar mass

Now that you have 2.50 moles of CH3OH, you can calculate the mass in g

2.50molCH3OH * (32.042g CH3OH / 1 mol CH3OH) = 80.105

32.042g / 1 mol is the same as 32.042 g/mol

Since there are 3 sig figs in the problem (2.50 has 3 sig figs), you round to 80.1 g CH3OH

You might be interested in

CH4 and CO₂ are the chemical formulas for:

nadya68 [22]

Answer:

CH4 = Methane

CO2 = Carbon dioxide

<h3>Hope this Helps!!</h3><h3>Hope this Helps!!:)</h3>

6 0

3 years ago

Lighting a match changes ch and energy into heat.

otez555 [7]

Answer:

The answer remains the same. The total amount of energy stays the same because the 1st Law of Thermodynamics states that energy can neither be created nor destroyed, it can only change forms. So the chemical energy is just being converted into heat and light.

Explanation:

hope this helps...

7 0

2 years ago

The solubility of silver(I)phosphate at a given temperature is 1.02 g/L. Calculate the Ksp at this temperature. After you get yo

Snezhnost [94]

<u>Answer:</u> The solubility product of silver (I) phosphate is $9.57\times 10^{-10}$

<u>Explanation:</u>

We are given:

Solubility of silver (I) phosphate = 1.02 g/L

To convert it into molar solubility, we divide the given solubility by the molar mass of silver (I) phosphate:

Molar mass of silver (I) phosphate = 418.6 g/mol

$\text{Molar solubility of silver (I) phosphate}=\frac{1.02g/L}{418.6g/mol}=2.44\times 10^{-3}mol/L$

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The chemical equation for the ionization of silver (I) phosphate follows:

$Ag_3PO_4(aq.)\rightleftharpoons 3Ag^{+}(aq.)+PO_4^{3-}(aq.)$

3s s

The expression of $K_{sp}$ for above equation follows:

$K_{sp}=(3s)^3\times s$

We are given:

$s=2.44\times 10^{-3}M$

Putting values in above expression, we get:

$K_{sp}=(3\times 2.44\times 10^{-3})^3\times (2.44\times 10^{-3})\\\\K_{sp}=9.57\times 10^{-10}$

Hence, the solubility product of silver (I) phosphate is $9.57\times 10^{-10}$

4 0

2 years ago

Explain what the terms "Pyranose" and "Furanose" represent

Illusion [34]

Answer:

hhh

Explanation:

Pyranose is collective term for the saccharides which have chemical structure which includes six-membered ring that consists of one oxygen atom and five carbon atoms.

Furanose is collective term for carbohydrates which have chemical structure which includes five-membered ring system that consists of one oxygen atom and four carbon atoms.

Glucose exists both in pyranose and furanose form. It's structure is shown in image.

5 0

2 years ago

23<br> How many moles are in 500g of Calcium Nitrate? 10 M

Harman [31]

500 grams Calcium Nitrate to mol= 3.04715

6 0

3 years ago