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3 years ago

Hydrogen peroxide can decompose to water and oxygen by the following reaction

1 answer:

6 0

To begin calculating, there is one thing you need to remember : $1 mole of H2O2=34.0148 g$
Then we have $5.00g of H2O2=5/34.0148=0.146995$
As you know decomposition of 2moles now has prodused 196kj
So, q is made due $0.146995 moles of H2O2$ $=(-196/2)*0.146995=-14.40551Kj$
I'm sure it will help.

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One of the several oxides of tin found in the earthâs crust is 78.77% by mass tin. what is its empirical formula and name?

Zanzabum

To determine the empirical formula of the compound given, we need to determine the ratio of each element in the compound. To do that we assume to have 100 grams sample of the compound with the given composition. Then, we calculate for the number of moles of each element. We do as follows:
mass moles
Sn 78.77 0.6636
O 21.23 1.3269

Dividing the number of moles of each element with the smallest value, we will have the empirical formula:

moles ratio
Sn 0.6636 0.6636/0.6636 = 1
O 1.3269 1.3269/0.6636 = 2

The empirical formula would be SnO2.

3 0

4 years ago

How many grams are in 3.0 moles of KCI (potassium chloride)?

Ganezh [65]

Answer: I thinc 223.6539

Explanation:

7 0

3 years ago

Which are precautions to take when working with heat and fire? Check all that apply.

Oduvanchick [21]

Wear face mask and holy the conical flask or test tube away from your face and body and dont put it on directly on fire or it

3 0

3 years ago

A solution is prepared by dissolving ammonium sulfate in enough water to make of stock solution. A sample of this stock solution

Pani-rosa [81]

Answer: molarity of ammonium ions = 0.274mol/L

molarity of sulfate ions = 0.137mol/L

Note: The complete question is given below

A solution is prepared by dissolving 10.8 g ammonium sulfate in enough water to make 100.0 mL of stock solution. A 10.00-mL sample of this stock solution is added to 50.00 mL of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.

Explanation:

Molar concentration = no of moles/volume in liters

no of moles = mass/molar mass

mass of ammonium sulfate = 10.8g, molar mass of ammonium sulfate, (NH₄)₂SO₄ = (14+4)*2 + 32+ (16)*4 = 132g/mol

no of moles = 10.8g/132g/mol = 0.0820moles

Molarity of stock solution = 0.0820mol/(100ml/1000ml* 1L) = 0.0820mol/0.1L Molarity of stock solution = 0.820mol/L

Concentration of final solution is obtained from the dilution formula,

C1V1 = C2V2

C1 = 0.820M, V1 = 10mL, C2 = ?, V2 = 60mL

C2 = C1V1/V2

C2 = 0.820*10/60 = 0.137mol/L

molar concentration of ions = molarity of solution * no of ions

molarity of ammonium ions = 0.137mol/L * 2 = 0.274mol/L

molarity of sulfate ions = 0.137 mol/L * 1 = 0.137mol/L

4 0

4 years ago

Explain why is it better to use images on hazard labels, rather than words

enot [183]

In the case of an emergency where you might not have enough time to read several lines of writing, not to mention trying to find the hazard warnings when the whole bottle is probably covered in writing, it is much easier to locate and read universal hazard symbols.

5 0

3 years ago

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