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3 years ago

Can vinegar decompose into other harmful chemicals

Chemistry

1 answer:

Zolol [24]3 years ago

6 0

Answer:

True/Yes

Explanation:

Hydrogen peroxide and vinegar: A component of vinegar is acetic acid, and according to Lu, this acetic acid will form a chemical called peracetic acid when mixed with hydrogen peroxide. Peracetic acid is toxic and corrosive, meaning it can damage or break down the surface it's applied to.

You might be interested in

Use the balanced chemical equation below to answer the following: How many moles of

Arada [10]

hmm, i not sure i come back later with the answer.

4 0

3 years ago

Calculate the number of moles in 1.8gram of h2o<br>

Rzqust [24]

Answer:

~.1058 Moles

Explanation:

The formula for this question is the following.

1 mole of a compound/molar mass of the compound.

First we need to find the molar mass of H2O, which is the atomic mass on the periodic table. Hydrogen is 1.01, Oxygen is 16.00. Add those together to get the molar mass of the compound and you'll get an equation that looks like this.

1 mole of H2O/ 17.01 g/mol H2O

We now know that in 1 mole of H20 there is 17.01 g.

Take 1.8g and divide it by 17.01, you get your answer.

7 0

3 years ago

Ammonium phosphate NH43PO4 is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid H3PO4 with

mestny [16]

Answer:

The weight of ammonium phosphate is 10 gm

Explanation:

The chemical reaction of ammonium phosphate formation is as follows.

$3NH_{3}+H_{3}PO_{4} \rightarrow [(NH_{4})_{3}PO_{4}]$

From the reaction one mole of phosphoric acid is equivalent to the one mole of ammonium phosphate.

So, "n" mole of phosphoric acid is equivalent to the "n" mole of ammonium phosphate.

Number of moles in 7.1 grams of phosphoric acid = $\frac{7.1}{Molarmass\,of phosphoric\,acid}$

$\frac{7.1}{98}= \frac{1}{13.8}$

<u>Formation weight of ammonium phosphate :</u>

Weight of ammonium phosphate = $= \frac{1}{13.8}\times Molarmass\,of\,\,ammonium\,phosphate$

$=\frac{149}{13.8} = 10.79gm$

Round the answer to 2 significant digits

Weight of ammonium phosphate = 10 gm

7 0

3 years ago

Explain step by step, please!

konstantin123 [22]

Answer:

10. Stacey

11. 7.86 g/cm^3 (3dp)

OR 786 kg/m^3 (SI units)

Explanation:

10. The correct exact answer for 10 is 26.169 (6.71x3.9). While Sam was correct in the precision of the number (2dp just like the given measurements), he was incorrect in the number itself. While Stacy's number was not as precise as Sam's (0dp), it is correctly rounded to the nearest whole number, therefore Stacy is correct.

11. The formula for density is mass/volume. The mass here is 264g and the volume is 33.6ml, therefore the density is 7.86 (3 sig figs because given values had 3) g/cm^3, because the mass was given in grams and 1ml = 1cm^3. Converted to SI units it is 786 kg/m^3.

Hope this helped!

4 0

3 years ago

The rate of formation of C in the reaction 2 A + B →2 C + 3 D is 2.7 mol dm−3 s −1 . State the reaction rate, and the rates of f

Zielflug [23.3K]

Answer:

Rate of reaction = $1.35mol.dm^{-3}.s^{-1}$

Rate of consumption of A = $2.7mol.dm^{-3}.s^{-1}$

Rate of consumption of B = $1.35mol.dm^{-3}.s^{-1}$

Rate of formation of D = $4.15mol.dm^{-3}.s^{-1}$

Explanation:

According to laws of mass action for the given reaction,

$Rate= -\frac{1}{2}\frac{\Delta [A]}{\Delta t}=-\frac{\Delta [B]}{\Delta t}=\frac{1}{2}\frac{\Delta [C]}{\Delta t}=\frac{1}{3}\frac{\Delta [D]}{\Delta t}$

where, $-\frac{\Delta [A]}{\Delta t}$ is rate of consumption of A, $-\frac{\Delta [B]}{\Delta t}$ is rate of consumption of B, $\frac{\Delta [C]}{\Delta t}$ is rate of formation of C and $\frac{\Delta [D]}{\Delta t}$ is rate of formation of D

Here $\frac{\Delta [C]}{\Delta t}=2.7mol.dm^{-3}.s^{-1}$

So, Rate of reaction = $(\frac{1}{2}\times 2.7mol.dm^{-3}.s^{-1})=1.35mol.dm^{-3}.s^{-1}$

Rate of formation of D = $(\frac{3}{2}\times \frac{\Delta [C]}{\Delta t})=(\frac{3}{2}\times 2.7mol.dm^{-3}.s^{-1})=4.15mol.dm^{-3}.s^{-1}$

Rate of consumption of A = $(\frac{2}{2}\times \frac{\Delta [C]}{\Delta t})=(\frac{2}{2}\times 2.7mol.dm^{-3}.s^{-1})=2.7mol.dm^{-3}.s^{-1}$

Rate of consumption of B = $(\frac{1}{2}\times \frac{\Delta [C]}{\Delta t})=(\frac{1}{2}\times 2.7mol.dm^{-3}.s^{-1})=1.35mol.dm^{-3}.s^{-1}$

3 0

3 years ago