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3 years ago

How do you think you results would be affected if using K2Cr2O7/H2SO4 as the oxidizing agent instead of KMnO4

Chemistry

1 answer:

makkiz [27]3 years ago

3 0

Answer:

Due to weak oxidizing agent.

Explanation:

In my opinion the results would be affected if using K2Cr2O7/H2SO4 as the oxidizing agent instead of KMnO4 because K2Cr2O7/H2SO4 is weak oxidizing agent as compared to KMnO4. An oxidizing agent is a substance that has the ability to oxidize other substances means that accept their electrons so that's why the results of strong oxidizing agent is different than weak oxidizing agent.

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When adding the measurements 42.1014 g + 190.5 g, the answer has ___ significant figures.

andrezito [222]

Answer: The answer has 7 significant figures

Explanation:

The addition of 190.5 and 42.1014 will give 232.6014. Counting the digits will give 7 significant figures.

4 0

3 years ago

The concentration of copper(II) sulfate in one brand of soluble plant fertilizer is 0.070% by weight. If a 21.5 g sample of this

Semmy [17]

Answer:

The molar concentration of $Cu^{2+}$ ions in the given amount of sample is $4.73\times 10^{-5}M$

Explanation:

Given that,

Mass of sample = 21.5 g

0.07 % (m/m) of copper (II) sulfate in plant fertilizer

This means that, in 100 g of plant fertilizer, 0.07 g of copper (II) sulfate is present

So, in 20 g of plant fertilizer

, $=\frac{0.07}{100}\times 21.5}\\=0.0151g$ of copper (II) sulfate is present.

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Mass of solute (copper (II) sulfate) = 0.0151 g

Molar mass of copper (II) sulfate = 159.6 g/mol

Volume of solution = 2.0 L

$\text{Molarity of solution}=\frac{0.0151g}{159.6g/mol\times 2.0L}\\\\\text{Molarity of solution}=4.73\times 10^{-5}M$

The chemical equation for the ionization of copper (II) sulfate follows:

$CuSO_4\rightarrow Cu^{2+}+SO_4^{2-}$

1 mole of copper (II) sulfate produces 1 mole of copper (II) ions and sulfate ions

Molarity of copper (II) ions = $4.73\times 10^{-5}M$

Hence, the molar concentration of $Cu^{2+}$ ions in the given amount of sample is $4.73\times 10^{-5}M$

4 0

3 years ago

What is the mass of 3.0 x 10^23 atoms of neon

anzhelika [568]

3.0e23 atoms Ne

"E" means 10^

Then we multiply it by a mole of Ne. By the definetion of a mole, it is always 6.022e23 atoms of an element.

So now, we do this:

3.0e23 atoms Ne x (1 mol Ne / 6.022e23 atoms Ne)

After that, we use molar mass. A mole of Neon is equal, in terms of grams, to its avg. atomic mass. This goes true for any element.

It ends up like this:

3.0e23 atoms Ne x (1 mol Ne / 6.022e23 atoms Ne) x (20.1797 g Ne / 1 mol Ne)

Now cancel out the "atoms Ne" and "1 mol Ne"

You end up with a grand total of...
*plugs everything into a calculator*

10.05298... g Ne.

We need to round to 2 sig. figs. (3.0) so now it's....

10 g Ne.

Note that this method can only be used for converting atoms of an element to mass in grams.
Source(s):
A periodic table for the atomic mass of neon.
A chemistry textboook
A chemistry class.

8 0

3 years ago

You mix a blue and a clear liquid together and they turn bright green. Has a chemical reaction occurred?

Margarita [4]

Yes because color change is a sign of a chemical reaction.

7 0

3 years ago

Read 2 more answers

How many moles of CO2 are produced

Sergio039 [100]

Answer:

30.0 mol CO₂

Explanation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

To answer this problem we need to convert moles of C₃H₈ into moles of CO₂: We'll do that by using the <u>stoichiometric coefficients</u>, using a conversion factor that has C₃H₈ moles in the denominator and CO₂ moles in the numerator:

10.0 mol C₃H₈ * $\frac{3molCO_2}{1molC_3H_8}$ = 30.0 mol CO₂

6 0

3 years ago