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3 years ago

Which is one way that Dalton's atomic theory has been shown to be incorrect?

Chemistry

1 answer:

KATRIN_1 [288]3 years ago

6 0

The answer is D. His belief is was that atoms could not be split. That is what was disproved really fast ;)

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Given the partial equation: VO2++ Mn2+ → VO2++ MnO4−, balance the reaction in acidic solution using the half-reaction method and

bezimeni [28]

Answer:

Mn^2+(aq) + 4H2O(l) + 5[VO2]^+(aq) + 10H^+(aq) ---------->MnO4^-(aq) + 8H^+(aq) + 5[VO]^2+(aq) + 5H2O(l)

Explanation:

Oxidation half equation:

Mn^2+(aq) + 4H2O(l) ------------> MnO4^-(aq) + 8H^+(aq) + 5e

Reduction half equation:

5[VO2]^+(aq) + 10H^+(aq) + 5e --------> 5[VO]^2+(aq) + 5H2O(l)

Overall redox reaction equation:

Mn^2+(aq) + 4H2O(l) + 5[VO2]^+(aq) + 10H^+(aq) ---------->MnO4^-(aq) + 8H^+(aq) + 5[VO]^2+(aq) + 5H2O(l)

6 0

3 years ago

When 1.2383 g of an organic iron compound containing Fe, C, H, and O was burned in O2, 2.3162 g of CO2 and 0.66285 g of H2O were

uysha [10]

Answer:

$\boxed{\text{C$_{15}$H$_{21}$FeO$_{6}$}}$

Explanation:

Let's call the unknown compound X.

1. Calculate the mass of each element in 1.23383 g of X.

(a) Mass of C

$\text{Mass of C} = \text{2.3162 g } \text{CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{0.632 07 g C}$

(b) Mass of H

$\text{Mass of H} = \text{0.66285 g }\text{H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g } \text{{H$_{2}$O}}} = \text{0.074 157 g H}$

(c)Mass of Fe

(i)In 0.4131g of X

$\text{Mass of Fe} = \text{0.093 33 g Fe$_{2}$O$_{3}$}\times \dfrac{\text{111.69 g Fe}}{\text{159.69 g g Fe$_{2}$O$_{3}$}} = \text{0.065 277 g Fe}$

(ii) In 1.2383 g of X

$\text{Mass of Fe} = \text{0.065277 g Fe}\times \dfrac{1.2383}{0.4131} = \text{0.195 67 g Fe}$

(d)Mass of O

Mass of O = 1.2383 - 0.632 07 - 0.074 157 - 0.195 67 = 0.336 40 g

2. Calculate the moles of each element

$\text{Moles of C = 0.63207 g C}\times\dfrac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.052 629 mol C}\\\\\text{Moles of H = 0.074157 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.073 568 mol H}\\\\\text{Moles of Fe = 0.195 67 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.845 g Fe}} = \text{0.003 5038 mol Fe}\\\\\text{Moles of O = 0.336 40} \times \dfrac{\text{1 mol O}}{\text{16.00 g O}} = \text{0.021 025 mol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{0.052629}{0.003 5038}= 15.021\\\\\text{H: } \dfrac{0.073568}{0.003 5038} = 20.997\\\\\text{Fe: } \dfrac{0.003 5038}{0.003 5038} = 1\\\\\text{O: } \dfrac{0.021025}{0.003 5038} = 6.0006$

4. Round the ratios to the nearest integer

C:H:O:Fe = 15:21:1:6

5. Write the empirical formula

$\text{The empirical formula is } \boxed{\textbf{C$_{15}$H$_{21}$FeO$_{6}$}}$

5 0

3 years ago

If an electrostatically charged object is placed near other objects, which of the following will occur? Select all that apply. I

Alja [10]

Answer:

As the electrostatically charged object is to be placed in the field of charged particles it will be attracted to those who would be of oppositely charged and repelled by the same charged particles. phenomenon of like charges repel and opposite charges attract each other will be carried out and no deflection will be shown by the charge towards the neutral charge.

5 0

2 years ago

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natulia [17]

Answer:

purrrrr cat

Explanation:

7 0

2 years ago

I’m a paragraph of no less than 125 words describe the steps you can take to effectively organize your writing

Tanya [424]