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3 years ago

What are some common mistakes students make with decomposition reactions?

1 answer:

6 0

I think because students get confused or forget which elements are diatomic
A decomposition usually is simple and straight forward but many students just don't know which elements should be diatomic and which shouldn't.

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URGENT!! Can someone please explain this to me?

djverab [1.8K]

Answer:

$CH _{4}+ 2O _{2} → CO _{2} + 2H _{2}O$

for the balanced equation

$from \: the \: equation \\ 1 \: mole \: of \: methane \: gives \: 1 \: mole \: of \: carbon \: dioxide \\ 7.4 \: moles \: of \: carbon \: dioxide \: will \: be \: given \: by \: (7.4 \times 1)moles \: of \: methane \\ = 7.4\: moles \: of \: methane \\ \\ since \: moles \: of \: oxygen \: double \: those \: of \: methane \\ moles \: of \: oxygen \: = 7.4\times 2 \\ = 14.8 \: moles \: of \: oxygen$

4 0

3 years ago

Dimensional analysis, Plz hell I don’t understand and plz explain

Anastasy [175]

Explanation:

30 lb is 480 ounces

34 mi/second is 54.718 kilometre/ second

455 lb/ gal is 54521.024 grams / litre

50 cl is 500 millilitres

55nm is 5.5 × 10^-6 centimetre

7 0

3 years ago

. Consider the following half-reactions:

Vedmedyk [2.9K]

d. Fe(s) and Al(s)

<h3>Further explanation</h3>

In the redox reaction, it is also known

Reducing agents are substances that experience oxidation

Oxidizing agents are substances that experience reduction

The metal activity series is expressed in voltaic series

The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent

The more to the right, the metal is less reactive (harder to release electrons) and the stronger oxidizing agent

So that the metal located on the left can push the metal on the right in the redox reaction

The electrodes which are easier to reduce than hydrogen (H), have E cells = +

The electrodes which are easier to oxidize than hydrogen have a sign E cell = -

So the above metals or metal ions will reduce Pb²⁺ (aq) will be located to the left of the Pb in the voltaic series or which have a more negative E cell value (greater reduction power)

The metal : d. Fe(s) and Al(s)

7 0

3 years ago

Changes in the composition of the atmosphere have caused gradual changes in earth's _______ throughout history, causing changes

bearhunter [10]

climate

Changes in the composition of the atmosphere have caused gradual changes in earth's <u>climate</u> throughout history, causing changes in plant and animal life that contributed to mass extinctions.

The following are some of the reasons:

UV light
climate
pollutants
hydrofluorocarbons

heat

The surface of the Earth warms up as sunlight strikes it.
Surface-emitted infrared light is absorbed in the atmosphere and transformed into heat.
The temperature close to the surface rises as a result of this heat being trapped in the atmosphere.

<h3>UV light:</h3>

indirect impacts of climate change on UV radiation from the surface.
By changing the concentrations of ozone, UV-absorbing tropospheric gases, aerosols, and clouds in the atmosphere, climate change may have indirectly affected UV radiation levels in the past.
These influences are probably going to persist in the future.

<h3>climate:</h3>

People are at risk from food and water shortages, greater flooding, high heat, an increase in disease, and economic loss due to climate change.
Conflict and human migration are potential outcomes.
Climate change is the top hazard to world health in the twenty-first century, according to the World Health Organization (WHO).

<h3>pollutants:</h3>

these are also resulting in the increase of temperature of the Earth and is also damaging ozone layer.

To learn more about the changes in earth visit:

brainly.com/question/13434833?

#SPJ4

6 0

1 year ago

3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 15

shutvik [7]

Answer:

C₅H₁₀O₅

Explanation:

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

$\text{Mass of C} = \text{5.13 g CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{1.400 g C}$

(b) Mass of H

$\text{Mass of H} = \text{2.10 g H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.2349 g H}$

(c) Mass of O

Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

$\text{Moles of C = 1400 mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{116.6}{116.6}= 1\\\\\text{H: } \dfrac{233.1}{116.6} = 1.999\\\\\text{O: } \dfrac{116}{116.6} = 1.00$

4. Round the ratios to the nearest integer

C:H:O = 1:2:1

5. Write the empirical formula

The empirical formula is CH₂O.

6. Calculate the molecular formula.

EF Mass = (12.01 + 2.016 + 16.00) u = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

$n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{150 u}}{\text{30.03 u}} = 5.00 \approx 5$

MF = (CH₂O)₅ = C₅H₁₀O₅

The molecular formula of X is C₅H₁₀O₅.

8 0

4 years ago