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2 years ago

T What type of bond is formed between the two nitrogen atoms in diatomic nitrogen, N2?

1 answer:

8 0

Nitrogen is a diatomic molecule in the VA family on the periodic table. Nitrogen has five valence electrons, so it needs three more valence electrons to complete its octet. A nitrogen atom can fill its octet by sharing three electrons with another nitrogen atom, forming three covalent bonds, a so-called triple bond.

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Bauxite must go through two processes to produce aluminum metal. The yield of the Bayer process, which extracts aluminum oxide f

Mashcka [7]

The yield of aluminium obtained from 1 m^3 of bauxite is 419810 g

<h3>What is a Percent yield</h3>

A percent yield of a substance measures the amount of the substance actually obtained as a percentage ratio of expected yield.

Percent yield = actual yield / expected yield × 100%

<h3>How to calculate the mass of aluminium obtained from bauxite </h3>

From the data given:

40 % of the bauxite is converted to aluminium oxide.

Volume of bauxite = 1 m^3

40 % of 1 m^3 = 0.4 m^3

volume of aluminium oxide = 0.4 m^3

density of aluminium oxide = 3965 kg/m^3

Using mass = density × volume

mass of aluminium oxide = 0.4 × 3965 kg

mass of aluminium oxide = 1586 kg

Formula of aluminium oxide is Al203

molar mass of aluminium oxide = 102 g

percentage mass of aluminium in one mole of aluminium oxide = mass of aluminium / mass of aluminium oxide × 100 %

Percentage mass of aluminium in aluminium oxide = 54/102 × 100

Percentage mass of aluminium in aluminium oxide = 52.94 %

Expected mass of aluminium from aluminium oxide = 52.94 × 1586

Expected mass of aluminium = 839.62 kg

Actual yield = 40 % × 839.62

Actual yield of aluminium = 419.81 kg

mass of aluminium in grams = 419810 g

Therefore, mass of aluminium obtained from 1 m^3 of bauxite is 419810 g

Learn more about percent yield at: brainly.com/question/8638404

3 0

2 years ago

Answer choices are A. Constant speed<br> B. Stopped<br> C.Moving sideways<br> D.Accelerating

Maurinko [17]

B.stopped
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7 0

2 years ago

What is the limiting reactant in a reaction where 10.0 mol of iron is treated with 12.0 mol of bromine? The product that forms i

hammer [34]

<u>Answer:</u> The limiting reagent in the reaction is bromine.

<u>Explanation:</u>

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

Given values:

Moles of iron = 10.0 moles

Moles of bromine = 12.0 moles

The chemical equation for the reaction of iron and bromine follows:

$2Fe+3Br_2\rightarrow 2FeBr_3$

By the stoichiometry of the reaction:

If 3 moles of bromine reacts with 2 moles of iron

So, 12.0 moles of bromine will react with = $\frac{2}{3}\times 12.0=8moles$ of iron

As the given amount of iron is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Hence, bromine is considered a limiting reagent because it limits the formation of the product.

Thus, the limiting reagent in the reaction is bromine.

3 0

2 years ago

Must show units and how they cancelli 1.) 175 km to um 3.) 385 nm to dm 5.) 492 um tom 7.) 52 x 103 dm to mm 9.) 321x 1035 mm to

morpeh [17]

Explanation:

1.) 175 km to μm

$1 km=10^9 \mu m$

$175 km=175\times 10^9\mu m=1.75\times 10^{11} \mu m$

3.) 385 nm to dm

$1 nm=10^{-8} dm$

$385 nm=385\times 10^{-8} dm=3.85\times 10^{-6} dm$

5.) 492 μm to m

1 μm = $10^{-6} m$

$492 \μm=492\times 10^{-6} m=4.92\times 10^{-4} m$

7.) $52\times 10^3$ dm to mm

1 dm = 100 mm

$52\times 10^3 dm=52\times 10^3\times 100 mm=5.2\times 10^{6}dm$

9.) $321\times 10^{35}$ mm to km

$1 mm = 10^{-6} km$

$321\times 10^{35} mm=321\times 10^{35}\times 10^{-6} km=3.21\times 10^{31} km$

11.) $456\times 10^3$ m to km

m = 0.001 km

$456\times 10^3m =456\times 10^3 m\times 0.001 km=456 km$

13.) $422\times 10^3$ m to nm

$1 m = 10^{9} nm$

$422\times 10^3 m=422\times 10^3\times 10^{9} nm=4.22\times 10^{14} nm$

15.) $4.87\times 10^{30}$ m to pm

$1 m = 10^{12} pm$

$4.87\times 10^{30} m=4.87\times 10^{30}\times 10^{12} pm=4.82\times 10^{42} pm$

17.) $5.26\times 10^3$ m to um

1 m = $10^{6} \mu m$

$5.26\times 10^3 m=5.26\times 10^3\times 10^6 \mu m=5.26\times 10^{9} \mu m$

19.) $1.25\times 10^{35}$ m to Mm

1 m = $10^{-6} Mm$

$1.25\times 10^{35} m=1.25\times 10^{35}\times 10^{-6} Mm=1.25\times 10^{-29} Mm$

21.) $4.22\times 10^3$ Tm to nm

$1 Tm = 10^{21} nm$

$4.22\times 10^3 Tm=4.22\times 10^3\times 10^{21} nm=4.22\times 10^{24} nm$

6 0

3 years ago

If a 250.0 mL sealed plastic bag starts out at a temperature of 19.0 °C and after sitting in the sun reaches a temperature of 60

Marizza181 [45]

Answer:

Explanation:

If the pressure remains constant then the temperature and Volume are all that you have to consider.

Givens

T1 = 19oC = 19 + 273 = 292o K

T2 = 60oC = 60 + 273 = 333oK

V1 = 250 mL

V2 = x

Formula

V1/T1 = V2/T2

250/292 = x/333

Solution.

The solves rather neatly. Multiplly both sides by 333

250*333 / 292 = 333 *x / 333

Do the multiplication

250 * 333 / 292 = x

83250 / 292 = x

Divide by 292

x = 285.1 mL

The answer is D

8 0

2 years ago