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3 years ago

Which type of bond is formed by two atoms that equally share one pair of electrons? (1 point) ionic covalent metallic un?

Chemistry

2 answers:

babymother [125]3 years ago

7 0

The answer is covalent (these are exta words)

GenaCL600 [577]3 years ago

4 0

It's B Covalent.Now I'll add more words since the thing says I need to write at least 20 characters.

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A compound of a transition metal and iodine is 56.7% metal by mass.How many grams of the metal can be obtained from 630 g of thi

erma4kov [3.2K]

Answer:

357 g of the transition metal are present in 630 grams of the compound of the transition metal and iodine

Explanation:

In any sample of the compound, the percentage by mass of the transition metal is 56.7%. This means that for a 100 g sample of the compound, 56.7 g is the metal while the remaining mass, 43.3 g is iodine.

Given mass of sample compound = 630 g

Calculating the mass of iodine present involves multiplying the percentage by mass composition of the metal by the mass of the given sample;

56.7 % = 56.7/100 = 0.567

Mass of transition metal = 0.567 * 630 = 357.21 g

Therefore, the mass of the transition metal present in 630 g of the compound is approximately 357 g

4 0

3 years ago

List four greenhouse gases in the atmosphere. For each gas, describe its prevalence in the atmosphere, its natural sources, its

lakkis [162]

Answer: Carbon Dioxide, Methane, Nitrous Oxide and Fuorinated Gases

Explanation:Carbon dioxide (CO2): Carbon dioxide enters the atmosphere through burning fossil fuels (coal, natural gas, and oil), solid waste, trees and other biological materials, and also as a result of certain chemical reactions (e.g., manufacture of cement). Carbon dioxide is removed from the atmosphere (or "sequestered") when it is absorbed by plants as part of the biological carbon cycle.

Methane (CH4): Methane is emitted during the production and transport of coal, natural gas, and oil. Methane emissions also result from livestock and other agricultural practices and by the decay of organic waste in municipal solid waste landfills.

Nitrous oxide (N2O): Nitrous oxide is emitted during agricultural and industrial activities, combustion of fossil fuels and solid waste, as well as during treatment of wastewater.

Fluorinated gases: Hydrofluorocarbons, perfluorocarbons, sulfur hexafluoride, and nitrogen trifluoride are synthetic, powerful greenhouse gases that are emitted from a variety of industrial processes. Fluorinated gases are sometimes used as substitutes for stratospheric ozone-depleting substances (e.g., chlorofluorocarbons, hydrochlorofluorocarbons, and halons). These gases are typically emitted in smaller quantities, but because they are potent greenhouse gases, they are sometimes referred to as High Global Warming Potential gases ("High GWP gases").

5 0

3 years ago

The conversion of 37 ˚C to K

77julia77 [94]

Answer:

310.15

Explanation:

c + 273.15 = k

37+ 273.15 = 310.15

8 0

3 years ago

Read 2 more answers

When FeC13 is ignited in an atmosphere of pure oxygen, this reaction takes place. 4FeCl3(sJ 30lgJ ~ 2F~0 (sJ 6Cl2(gJ If 3.00 mol

oksano4ka [1.4K]

Answer : The reagent present in excess and remains unreacted is, $O_2$

Solution : Given,

Moles of $FeCl_3$ = 3.00 mole

Moles of $O_2$ = 2.00 mole

Excess reagent : It is defined as the reactants not completely used up in the reaction.

Limiting reagent : It is defined as the reactants completely used up in the reaction.

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2FeCl_3(s)+O_2(g)\rightarrow 2FeO(s)+3Cl_2(g)$

From the balanced reaction we conclude that

As, 2 moles of $FeCl_3$ react with 1 mole of $O_2$

So, 3.00 moles of $FeCl_3$ react with $\frac{3.00}{2}=1.5$ moles of $O_2$

From this we conclude that, $O_2$ is an excess reagent because the given moles are greater than the required moles and $FeCl_3$ is a limiting reagent and it limits the formation of product.

Hence, the reagent present in excess and remains unreacted is, $O_2$

4 0

3 years ago

1, 2 and 3 ........

Orlov [11]

2.) Average atomic mass =Σ (abundance x molar mass) /100 = (12.64 x 302.04 + 18.23 x 304.12 + 69.13 x 305.03) /100 =304.486 u(Dalton) . This is avg atomic mass.

8 0

3 years ago