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tino4ka555 [31]

2 years ago

10

Which of the following would be most useful in trying to obtain procedural information to replicate an experiment previously pub

lished?

1 answer:

Jet001 [13]2 years ago

7 0

Answer: Peer-reviewed journal article is the most useful because the information in them had been carefully scrutinized and aproved by people who are experts in that particular field.

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Which of the following is a product formed when Ag2O decomposes?

ASHA 777 [7]

The product formed when Ag2O decompose is O2 ( answer c)

explanation

Decomposition involve separation of a single chemical compound into two ore more elemental parts Ag2O decomposes to yield Ag and O2

that is 2Ag2O→ 4Ag + O2

2 moles of Ag2O decomposes to yield 4 moles of Ag and 1 mole of O2

6 0

3 years ago

Read 2 more answers

Which of the following is an example of an element? a brass b oxygen c rust d water

Anna [14]

Answer:Water because it is an element

8 0

2 years ago

Read 2 more answers

A certain alcohol contains only three elements, carbon, hydrogen, and oxygen. Combustion of a 60.00 gram sample of the alcohol p

marin [14]

Answer:

$C_2H_6O$

Explanation:

The first step is the <u>calculation of the moles</u> of $H_2O$ and $CO_2$ , so:

$114.6~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=2.6~mol~of~CO_2$

$70.44~g~H_2O\frac{1~mol~H_2O}{18~g~H_2O}=~3.9~mol~H_2O$

Now, in 1 mol of CO2 we have 1 mol of C and in 1 mol of $H_2O$ we have 1 mol of H. Additionally, if we want to calculate the moles of oxygen we need to <u>calculate the grams of C and O</u> and then do the <u>substraction</u> form the initial amount, so:

$2.6~mol~CO_2\frac{1~mol~C}{1~mol~CO_2}\frac{12~g~C}{1~mol~C}=31.25~g~of~C$

$3.9~mol~H_2O\frac{2~mol~H}{1~mol~H_2O}\frac{1~g~H}{1~mol~H}=7.82~g~of~H$

$Total~grams=~31.25~+~7.82=39.08~g$

$grams~of~O=60.00~g-~39.08~g=20.92~g~of~O$

Now we can <u>convert the grams</u> of O to moles, so:

$20.92~g~of~O\frac{1~mol~O}{16~g~O}=1.30~mol~O$

The next step is to divide all the mol values by the <u>smallest one</u>:

$O=\frac{1.30~mol~O}{1.30~mol~O}=~1$

$C=\frac{2.6~mol~C}{1.30~mol~O}=~2$

$H=\frac{7.82~mol~H}{1.30~mol~O}=6$

Therefore the formula is $C_2H_6O$

6 0

3 years ago

nikitadnepr [17]

Answer:

$a)\ 2FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3\\b)\ Decomposition\ Reaction$

Explanation:

<em>Ferrous Sulphate</em> $(FeSO4)$ <em> is generally found as Lime-Green Crystals. On heating, these crystals almost immediately turn white-yellow. They then, break down to produce an anhydrous mixture of Sulphur Trioxide </em> $(SO_3)$ <em>, Sulphur Dioxide </em> $(SO_2)$ <em> as well as Ferric Oxide </em> $(Fe_2O_3)$ <em>.</em>

<em>We can hence, frame a skeletal equation of this reaction and try to balance it.</em>

<em>Hence,</em>

$FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3$

<em>Now,</em>

<em>a)In order to balance it through the 'Hit &Trial Method', we'll follow a series of </em><em>steps</em><em>:</em>

<em>1. First, lets compare the number of Fe (Iron) atoms on the RHS and LHS. We find that, the no. of Fe Atoms on the RHS is twice the number of Fe Atoms on the LHS. We hence, add a co-effecient 2 beside </em> $FeSO_4$ .

<em>2. Now, Iron atoms, Sulphur Atoms and Oxygen atoms occur 2, 2, 8 respectively on both the sides:</em>

<em> Hence, As all the other elements as well as iron, balance, we've arrived upon our Balanced Equation :</em>

<em> </em> $2FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3$

<em>b) We know that, decomposition reactions are [generally] endothermic reactions in which Large Compounds </em><em>decompose </em><em>into smaller elements and compounds. Here, as Ferrous Sulphate </em><em>decomposes </em><em>into Sulphur Dioxide, Sulphur Trioxide and Ferric Oxide, the reaction that occurs here is </em><em>Decomposition Reaction.</em>

7 0

2 years ago

The lithosphere, hydrosphere, and ______________ are the abiotic parts of the planet.

VladimirAG [237]

The lithosphere, hydrosphere, and atmosphere are the abiotic parts of the planet.

8 0

2 years ago