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deff fn [24]
3 years ago
6

How many grams of H 2O are produced from 28.8 g of O 2? (Molar Mass of H 2O = 18.02 g) (Molar Mass of O 2=32.00 g) 4 NH 3 (g) +

7 O 2 (g) → 4 NO 2 (g) + 6 H 2O (g)
Chemistry
1 answer:
krek1111 [17]3 years ago
5 0

Answer:  13.9 g of H_2O will be produced from the given mass of oxygen

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} O_2=\frac{28.8g}{32.00g/mol}=0.900moles

The balanced chemical reaction is:

4NIO_2(g)+7O_2(g)\rightarrow 4NO_2(g)+6H_2O(g)

According to stoichiometry :

7 moles of O_2 produce =  6 moles of H_2O

Thus 0.900 moles of O_2 will produce =\frac{6}{7}\times 0.900=0.771moles  of H_2O

Mass of H_2O=moles\times {\text {Molar mass}}=0.771moles\times 18.02g/mol=13.9g

Thus 13.9 g of H_2O will be produced from the given mass of oxygen

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A sample of the mineral hematite Iron (III) oxide has a mass of 12.4g. How many moles of the mineral are present?
8090 [49]

Thus problem is providing us with the mass of iron (III) oxide as 12.4 g so the moles are required and found to be 0.0776 mol after the calculations:

<h3>Mole-mass relationships:</h3>

In chemistry, we use mole-mass relationships in order to calculate grams from moles and vice versa. In this case, since we are given the mass of iron (III) oxide as 12.4 g one can calculate the moles by firstly quantifying its molar mass:

Fe_2O_3\rightarrow 2*55.85 g/mol+3*16.00 g/mol=159.7g/mol

Then, we prepare a conversion factor in order to cancel out the grams and thus, get moles:

12.4gFe_2O_3*\frac{1molFe_2O_3}{159.7gFe_2O_3} \\\\=0.0776molFe_2O_3

Learn more about mole-mass relationships: brainly.com/question/18311376

8 0
2 years ago
Calculate the molar concentration of the Cl⁻ ions in 0.73 M MgCl2(aq).
Marta_Voda [28]
MgCl2(aq) is an ionic compound which will have the releasing of 2 Cl⁻ ions ions in water for every molecule of MgCl2 that dissolves.
MgCl2(s) --> Mg+(aq) + 2 Cl⁻(aq)
            [Cl⁻] = 0.73 mol MgCl2/1L × 2 mol Cl⁻ / 1 mol MgCl2 = 1.5 M
The answer to this question is [Cl⁻] = 1.5 M
7 0
3 years ago
Classify each element. Note that another term for main group is representative, another term for semimetal is metalloid, and the
NikAS [45]

The question is incomplete, here is the complete question:

Classify each element. Note that another term for main group is representative, another term for semi-metal is metalloid, and the inner transition metals are also called the lanthanide and actinide series.

Hf, Am, In, Ta, As, Se, Rn

<u>Answer:</u>

Hafnium and tantalum are transition elements.

Americium is a inner transition element.

Indium, Selenium and Radon are main group elements.

Arsenic is a metalloid.

<u>Explanation:</u>

Main group elements are the elements which belong to s block and p block. They are also known as representative elements.

S-block elements are defined as the elements whose last electron enters s-sub shell. The general electronic configuration of these elements is ns^{1-2}

P-block elements are defined as the elements whose last electron enters p-sub shell. The general electronic configuration of these elements is np^{1-6}

Metalloids are defined as the elements which show intermediate properties between metals and non-metals. There are 7 metalloids in the periodic table. They are: Boron, Silicon, germanium, Arsenic, Antimony, Tellurium and Polonium.

Transition elements are known as d-block elements. D block elements are defined as the elements whose last electron enters d sub shell. The general electronic configuration of these elements is [(n-1)d^{1-10}ns^{0-2}]

Inner transition elements are known as (f block) elements. (F block) elements are defined as the elements whose last electron enters (f subshell). The general electronic configuration of these elements is [(n-2)f^{1-14}(n-1)d^{0-1}ns^{2}]. They are also known as lanthanide and actinide series.

For the given elements:

  • <u>Option 1:</u> Hf

Hafnium is the 72nd element of the periodic table having electronic configuration of [Xe]4f^{14}5d^26s^2

As, the last electron is entering the d subshell, it is a transition element.

  • <u>Option 2:</u> Am

Americium is the 95th element of the periodic table having electronic configuration of [Rn]5f^{7}6d^07s^2

As, the last electron is entering the (f subshell), it is a inner transition element.

  • <u>Option 3:</u> In

Indium is the 49th element of the periodic table having electronic configuration of [Kr]5s^25p^1

As, the last electron is entering the p subshell, it is a main group element.

  • <u>Option 4:</u> Ta

Tantalum is the 73rd element of the periodic table having electronic configuration of [Xe]4f^{14}5d^56s^2

As, the last electron is entering the d subshell, it is a transition element.

  • <u>Option 5:</u> As

Arsenic is the 33rd element of the periodic table having electronic configuration of [Ar]4s^24p^3

As, the last electron is entering the p subshell, it is a main group element. It shows an intermediate property of metal and non-metal. Thus, it is a metalloid.

  • <u>Option 6:</u> Se

Selenium is the 34th element of the periodic table having electronic configuration of [Ar]4s^24p^4

As, the last electron is entering the p subshell, it is a main group element.

  • <u>Option 7:</u> Rn

Radon is the 86th element of the periodic table having electronic configuration of [Xe]4f^{14}5d^{10}6s^26p^6

As, the last electron is entering the p subshell, it is a main group element.

5 0
3 years ago
Giving brainliest and thanks to best answer &lt;333 :&gt;
valkas [14]

Answer:

C

Explanation:

radiation from the sun first warms the outer atmosphere (trophosphere)

convection(aka just heat moving through gas or liquid) brings the warmth down lower

conduction heats the ground

gl lol :))

7 0
3 years ago
Read 2 more answers
Calculate the density of carbon dioxide at STP
Artist 52 [7]

Answer:

Density = mass/volume

= 44/22.4

= 1.96 gram/liter

The density of the Carbon Dioxide at S.T.P. (Standard Temperature and Volume) is 1.96 gram/liter.

5 0
3 years ago
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