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2 years ago

which of the following is a product formed whrb NO2 and H2O react together A O2 B HNO3 C H2 D P(OH)2

Chemistry

2 answers:

marshall27 [118]2 years ago

8 0

The answer to your question is HNO3

tino4ka555 [31]2 years ago

8 0

Answer : The correct option is, (B) $HNO_3$

Explanation :

Balanced chemical reaction : It is defined as the number of atoms of individual elements present on reactant side must be equal to the product side.

The balanced chemical reaction will be,

$3NO_2+H_2O\rightarrow 2HNO_3+NO$

By the stoichiometry, we can say that 2 moles of $NO_2$ react with 1 mole of $H_2O$ to give 1 moles of $HNO_3$ and 1 mole of $NO$ .

In this reaction, $NO_2$ and $H_2O$ are the reactants and $HNO_3$ and $NO$ are the products.

Hence, the correct option is, (B) $HNO_3$

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1 year ago

Dihydrogen dioxide decomposes into water and oxygen gas. Calculate the amounts requested if 1.34 moles of dihydrogen dioxide rea

a_sh-v [17]

Answers:

a. Moles of oxygen formed: 0.670 mol

b. Moles of water formed: 1.34 mol

c. Mass of water formed: 24.1 g

d. Mass of oxygen formed: 21.4 g

Explanation:

Dihdyrogen dioxide is the chemical name for a compound made of two hydrogen atoms and two oxide atoms, i.e. H₂O₂, which is also known as hydrogen peroxide or oxygenated water.

The decomposition reaction of dihydrogen dioxide into water and oxygen gas is represented by the balanced chemical equation:

$2H_2O_2(l)\rightarrow 2H_2O(l)+O_2(g)$

The mole ratios derived from that balanced chemical equation are:

2 mol H₂O₂ : 2 mol H₂O : 1 mol O₂

a. Moles of oxygen formed

Set the proportion using the theoretical mole ratio of H₂O₂ to O₂ and the amount of moles of dyhydrogen dioxide that react:

$2\text{ mol }H_2O_2/1\text{ mol }O_2=1.34\text{ mol }H_2O_2/x$

When you solve for x, you get:

x = 1.34 mol H₂O₂ × 1 mol O₂ / 2 mol H₂O₂ = 0.670 mol O₂

b. Moles of water formed

Set the proportion using the theoretical mole ratio of H₂O₂ to H₂O and the amount of moles of dyhydrogen dioxide that react:

$2\text{ mol }H_2O_2/2\text{ mol }H_2O=1.34\text{ mol }H_2O_2/x$

When you solve for x, you get:

x = 1.34 mol H₂O₂ × 2 mol H₂O / 2 mol H₂O₂ = 1.34 mol H₂O

c. Mass of water formed

Using the number of moles of water calculated in the part b., you calculate the mass of water formed, in grams, using the molar mass of water:

Molar mass of water = 18.015 g/mol

Number of moles = mass in grams / molar mass

⇒ mass in grams = number of moles × molar mass

⇒ mass in grams = 1.34 mol × 18.015 g/mol = 24.1 g

d. Mass of oxygen formed

Using the number of moles of oxygen determined in the part a., you calculate the mass in grams using the molar mass of O₂.

Molar mass of O₂ = 32.00 g/mol

mass = molar mass × number of moles

mass = 32.00 g/mol × 0.670 mol = 21.4 g.

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2 years ago

A. Calculations for the Determination of Ammonium Chloride The data from the data entry portion of the report has been copied in

Vladimir79 [104]

Answer:

Explanation:

Considering question A

Mass of original sample is $m_o = 0.945 \ g$

Mass of NH4Cl is $m_n = 0.116 \ g$

Percent of NH4Cl is $k = 12.275 \%$

Mass of NaCl $m_k = 0.359 \ g$

Mass of SiO2 $m_e = 0.46$

Mass of original sample $m_o = 0.945 \ g$

Differences in these weights (g) (use the absolute value of the difference)

recovery of matter $G = 0.01 \ g$

The correct option is C

From the question we are told that

The mass of evaporating dish on #1 is $m_1 = 38.646 \ g$

The mass of evaporating dish and original sample $m_2 = 39 591 \ g$

The mass of evaporating dish after subliming $NH_4Cl$ is $m_3 = 39.4750 \ g$

Generally the mass of the original sample is mathematically represented as

$m_o = m_2 - m_1$

=> $m_o = 39 591 - 38.646$

=> $m_o = 0.945 \ g$

Generally the mass of $NH_4Cl$ is mathematically represented as

$m_n = m_2 - m_3$

=> $m_n = 39 591 - 39.4750$

=> $m_n = 0.116 \ g$

The Percent $NaH_4 Cl (g)$

$k = \frac{ m_n}{m_o} *100$

=> $k = \frac{0.116 }{0.945} *100$

=> $k = 12.275 \%$

Considering question B

The mass of evaporating dish #2 is $m_g = 38700\ g$

The mass of watch glass is $m_a = 28 299 \ g$

The mass of evaporating dish #2, watch glass and NaCl $m_b = 67,355 \ g$

Generally the mass of NaCl is

$m_k = m_b -[m_g + m_a]$

=> $m_k = 67,355 -[38700 + 28 299]$

=> $m_k = 0.359 \ g$

Considering question C

The mass of evaporating dish is $m_p= 38.645$

The mass of evaporating dish and SiO2 $m_s = 39.105 \ g$

Generally the mass of SiO2 is mathematically represented as

$m_e = 39.105 - 38.645$

=> $m_e = 0.46$

Considering D

The mass of the original sample is $m_o = 0.945 \ g$

Generally the experimental mass recovered (NH_4Cl,NaCl, SiO2 ) is mathematically evaluated as

$M =0.116 + 0.46 + 0.359$

$M = 0.935 \ g$

Generally the differences in these weights (g) of recovery of matter is mathematically represented as.

$G =0.945- 0.935$

=> $G = 0.01 \ g$

While drying the NaCl, the liquid boiled and some splattered out of the evaporating dish, causing the recovered mass to be lower.

6 0

2 years ago