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3 years ago

Science 20 Chapter 2 Quiz

Chemistry

1 answer:

vichka [17]3 years ago

3 0

Answer:

The starting reactant in a single displacement reaction is a pure element such as solid zinc metal as well as an aqueous compound. Aqueous implies that the compound is water-soluble. The reaction will result in products that include a different pure element as well as a new aqueous solution.

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Combustion analysis of 0.600 g of an unknown compound containing carbon, hydrogen, and oxygen produced 1.043 g of CO2 and 0.5670

Aliun [14]

Answer : The empirical formula of the compound is $C_3H_8O_2$

Explanation :

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=1.043g$

Mass of $H_2O=0.5670g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 1.043 g of carbon dioxide, $\frac{12}{44}\times 1.043=0.284g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.5670 g of water, $\frac{2}{18}\times 0.5670=0.063g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.600) - (0.284 + 0.063) = 0.253 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.284g}{12g/mole}=0.0237moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.063g}{1g/mole}=0.063moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.253g}{16g/mole}=0.0158moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0158 moles.

For Carbon = $\frac{0.0237}{0.0158}=1.5$

For Hydrogen = $\frac{0.063}{0.0158}=3.98\approx 4$

For Oxygen = $\frac{0.0158}{0.0158}=1$

The ratio of C : H : O = 1.5 : 4 : 1

To make in a whole number we are multiplying the ratio by 2, we get:

The ratio of C : H : O = 3 : 8 : 2

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 8 : 2

Hence, the empirical formula for the given compound is $C_3H_8O_2$

6 0

3 years ago

g Reduction involves the A) loss of neutrons, gain of electrons, and an increase in oxidation state. B) loss of neutrons. C) inc

artcher [175]

Answer:

E. Gain of electrons

Explanation:

A reduction reaction is one part of the two concurrent reactions that take place in a redox (reduction-oxidation) reaction.

During reduction, an atom gains electrons from a donor atom, and it's oxidation number becomes smaller.

Option A is wrong because reduction does not increase oxidation state nor are neutrons involved

Option B is wrong because reduction is not a nuclear reaction (does not involve the nucleons)

Option C is wrong because reduction leads to reduction in oxidation state

Option D is wrong leads to a reduction in oxidation state when electrons are gained

Option E is correct because reduction involves gain of electrons

4 0

3 years ago

Complete combustion of 4.40 g of a hydrocarbon produced 14.1 g of co2 and 5.04 g of h2o. what is the empirical formula for the h

andrew11 [14]

For a hydrocarbon, the combustion reactions are the following:

C + O₂ --> CO₂
H₂ + 1/2 O₂ --> H₂O

The molar mass of CO₂ is 44 g/mol while C is 12 g/mol. Let's solve for amount of C in hydrocarbon.

Mass of C = (14.1 g CO₂)(1mol/44g)(1 mol C/1 mol CO₂)(12 g/mol) = 3.845 g C
So, that means that the mass of hydrogen is:
Mass of H = 4.4 - 3.845 = 0.555 g

Moles C = 3.845/12 = 0.32042
Moles H = 0.555/1 = 0.555
Divide both by the smaller value, 0.32042.

C: 0.32042/0.32042 = 1
H: 0.555/0.32042 = 1.732
We have to get an answer that is closest to a whole number. Let's try multiplying both with 4.
C: 1*4 = 4
H: 1.732*4 = 6.93≈7

<em>Thus, the empirical formula is C₄H₇.</em>

3 0

3 years ago

The compound chloral hydrate, known in detective stories as knockout drops, is composed of 14.52% C, 1.83% H, 64.30% Cl, and 19.

maria [59]

Answer:

C₂H₃Cl₃O₂

Explanation:

Given parameters:

Percentage composition:

14.52% C

1.83% H

64.30% Cl

19.35% O

Molar mass = 165.4 g/mo

Unknown:

Empirical formula = ?

Solution:

The empirical formula of a compound is the simplest formula of a compound. To solve this problem, let us follow the process below:

C H Cl O

%composition 14.52 1.83 64.3 19.35

Molar mass 12 1 17 16

number of

moles 14.52/12 1.83/1 64.3/3.5 19.35/16

1.21 1.83 1.81 1.21

Divide by

the smallest 1.21/1.21 1.83/1.21 1.81/1.21 1.21/1.21

1 1.51 1.51 1

multiply

by 2 2 3 3 2

Empirical formula C₂H₃Cl₃O₂

5 0

4 years ago

The movement of Earth's tectonic plates is related to A. seasonal changes in weather systems. B. cycles of sunspots. C. movement

Ksenya-84 [330]

c) the movement of the materials in the mantle

8 0

3 years ago