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2 years ago

The image is a model of sulfuric acid that has a chemical formula of H2SO4.

Chemistry

1 answer:

gladu [14]2 years ago

4 0

Models help us to understand the properties of molecules and help us to predict the kind of reactions they undergo.

A molecular model is a three dimensional representation of a molecule. Usually, molecular models could either be;

Ball and stick models or
Space filling models.

In either case, the model gives us a vivid representation of the molecule and enables us to predict its properties and reactivity based on the the inter-connectivity of atoms in the structure as shown in the model. The model of H2SO4 as shown gives us an idea of its properties and its reactivity.

Learn more: brainly.com/question/14072179

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An object has a mass of 2800g and a volume of 3cm3. What is its density?<br>

jeka57 [31]

Answer:

933.33 g/cm³

Explanation:

Density = mass / volume

Density = 2800/3 = 933.3 g/cm³.

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3 years ago

Hydrates are formed with _______ compounds

Triss [41]

Answer:

ionic compounds

Explanation:

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3 years ago

A metallic conductor moving at a constant speed in a magnetic field may develop a voltage across it. this is an example of _____

marysya [2.9K]

A metallic conductor moving at a constant speed in a magnetic field may develop a voltage across it. This is an example of Motional emf
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3 0

3 years ago

(a) What is the total volume (in L) of gaseous products, measured at 350°C and 735 torr, when an automobile engine burns 100. g

Anarel [89]

Answer:

Part A

The volume of the gaseous product is $V = 787L$

Part B

The volume of the the engine’s gaseous exhaust is $V_e = 2178 \ L$

Explanation:

Part A

From the question we are told that

The temperature is $T = 350^oC = 350 +273 =623K$

The pressure is $P = 735 \ torr = \frac{735}{760} = 0.967\ atm$

The of $C_8 H_{18} = 100.0g$

The chemical equation for this combustion is

$2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}$

The number of moles of $C_8 H_{18}$ that reacted is mathematically represented as

$n = \frac{mass \ of \ C_8H_{18} }{Molar \ mass \ of C_8H_{18} }$

The molar mass of $C_8 H_{18}$ is constant value which is

$M = 114.23 \ g/mole$

So $n = \frac{100 }{114.23} }$

$n = 0.8754 \ moles$

The gaseous product in the reaction is $CO_2_{(g)}$ and water vapour

Now from the reaction

2 moles of $C_8 H_{18}$ will react with 25 moles of $O_2$ to give (16 + 18) moles of $CO_2_{(g)}$ and $H_2 O_{(g)}$

1 mole of $C_8 H_{18}$ will react with 12.5 moles of $O_2$ to give 17 moles of $CO_2_{(g)}$ and $H_2 O_{(g)}$

This implies that

0.8754 moles of $C_8 H_{18}$ will react with (12.5 * 0.8754 ) moles of $O_2$ to give (17 * 0.8754) of $CO_2_{(g)}$ and $H_2 O_{(g)}$

So the no of moles of gaseous product is

$N_g = 17 * 0.8754$

$N_g = 14.88 \ moles$

From the ideal gas law

$PV = N_gRT$

making V the subject

$V = \frac{N_gRT}{P}$

Where R is the gas constant with a value $R = 0.08206 \ L\cdot atm /K \cdot mole$

Substituting values

$V = \frac{14.88* 0.08206 *623}{0.967}$

$V = 787L$

Part B

From the reaction the number of moles of oxygen that reacted is

$N_o = 0.8754 * 12.5$

$N_o = 10.94 \ moles$

The volume is

$V_o = \frac{10.94 * 0.08206 *623}{0.967}$

$V_o = 579 \ L$

No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as

$V_e = V_o * \frac{0.79}{0.21}$

Substituting values

$V_e = 579 * \frac{0.79}{0.21}$

$V_e = 2178 \ L$

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3 years ago

An atom x contains 14 protons, 13 electrons, and 12 neutrons, and atom y contains 14 protons, 14 electrons, and 12 neutrons. wha

Vika [28.1K]

<span>a. x and y are atoms of the same element.

If both atoms contain the same amount of protons, they are always the same element.
</span>

7 0

3 years ago

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