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3 years ago

The structure of carbon dioxide using dots and crosses

Chemistry

1 answer:

Gelneren [198K]3 years ago

4 0

Answer:

on image

Explanation:

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A student places three ice cubes in a beaker and allows them to partially melt. If she measures the temperature of the water in

Sonja [21]

<h2>Answers to the whole assignment:</h2>

**Check for proof photos at the bottom.**

**Answers are in bold.**

__________________________________________________________

A student places three ice cubes in a beaker and allows them to partially melt. If she measures the temperature of the water in the beaker, what will she most likely observe? Explain your answer.

She will see that the temperature of the water is 0 degrees Celsius and it stays at that temperature until all the ice cubes completely melt.

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Use the table on the right to calculate each required quantity. Record values to 3 significant digits.

First answer: 35.8 kJ

Second answer: 871 g

Third answer: Fe

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An iron sphere with a mass of 75.00 g is heated to a temperature of 385.0°C. It is then placed in a beaker containing 150.0 g H2O at 100.0°C.

First answer: 339 kJ

Second answer: 9.02 kJ

Third answer: No

__________________________________________________________

A sample of water has a mass of 100.0 g. Calculate the amount of heat required to change the sample from ice at –45.00°C to water vapor at 175.0°C. Relevant constant values are given to the right.

First answer: 325 kJ

Second answer: boiling the water

__________________________________________________________

Identify the correct description for each part labeled on the diagram.

A: ✔ heating solid

B: ✔ melting

C: ✔ heating liquid

D: ✔ boiling

E: ✔ heating gas

F: ✔ melting point

G: ✔ boiling point

Where are the particles closest together?

✔ A

<h2>Explanation:</h2>

When ice partially melts, the surrounding water remains at 0° Celsius. This is because ice melting only involves the water molecules overcoming intermolecular forces, or forces holding molecules together. The water molecules themselves aren't increasing in kinetic energy when turning into liquid. Since temperature measures kinetic energy, the temperature in this case, remains at 0°.

To find how much energy it takes to cause a phase change in a substance, you must find the number of moles, and determine whether to use ΔHvap and ΔHfus. ΔHvap is used to show how much energy is involved in evaporation and condensation, and ΔHfus is the same but used for melting and freezing. To find number of moles of a given substance, use the formula: mass of given substance divided by atomic mass. Both numbers have to use the same mass unit. Then, multiply number of moles by ΔHvap or ΔHfus to get kJ.

In a heating curve graph, the flat lines represent no change in energy, or the phase changes. The steep lines are change in energy, which are either heating or cooling. The bottom line is a heating/cooling solid. A heating/cooling solid is the coldest phase, which is why the line is at the bottom.

Here are photos of Edge to make things easier.

7 0

2 years ago

What is discriminate?

vodka [1.7K]

Answer:

Explanation:

To make a distinction in favor of or against a person or thing on basic or group, class or category, to which the person or thing(s) belong

7 0

3 years ago

The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute th

MrRissso [65]

The question is incomplete, complete question is :

The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25°C.

$\frac{K_1}{K_2}=?$

Activation energy of the reaction 1 , $Ea_1$ = 14.0 kJ/mol

Activation energy of the reaction 2, $Ea_1$ = 11.9 kJ/mol

Answer:

0.4284 is the ratio of the rate constants.

Explanation:

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

The expression used with catalyst and without catalyst is,

$\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}$

$\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}$

where,

$K_2$ = rate constant reaction -1

$K_1$ = rate constant reaction -2

Activation energy of the reaction 1 , $Ea_1$ = 14.0 kJ/mol = 14,000 J

Activation energy of the reaction 2, $Ea_1$ = 11.9 kJ/mol = 11,900 J

R = gas constant = 8.314 J/ mol K

T = temperature = $25^oC=273+25=298 K$

Now put all the given values in this formula, we get

$\frac{K_1}{K_2}=e^{\frac{11,900- 14,000Jl}{8.314 J/mol K\times 298 K}}=2.3340$

0.4284 is the ratio of the rate constants.

7 0

3 years ago

I AM GIVING BRAINLIEST PLEASEE HELPPPP I NEED HELPPPPPP PLEAEEEEEE

True [87]

Answer:

0.479 M or mol/L

Explanation:

So Molarity is moles/litres of solution...often written as M=mol/L

So here we are given grams of BaCl2 which we have to convert to moles. To convert to moles of BaCl2 we have to divide 63.2 g BaCl2 by molar mass of BaCl2 which is 208.23 g/mol so you get 63.2/208.23 = 0.3035 moles of BaCl2

Second step is converting the 634mL to litres by simply dividing by 1000 because we know 1 litre has 1000ml so 634/1000 = 0.634L

Now we just plug these guys in our molarity formula M=mol/L

M= 0.3035/0.634 = 0.479 M or mol/L

3 0

3 years ago

Read 2 more answers

A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2

a_sh-v [17]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

<u>Explanation:</u>

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 $CO_2=13.00g$

Mass of $H_2O=2.662g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

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

So, in 13.00 g of carbon dioxide, $\frac{12}{44}\times 13.00=3.54g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

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

So, in 2.662 g of water, $\frac{2}{18}\times 2.662=0.296g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

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

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles$

<u>Step 2:</u> 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 0.295 moles.

For Carbon = $\frac{0.295}{0.295}=1$

For Hydrogen = $\frac{0.296}{0.295}=1$

For Oxygen = $\frac{0.603}{0.295}=2.044\approx 2$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is $CHO_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

$n=\frac{90.04g/mol}{45g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4$

Hence, the empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

3 0

3 years ago

The structure of carbon dioxide using dots and crosses​

The structure of carbon dioxide using dots and crosses