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2 years ago

Which two geographic features form as a result of continental-continetal convergent plates?

Chemistry

1 answer:

Kaylis [27]2 years ago

5 0

Answer:

Deep ocean trenches, volcanoes, island arcs, submarine mountain ranges, and fault lines are examples of features that can form along plate tectonic boundaries. Volcanoes are one kind of feature that forms along convergent plate boundaries, where two tectonic plates collide and one moves beneath the other.

Explanation:

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A hydrocarbon contains 92.3 grams of carbon and the rest is hydrogen. Calculate its empirical formula and molecular formula. (Mr

mote1985 [20]

To find the Empirical Formula and Empirical Formula:

First note down the percentages of the elements in the given compound.

Here, as the compound is a hydrocarbon, we know it involves Carbon and Hydrogen bonded to each other.

Carbon's atomic mass = 12

Hydrogen's atomic mass = 1.008

The compound contains 92.3g of Carbon. It implies it has;

mass of hydrogen = 100 - 92.3 = 7.7g

To find the no.of gram atoms of an element we divide the percentage mass by atomic mass of the said element.

no. of gram atoms of Carbon = 92.3/12 = 7.6916 gram atoms of Carbon

Similarly for Hydrogen,

no. of gram atoms of hydrogen = 7.7/1.008 = 7.63 gram atoms of Hydrogen

Divide all the value of gram atoms by the smallest value among them.

Thus, to determine ratios:

7.69/7.63 = 1.008

Carbon and Hydrogen are in 1:1 ratio.

The ratio between number of atoms gives us the empirical formula. Hence, empirical formula of given compound is: CH.

Molecular mass of a compound = n(empirical formula mass)

Empirical Formula mass = (12 * 1 ) + (1.008 * 1) = 13.008

Molecular mass is given in the question. Therefore, the equation becomes:

78 = n(13.008)

n = 78/13.008

n = 5.99 ≅ 6

Therefore, the molecular formula of the given hydrocarbon is: C₆ H₆

3 0

3 years ago

How is the periodic table useful today?

S_A_V [24]

Answer:

The periodic table is useful today for finding out all the elements that exists on Earth. The elements there can be used to write down chemical formulas, calculate things like molar mass/atomic mass of each element, atomic number for each, number of valence electrons each element has, the oxidation number for each, etc. Lastly, it can be used to predict the properties of elements yet to be discovered.

Explanation:

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3 0

3 years ago

Read 2 more answers

G Calculate the mass, in grams, of 225.0 atoms of cadmium, Cd (1 mol of Cd has a mass of 112.41 g).

serious [3.7K]

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2 years ago

A gas contains 75.0 wt% methane, 10.0% ethane, 5.0% ethylene, and the balance water. (a) Calculate the molar composition of this

NeTakaya

Answer:

a) molar composition of this gas on both a wet and a dry basis are

5.76 moles and 5.20 moles respectively.

Ratio of moles of water to the moles of dry gas =0.108 moles

b) Total air required = 68.51 kmoles/h

So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.

Explanation:

Let assume we have 100 g of mixture of gas:

Given that :

Mass of methane =75 g

Mass of ethane = 10 g

Mass of ethylene = 5 g

∴ Mass of the balanced water: 100 g - (75 g + 10 g + 5 g)

Their molar composition can be calculated as follows:

Molar mass of methane $CH_4}= 16 g/mol$

Molar mass of ethane $C_2H_6= 30 g/mol$

Molar mass of ethylene $C_2H_4 = 28 g/mol$

Molar mass of water $H_2O=18g/mol$

number of moles = $\frac{mass}{molar mass}$

Their molar composition can be calculated as follows:

$n_{CH_4}= \frac{75}{16}$

$n_{CH_4}=$ 4.69 moles

$n_{C_2H_6} = \frac{10}{30}$

$n_{C_2H_6} =$ 0.33 moles

$n_{C_2H_4} = \frac{5}{28}$

$n_{C_2H_4} =$ 0.18 moles

$n_{H_2O}= \frac{10}{18}$

$n_{H_2O}=$ 0.56 moles

Total moles of gases for wet basis = (4.69 + 0.33 + 0.18 + 0.56) moles

= 5.76 moles

Total moles of gas for dry basis = (5.76 - 0.56)moles

= 5.20 moles

Ratio of moles of water to the moles of dry gas = $\frac{n_{H_2O}}{n_{drygas}}$

= $\frac{0.56}{5.2}$

= 0.108 moles

b) If 100 kg/h of this fuel is burned with 30% excess air(combustion); then we have the following equations:

$CH_4 + 2O_2_{(g)} ------> CO_2_{(g)} +2H_2O$

4.69 2× 4.69

moles moles

$C_2H_6+ \frac{7}{2}O_2_{(g)} ------> 2CO_2_{(g)} + 3H_2O$

0.33 3.5 × 0.33

moles moles

$C_2H_4+3O_2_{(g)} ----->2CO_2+2H_2O$

0.18 3× 0.18

moles moles

Mass flow rate = 100 kg/h

Their Molar Flow rate is as follows;

$CH_4 = 4.69 k moles/h\\C_2H_6 = 0.33 k moles/h\\C_2H_4=0.18kmoles/h$

Total moles of $O_2$ required = (2 × 4.69) + (3.5 × 0.33) + (3 × 0.18) k moles

= 11.075 k moles.

In 1 mole air = 0.21 moles $O_2$

Thus, moles of air required = $\frac{1}{0.21}*11.075$

= 52.7 k mole

30% excess air = 0.3 × 52.7 k moles

= 15.81 k moles

Total air required = (52.7 + 15.81 ) k moles/h

= 68.51 k moles/h

So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.

5 0

3 years ago

Calculate the number of calories needed to increase the temperature of 50.0 g of copper metal from 21.0 degrees C to 75.0 degree

KonstantinChe [14]

<h3>Answer:</h3>

1031.4 Calories.

<h3>Explanation:</h3>

We are given;

Mass of the copper metal = 50.0 g

Initial temperature = 21.0 °C

Final temperature, = 75°C

Change in temperature = 54°C

Specific heat capacity of copper = 0.382 Cal/g°C

We are required to calculate the amount of heat in calories required to raise the temperature of the copper metal;

Quantity of heat is given by the formula,

Q = Mass × specific heat capacity × change in temperature

= 50.0 g × 0.382 Cal/g°C × 54 °C

= 1031.4 Calories

Thus, the amount of heat energy required is 1031.4 Calories.

4 0

3 years ago