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3 years ago

What is the thinnest type of crust?

Chemistry

1 answer:

Trava [24]3 years ago

8 0

The correct answer is oceanic crust, 80 km, Hope this helps let me know.

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How does the compound shown in part 5a of the transparency differ from the elements that comprise it?

Tresset [83]

<span>A compound is ''composed'' of elements. The periodic table is made up of elements. Atoms makes up elements and elements when reacted together make compounds. Na+ and Cl- makes NACL....salt. a compound</span>

5 0

3 years ago

A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

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

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{180.6}{84}=2$

Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

3 0

3 years ago

This is agree or disagree I need help ASAP!!

PIT_PIT [208]

All of them are agree :)

8 0

3 years ago

Read 2 more answers

How much heat in kj is released by burning 9.5 grams of methane?

tatyana61 [14]

Methane is the compound CH4, and burning it uses the reaction:

CH4 + O2 -> CO2 + H2O, which is rather exothermic. To find the heat released by burning a certain amount of the substance, you should look at the bond enthalpy of each compound, and then compare the values before and after the reaction. In methane, there are 4 C-H bonds, which have bond energy of 416 kj/mol, resulting in a total bond energy of 1664 kj/mol. O2 is 494 kj/mol. Therefore we have a total of 2080 kj/mol on the left side. On the right side we have CO2, which has 2 C=O bonds, each at 799 kj/mol each, resulting in 1598 kj/mol, and H2O has 2 O-H bonds, at 459kj/mol each, resulting in a total of 2516 kj/mol on the right hand side. Now, this may be confusing because the left hand side seems to have less heat than the right, but you just need to remember: making minus breaking, which results in a total change of 436kj/mol heat evolved.

Now it is a simple matter of find the mols of CH4 reacted, using n=m/mr.

n = 9.5/16.042 = 0.592195 mol

Therefore, if we reacted 0.592195 mol, and we produced 436 kj for one mol, the total amount of energy evolved was 436*<span>0.592195 kj, or 258.197 kj.</span>

7 0

3 years ago

The equilibrium constant K changes with changes in<br>the temperature.

chubhunter [2.5K]

<h3>Answer:</h3><h2>Equilibrium constants are changed if you change the temperature of the system. Kc or Kp are constant at constant temperature, but they vary as the temperature changes. You can see that as the temperature increases, the value of Kp falls.</h2>

4 0

2 years ago