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3 years ago

Wind transports large rocks by _____.

1 answer:

8 0

Wind transports large rocks by rock erosion. Erosion should be distinguished from weathering. Erosion is the movement of rocks from one place to other location due to external factors like wind, water or gravity. Weathering, on the other hand, is the breaking of rock into smaller pieces by processes.

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nika2105 [10]

Answer:

Explanation:k

4 0

3 years ago

Read 2 more answers

What is the name of this hydrocarbon

slamgirl [31]

Answer:

the IUPAC name of the following compound is 2,3 dimethyl pentane!

5 0

3 years ago

You have 28.5 g of iron shot that has a volume of 3.60 mL. From this information, calculate the density of iron.

laila [671]

Answer:

$\boxed {\boxed {\sf \rho \approx 7.917 \ g/mL}}$

Explanation

Density can be found by dividing the mass by the volume.

$\rho=\frac{m}{v}$

We know the iron sheet has a mass of 28.5 grams and the volume is 3.60 milliliters.

$m= 28.5 \ g \\d= 3.60 \ mL$

Substitute the values into the formula.

$\rho=\frac{28.5 \ g}{ 3.60 \ mL}$

Divide.

$\rho = 7.91666667 \ g/mL$

Let's round to the nearest thousandth.

The 6 in the ten-thousandth tells us to round the 6 to a 7.

$\rho \approx 7.917 \ g/mL$

The density of iron is about 7.917 grams per milliliter.

8 0

2 years ago

For the reaction 3h2(g) + n2(g) 2nh3(g), kc = 9.0 at 350°c. calculate g° at 350°c.

miss Akunina [59]

When ΔG° is the change in Gibbs free energy

So according to ΔG° formula:

ΔG° = - R*T*(㏑K)

here when K = [NH3]^2/[N2][H2]^3 = Kc

and Kc = 9

and when T is the temperature in Kelvin = 350 + 273 = 623 K

and R is the universal gas constant = 8.314 1/mol.K

So by substitution in ΔG° formula:

∴ ΔG° = - 8.314 1/ mol.K * 623 K *㏑(9)

= - 4536

4 0

3 years ago

The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of

gavmur [86]

Answer: The empirical formula for the given compound is $C_3H_6O$

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:

Conversion factor: 1 g = 1000 mg

Mass of $CO_2=6.32mg=0.00632g$

Mass of $H_2O=2.58g=0.00258g$

Mass of compound = 2.78 mg = 0.00278 g

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

So, in 0.00632 g of carbon dioxide, $\frac{12}{44}\times 0.00632=0.00172g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 0.00258 g of water, $\frac{2}{18}\times 0.00258=0.000286g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.00278) - (0.00172 + 0.000286) = 0.000774 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.00172g}{12g/mole}=1.43\times 10^{-4}moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.000286g}{1g/mole}=2.86\times 10^{-4}moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.000774g}{16g/mole}=4.83\times 10^{-5}moles$

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 $4.83\times 10^{-5}mol$

For Carbon = $\frac{1.43\times 10^{-4}}{4.83\times 10^{-5}}=2.96\approx 3$

For Hydrogen = $\frac{2.86\times 10^{-4}}{4.83\times 10^{-5}}=5.92\approx 6$

For Oxygen = $\frac{4.83\times 10^{-5}}{4.83\times 10^{-5}}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 6 : 1

Hence, the empirical formula for the given compound is $C_3H_{6}O_1=C_3H_6O$

3 0

3 years ago