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3 years ago

The younger layers of rock in the Grand Canyon are on top of the older layers.

Chemistry

2 answers:

almond37 [142]3 years ago

8 0

Your answer is principles of superposition

laiz [17]3 years ago

4 0

The next layer is about 30 million years older and contains plant fossils in its mixed shale, limestone, and sandstone rock layers. The third to last layer is full of brachiopods, trilobites, and trilobite tunnels. You believe it is the oldest layer above surface in the Grand Canyon.

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Which functional group is found in an ester?

kherson [118]

The functional group found in an ester is a carbonyl group with an attached second oxygen atom that is bonded to a carbonyl carbon substituent by a single bond, usually a pair of alkyl or aromatic groups. In addition, it can be shown in text as RCOOR or occasionally as ROCOR, which made them responsible in the <span>distinctive and attractive odors and flavors of many flowers, perfumes, and ripe fruits. </span>

3 0

3 years ago

Read 2 more answers

Butane (C4H10) has a heat of vaporization of 22.44 kJ/mol and a normal boiling point of -0.4 ∘C. A 250 mL sealed flask contains

erastova [34]

Given that:

The heat of vaporization = 22.44 kJ/mol = 22440 J/mol
normal boiling point which is the initial temperature = 0.4° C = (273 + (-0.4))K = 272.6 K
volume = 250 mL = 0.250 L
Mass of butane = 0.8 g
the final temperature = -22° C = (273 + (-22)) K = 251 K

The first step is to determine the vapor pressure at the final temperature of 251K by using the Clausius-Clapeyron equation. This is following by using the ideal gas equation to determine the numbers of moles of butane gas. After that, the mass of butane present in the liquid is determined by using the relation for the number of moles.

Using Clausius-Clapeyron Equation:

$\mathbf{In (\dfrac{P_2}{P_1} )= -\dfrac{\Delta H_{vap}}{R}(\dfrac{1}{T_2} - \dfrac{1}{T_1})}$

where;

P1 and P2 correspond to the temperature at T1 and T2.

∴

replacing the values into the given equation, we have;

$\mathbf{In \dfrac{P_2}{1\ atm} = -\dfrac{22440 \ J/mol}{8.314 \ J/mol.K}(\dfrac{1}{251 \ K} - \dfrac{1}{272.6 \ K})}$

$\mathbf{In \dfrac{P_2}{1\ atm} =-(0.852053785)}$

$\mathbf{P_2=0.427 \ atm}$

As such, at -22° C; the vapor pressure = 0.427 atm

Now, using the ideal gas equation:

PV = nRT

where:

P = Pressure
V = volume
n = number of moles of butane
R = universal gas constant
T = temperature

∴

Making (n) the subject of the formula:

$\mathbf{n = \dfrac{PV}{RT}}$

$\mathbf{n = \dfrac{0.427 atm \times 0.250 L}{(0.08206 \ L.atm/k.mol) \times 251}}$

$\mathbf{n =0.00518 mol}$

We all know that the standard molecular weight of butane = 58.12 g/mol

∴

Using the relation for the number of moles which is:

$\mathbf{number \ of \ moles = \dfrac{mass}{molar mass}}$

mass = 0.00518 mole × 58.12 g/mol

mass = 0.301 g

∴

The mass of butane in the flask = 0.301 g

But the mass of the butane present as a liquid in the flask is

= 0.8 g - 0.301 g

= 0.499 g

In conclusion, the mass of the butane present as a liquid in the flask is 0.499 g

Learn more about vapourization here:

brainly.com/question/17039550?referrer=searchResults

7 0

3 years ago

_________ is a preparation of raw seafood marinated in an acidic mixture.

Effectus [21]

Mmm this sounds like ceviche to me! It's a Latin American dish with seafood marinated in lemon/lime, cilantro, peppers, and other good stuff!

3 0

3 years ago

Convert 0.0054 Megasec (Msec) to Hectosec (Hsec)

11111nata11111 [884]

One megasecond is equal to 10000 hectoseconds

so, if you want to put it in a ratio, it would be

1:10000
0.0054:x

cross multiply

x= 0.0054*10000

x= 54 hectoseconds

4 0

3 years ago

II. Ionic Equations

mario62 [17]

Answer:

Complete ionic: $\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}$ .

Net ionic: $\begin{aligned}& \rm Ag^{+}\, (aq) + Cl^{-}\, (aq) \to AgCl\, (s)\end{aligned}$ .

Explanation:

Start by identifying species that exist as ions. In general, such species include:

Soluble salts.
Strong acids and strong bases.

All four species in this particular question are salts. However, only three of them are generally soluble in water: $\rm AgNO_3$ , $\rm CaCl_2$ , and $\rm Ca(NO_3)_2$ . These three salts will exist as ions:

Each $\rm AgNO_3\, (aq)$ formula unit will exist as one $\rm Ag^{+}$ ion and one $\rm {NO_3}^{-}$ ion.
Each $\rm CaCl_2$ formula unit will exist as one $\rm Ca^{2+}$ ion and two $\rm Cl^{-}$ ions (note the subscript in the formula $\rm CaCl_2\!$ .)
Each $\rm Ca(NO_3)_2$ formula unit will exist as one $\rm Ca^{2+}$ and two $\rm {NO_3}^{-}$ ions.

On the other hand, $\rm AgCl$ is generally insoluble in water. This salt will not form ions.

Rewrite the original chemical equation to get the corresponding ionic equation. In this question, rewrite $\rm AgNO_3$ , $\rm CaCl_2$ , and $\rm Ca(NO_3)_2$ (three soluble salts) as the corresponding ions.

Pay attention to the coefficient of each species. For example, indeed each $\rm AgNO_3\, (aq)$ formula unit will exist as only one $\rm Ag^{+}$ ion and one $\rm {NO_3}^{-}$ ion. However, because the coefficient of $\rm AgNO_3\, (aq)\!$ in the original equation is two, $\!\rm AgNO_3\, (aq)$ alone should correspond to two $\rm Ag^{+}\!$ ions and two $\rm {NO_3}^{-}\!$ ions.

Do not rewrite the salt $\rm AgCl$ because it is insoluble.

$\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}$ .

Eliminate ions that are present on both sides of this ionic equation. In this question, such ions include one unit of $\rm Ca^{2+}$ and two units of $\rm {NO_3}^{-}$ . Doing so will give:

$\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, Cl^{-}\, (aq) \to 2\, AgCl\, (s)\end{aligned}$ .

Simplify the coefficients:

$\begin{aligned}& \rm Ag^{+}\, (aq) + Cl^{-}\, (aq) \to AgCl\, (s)\end{aligned}$ .

7 0

3 years ago