Obtaining crude oil for refining is an example of

How many elements in C6H12O6

zysi [14]

Answer:

3

Explanation:

there are 3 different elements in sugar. just add up all the letters you see

8 0

3 years ago

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Which example best illustrates a method to show differences among Earth’s layers?

hoa [83]

Answer:

Carson models how the continental crust varies in thickness. Marisol records atmospheric and oceanic temperatures at several beaches. Eliza analyzes seismic wave activity from an earthquake using a computer model.

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3 years ago

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In some reactions, Be behaves like a typical alkaline earth metal; in others, it does not. Complete and balance the following: (

mina [271]

The correct answer is BeCl_2(l)+2Cl^-(solvated)→BeCl_4^2-.

Evaluating be behavior to see :

how it differs from the other Group 2A (2) members.

In this reaction Be behaves like other alkaline earth metals

The complete equation can be given as

BeCl_2(l)+2Cl^-(solvated)→BeCl_4^2-

BeCl_2 tends to form a chloro bridged dimer in the vapour state, however at high temperatures of the order of 1200K, this dimer dissociates into the linear monomer.

BeCl_2 has a chain structure in its solid form. Each Be atom in this structure is surrounded by chlorine atoms, two of which are connected by conversion bonds and the remaining two by covalent coordinate connections. This chain structure is displayed.

To know more about BeCl₂(I) + Cl⁻ refer the link:

brainly.com/question/5017059

#SPJ4

4 0

2 years ago

Calculate the missing variables in each experiment below using Avogadro’s law.

blagie [28]

Answer:

The answer to your question is: letter c

Explanation:

Data

V1 = 612 ml n1 = 9.11 mol

V2 = 123 ml n2 = ?

Formula

$\frac{V1}{n1} = \frac{V2}{n2}$

$n2 = \frac{n1V2}{V1}$

$n2 = \frac{(9.11)((123)}{(612)}$

n2 = 1.83 mol

5 0

3 years ago

How many liters of water vapor can be produced if 108 grams of methane gas (CH4) are combusted at 312 K and 0.98 atm? Show all w

MA_775_DIABLO [31]

Answer:

Approximately 352 liters of water vapor will be produced.

Assumption: the water vapor here behaves like an ideal gas.

Explanation:

Relative atomic mass data from a modern periodic table:

Carbon: 12.011;
Hydrogen: 1.008.

Burning methane $\rm CH_4$ in excess oxygen will produce carbon dioxide and water. The balanced equation will be

$\rm CH_4\; (g) + 2\; O_2\;(g) \to CO_2\;(g) + 2\; H_2O\;(g)$ .

Hint: start by assign "1" to methane while balancing this equation.

How many moles of molecules in 108 gram of methane?

Molar mass of methane:

$M(\rm CH_4) = 12.011 + 4\time 1.008 =16.043\;g\cdot mol^{-1}$ .

$\displaystyle n(\rm CH_4) = \frac{m}{M} = \frac{108}{16.043} = 6.73191\;mol$ .

How many moles of water molecules will be produced?

Consider the ratio between the coefficient in front of water and that in front of methane in the equation:

$\displaystyle \frac{n(\mathrm{H_2O})}{n(\mathrm{CH_4})} = 2$ .

$\displaystyle n(\mathrm{H_2O}) = n(\mathrm{CH_4})\cdot \frac{n(\mathrm{H_2O})}{n(\mathrm{CH_4})} = \rm 2\times 6.73191\;mol = 13.4638\;mol$ .

Assume that water vapor here behaves like an ideal gas.

Ideal gas constant in liters and $\rm atm$ :

$R \rm \approx 0.0820573\;L\cdot atm \cdot K^{-1}\cdot mol^{-1}$ . (NIST.)

Make sure that the temperature here is in degrees Kelvins.

$\displaystyle \begin{aligned}V &= \frac{n\cdot R\cdot T}{P} \\ &= \rm \frac{13.4638\;mol\times 0.0820573\;L\cdot atm \cdot K^{-1}\cdot mol^{-1}\times 312\; K}{0.98\; atm}\\ &\rm \approx 352\; L\end{aligned}$ .

4 0

4 years ago