What is The compound with the formula SiCl4

Chapter 2.4.1. The ______ energy lost by the inward movement of nickel-iron within a differentiating molten planetary body would

Gwar [14]

Answer:b. gravitational, kinetic, thermal

Explanation:

The above explains the mechanism of the core forming process on earth/planet.

It is believed that this process might has contributed significantly to a planet's early stages heating. The time when these core-forming event happened is thought to have been mainly completed very early when Earth was young . The type of this event rather than it being seen as a single catastrophic event, it is likely to have been as a result of contractions on the earth severally.

The addition of partially differentiated material from another giant impact the rate of this spasm , and it increases each time the planet's mass is to increased.

This is a little on the history of planetary evolution.

8 0

3 years ago

Identify reagents necessary to convert cyclohexane into 1,3-cylohexadiene: notice that the starting material has no leaving grou

Vinvika [58]

<span>1,3-cylohexadiene i synthesized starting from cyclohexane in following 4 steps.

1) Free Radical Substitution Rxn: Halogenation of cyclohexane in the presence of UV yield chlorocyclohexane.

2) Elimination Rxn: Dehydrohalogenation of chlorocyclohexane yields cyclohexene.

3) Halogenation of Cyclohexene (Electrophillic Addition Rxn) gives 1,2-dihalocyclohexane.

4) Elemination Rxn: When dibromocyclohexane is treated with KOH and heated it gives 1,3-cyclohexadiene as shown below,</span>

4 0

3 years ago

using the equaation 2h2+o2-->2h2o if 10.0g of hydrogen are used in the presence of excess oxygen how many grams of water will

astra-53 [7]

Answer:

90g of H2O

Explanation:

2H2 + O2 —> 2H2O

First, we calculate the molar masses of H2 And H20.

Molar Mass of H2 = 2g/mol

Mass conc of H2 from the balanced equation = 2 x 2 = 4g

Molar Mass of H2O = 2 + 16 = 18g/mol

Mass conc of H2O from the balanced equation = 2x18 = 36g

From the equation,

4g of H2 produced 36g of H2O

Therefore, 10g of H2 will be produce = (10x36)/4 = 90g of H2O

7 0

3 years ago

How many moles of NH3 can be produced from 12.0 mol of H2 and excess N2? Express your answer numerically in moles. View Availabl

VladimirAG [237]

Answer:

A) 8.00 mol NH₃

B) 137 g NH₃

C) 2.30 g H₂

D) 1.53 x 10²⁰ molecules NH₃

Explanation:

Let us consider the balanced equation:

N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)

Part A

3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:

$12.0molH_{2}.\frac{2molNH_{3}}{3molH_{2}} =8.00molNH_{3}$

Part B:

1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:

$4.04molN_{2}.\frac{2molNH_{3}}{1molN_{2}} .\frac{17.0gNH_{3}}{1molNH_{3}} =137gNH_{3}$

Part C:

According to the <em>balanced equation</em> 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:

$13.02gNH_{3}.\frac{6.00gH_{2}}{34.0gNH_{3}} =2.30gH_{2}$

Part D:

6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:

$7.62 \times 10^{-4} gH_{2}.\frac{2molNH_{3}}{6.00gH_{2}} .\frac{6.02\times 10^{23}moleculesNH_{3} }{1molNH_{3}}=1.53\times10^{20}moleculesNH_{3}$

3 0

4 years ago

If acetic acid is the only acid that vinegar contains (ka=1.8×10−5), calculate the concentration of acetic acid in the vinegar.

kicyunya [14]

Ethanoic (Acetic) acid is a weak acid and do not dissociate fully. Therefore its equilibrium state has to be considered here.

$CH_{3}COOH \ \textless \ ---\ \textgreater \ H^{+} + CH_{3}COO^{-}$

In this case pH value of the solution is necessary to calculate the concentration but it's not given here so pH = 2.88 (looked it up)

pH = 2.88 ==> $[H^{+}]$ = $10^{-2.88}$ = 0.001 $moldm^{-3}$

The change in Concentration Δ $[CH_{3}COOH]$ = 0.001 $moldm^{-3}$

  CH3COOH H+ CH3COOH
Initial   $moldm^{-3}$      $x$ 0 0

Change $moldm^{-3}$ -0.001 +0.001 +0.001

Equilibrium $moldm^{-3}$ $x$ - 0.001 0.001 0.001


Since the $k_{a}$ value is so small, the assumption
$[CH_{3}COOH]_{initial} = [CH_{3}COOH]_{equilibrium}$ can be made.

$k_{a} = [tex]= 1.8*10^{-5} = \frac{[H^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]} = \frac{0.001^{2}}{x}$

Solve for x to get the required concentration.

note: 1.)Since you need the answer in 2SF don&t round up values in the middle of the calculation like I've done here.

2.) The ICE (Initial, Change, Equilibrium) table may come in handy if you are new to problems of this kind

Hope this helps!

8 0

3 years ago