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3 years ago

Fun fact! <3

Chemistry

2 answers:

aleksklad [387]3 years ago

7 0

✧ Yes, since one of our senses are missing we won't really detect the taste.

Explanation: <em>Think about it, we usually smell food and we can already tell if it's going to taste good or not. Looking or touching food doesn't always help to know if we will like the taste. ✧</em>

victus00 [196]3 years ago

3 0

Answer:

:O Fr? Bet let me go get them lol

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lidiya [134]

The correct awser is the D/the 4th one

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3 years ago

bp’s efforts to close the blowout preventer and install a containment dome following an explosion on the deepwater horizon drill

Eduardwww [97]

The British Petroleum (bp's) effort to close the blowout preventer and install a containment dome following an explosion on the deepwater horizon drilling rig is an example of <u>regulatory policy </u>change.

The deepwater horizon drilling rig is a semi-submersible, transportable, floating, flexibly oriented drilling rig that could work in sea depths of up to 10,000 feet which is approximately 3,000 meters.

During the explosion that occurred in April 2010, British Petroleum made several efforts to contain the damages made and to prevent further outbreaks of disasters.

Part of the changes was shifting from a blowout preventer that has a specialized valve to seal, manage, and monitor oil and gas wells in order to prevent blowouts to a containment dome (a crucial component of a system meant to control an oil well's underwater blowout).

Therefore, we can conclude that the British Petroleum (bp's) effort to close the blowout preventer and install a containment dome following an explosion on the deepwater horizon drilling rig is an example of <u>regulatory policy </u>change.

Learn more about deepwater horizon drilling rig here:

brainly.com/question/21103915?referrer=searchResults

6 0

3 years ago

How many hydrogen bonds can CH3NH2 make to water?

ikadub [295]

CH3NH2 can only have as many hydrogen bonds as hydrogen bonding sites in the molecule. CH3NH2 has two N−H bonds and a lone pair of electrons on the nitrogen atom. Therefore, CH3NH2 can form three hydrogen bonds with water.

4 0

3 years ago

Calculate Delta H in KJ for the following reactions using heats of formation:

lozanna [386]

Answer:

$\Delta H\textdegree = -2856.8\;\text{kJ}$ per mole reaction.

$\Delta H\textdegree = -22.3\;\text{kJ}$ per mole reaction.

Explanation:

What is the standard enthalpy of formation $\Delta H_f\textdegree{}$ of a substance? $\Delta H_f\textdegree{}$ the enthalpy change when one mole of the substance is formed from the most stable allotrope of its elements under standard conditions.

Naturally, $\Delta H_f\textdegree{} = 0$ for the most stable allotrope of each element under standard conditions. For example, oxygen $\text{O}_2$ (not ozone $\text{O}_3$ ) is the most stable allotrope of oxygen. Also, under STP $\text{O}_2$ is a gas. Forming $\text{O}_2\;(g)$ from itself does not involve any chemical or physical change. As a result, $\Delta H_f\textdegree{} = 0$ for $\text{O}_2\;(g)$ .

Look up standard enthalpy of formation $\Delta H_f\textdegree{}$ data for the rest of the species. In case one or more values are not available from your school, here are the published ones. Note the state symbols of the compounds (water/steam $\text{H}_2\text{O}$ in particular) and the sign of the enthalpy changes.

$\text{C}_2\text{H}_6\;(g)$ : $-84.0\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{CO}_2\;(g)$ : $-393.5\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{H}_2\text{O}\;{\bf (g)}$ : $-241.8\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{PbO}\;(s)$ : $-217.9\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{PbO}_2\;(s)$ : $-276.6\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{Pb}_3\text{O}_4\;(s)$ : $-734.7\;\text{kJ}\cdot\text{mol}^{-1}$

How to calculate the enthalpy change of a reaction $\Delta H_\text{rxn}$ (or simply $\Delta H$ from enthalpies of formation?

Multiply the enthalpy of formation of each product by its coefficient in the equation.
Find the sum of these values. Label the sum $\Sigma (n\cdot \Delta_f(\text{Reactants}))$ to show that this value takes the coefficients into account.
Multiply the enthalpy of formation of each reactant by its coefficient in the equation.
Find the sum of these values. Label the sum $\Sigma (n\cdot \Delta_f(\text{Products}))$ to show that this value takes the coefficient into account.
Change = Final - Initial. So is the case with enthalpy changes. $\Delta H_\text{rxn} = \Sigma (n\cdot \Delta_f(\textbf{Products})) - \Sigma (n\cdot \Delta_f(\textbf{Reactants}))$ .

For the first reaction:

$\Sigma (n\cdot \Delta_f(\text{Reactants})) = 4\times (-393.5) + 6\times (-241.8) = -3024.8\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\Sigma (n\cdot \Delta_f(\text{Products})) = 2\times (-84.0) + 7\times 0 = -168.0\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\begin{aligned}\Delta H_\text{rxn} &= \Sigma (n\cdot \Delta_f(\textbf{Products})) - \Sigma (n\cdot \Delta_f(\textbf{Reactants}))\\ &= (-3024.8\;\text{kJ}\cdot\text{mol}^{-1}) - (-168.0\;\text{kJ}\cdot\text{mol}^{-1})\\ &= -2856.8\;\text{kJ}\cdot\text{mol}^{-1} \end{aligned}$ .