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3 years ago

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Chemistry

1 answer:

Wittaler [7]3 years ago

8 0

Answer:

Here is a set of possible answers:

By how much can CO2 emissions be reduced?

How much is it going to cost the city to install the wind farms?

Will families have to pay for part of the wind farms?

What is the out-of-pocket cost of installing solar panels on a house after the assistance has been granted?

Are there any yearly operating costs to use solar panels?

How reliable is solar energy in this area? In an average year, how many months can solar energy be used?

Do we have a backup source of energy if the wind farm or solar panels don’t generate enough electricity?

Explanation:

This is the actucal answer so you might want to change it up

You might be interested in

What is the molar mass of magnesium

mina [271]

Hey there!

The molar mass of magnesium is 24.305 grams.

The molar mass of an element is the same as the atomic mass of an element, but measured in grams.

It's on the periodic table underneath the element symbol.

Hope this helps!

3 0

3 years ago

EXPLAIN why a skateboard coasting on a flat surface slows down and comes to a stop

ankoles [38]

A skateboard coasting on a flat surface slows down and then comes to a stop because the skateboard runs out of energy, and slows down, also, sense it is on a flat surface and not going downhill, it goes slower.

6 0

3 years ago

According to the following reaction, how many grams of sulfur are formed when 37.4 g of water are formed?

julia-pushkina [17]

Answer:

Mass = 100.8 g

Explanation:

Given data:

Mass of sulfur formed = ?

Mass of water formed = 37.4 g

Solution:

Chemical equation:

2H₂S + SO₂ → 3S + 2H₂O

Number of moles of water:

Number of moles = mass/molar mass

Number of moles = 37.4 g/ 18 g/mol

Number of moles = 2.1 mol

Now we will compare the moles of water and sulfur.

H₂O : S

2 : 3

2.1 : 3/2×2.1 = 3.15

Mass of sulfur:

Mass = number of moles × molar mass

Mass = 3.15 mol × 32 g/mol

Mass = 100.8 g

8 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:

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

Try these steps for the second reaction:

$\Delta H_\text{rxn} = -22.3\;\text{kJ}\cdot\text{mol}^{-1}$ .

6 0

4 years ago

6. Give reasons.

MArishka [77]

Answeryou understand but id ont if your know the answer plz help me

Explanation:

3 0

3 years ago