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3 years ago

Please help me for real

Chemistry

1 answer:

Bond [772]3 years ago

7 0

Answer:

I believe it is the replication fork. so B

Explanation:

I haven't done this section in quite some time. this was biology for me, and I'm in chem right now. so I hope that's right. I hope I could help you :)

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You and a friend move a grand piano across the living room, exerting 25 newtons of force and performing 175 joules of work. It t

agasfer [191]

<h3>Answer:</h3>

Distance = 7 m

Power = 12.5 J/s or watts

<h3>Explanation:</h3>

Work done = 175 Joules

Force exerted = 25 Newtons

Work done is the product of force exerted and the distance moved by a body.

Therefore;

Work done = Force × distance

Rearranging the formula we can work out the distance;

Distance = work done ÷ Force

= 175 J ÷ 25 N

= 7 m

Thus, the grand piano was moved by 7 m across the living room .

On the other hand,

Power is the rate at which the work is done.

Therefore;

Power = Work done ÷ time

Work done = 175 Joules

Time = 14 seconds

Thus;

Power generated = 175 J ÷ 14 sec

= 12.5 J/s or watts

Hence, a power of 12.5 J/s was generated in moving the grand piano.

8 0

4 years ago

When two immiscible liquids are mixed, what physical property determines which solvent layer will be on top?

Digiron [165]

Answer:

Density of the liquid.

Explanation:

Density is the mass per unit volume of a substance.

At specific volume of each immiscible substance, the substance with lower relative molecular mass has the lower density and such floats.

Thus, the physical properties that determines the solvent layer that will be on top is DENSITY.

3 0

4 years ago

Which are homogenous mixtures? Check all that apply.

AURORKA [14]

The following are considered to be a homogeneous mixture: vinegar, sugar water and soda pop in a sealed bottle.

7 0

4 years ago

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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

3 years ago

Relationship between galactose and mannose

BartSMP [9]

Answer:

Mannose is a simple sugar or monosaccharide that is found as part of some polysaccharides in plants and in some animal glycoproteins. Galactose is converted to glucose in the liver to serve as fuel for cells in the body.

Explanation:

6 0

3 years ago

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