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3 years ago

A plant uses energy from the Sun to make food. What kind of energy transformation is this?

Chemistry

2 answers:

Norma-Jean [14]3 years ago

8 0

I'm pretty sure it's photosynthesis

aniked [119]3 years ago

8 0

It is definitely photosynthesis

Hope that helped!!!

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Can anyone do balance equation?

miss Akunina [59]

47)1-6 and 5&6-8 hope this helps!

3 0

3 years ago

What mass of solid that has a molar mass 89.0 g/mol should be added to 100.0 g of benzene to raise the boiling point of benzene

taurus [48]

On the basis of given data:

Mass of solvent (benzene) = 100 gm or 0.1 Kg

Molar mass of solute (solid) = 89 g/mol

Increase in boiling point (deltaTb) = 2.42 degree C

The boiling point elevation constant k = 2.53 C.kg/mol

There is a need to determine the mass of the solid to be added, the elevation in boiling point is proportional to the m (molality) of the solute;

ΔTb = k.m

Here, m that is molality = moles of solute/kg of solvent

Therefore,

ΔTb = k (benzene) × moles of solid/kg of benzene

2.42 = 2.53 × moles of solid/0.1

moles of solid = 0.0956 moles

The molar mass of solid = 89 g/mol

Thus, the mass of solid = 0.0956 moles × 89 g/mol

= 8.508 g

Thus, the mass of solid to be added is 8.508 g.

3 0

2 years ago

Which temperature is warmer 0° F or 0°C

Y_Kistochka [10]

Answer:

0 degree C

Explanation:

0 degree C = 32 degree F

0 degree F = -17.7778 degree C

4 0

3 years ago

Read 2 more answers

A. Based on the activation energies and frequency factors, rank the following reactions from fastest to slowest reaction rate, a

deff fn [24]

Answer:

A) $E_{a} = 350KJ/mol$ , $E_{a} = 50KJ/mol$ , $E_{a} = 50KJ/mol$

A = 1.5× $10^{-7}s^{-1}$ , A = 1.9× $10^{-7} s^{-1}$ , A=1.5× $10^{-7} s^{-1}$

B) 4.469

Explanation:

From Arrhenius equation

$K=Ae^{\frac{E_{a} }{RT} }$

where; K = Rate of constant

A = Pre exponetial factor

$E_{a}$ = Activation Energy

R = Universal constant

T = Temperature in Kelvin

Given parameters:

$E_{a} =165KJ/mol$

$T_{1}=505K$

$T_{2}=525K$

$R=8.314JK^{-1}mol^{-1}$

taking logarithm on both sides of the equation we have;

$InK=InA-\frac{E_{a} }{RT}$

since we have the rate of two different temperature the equation can be derived as:

$In(\frac{K_{2} }{K_{1} } )=\frac{E_{a} }{R}(\frac{1}{T_{1} } -\frac{1}{T_{2} } )$

$In(\frac{K_{2} }{K_{1} } )=\frac{165000J/mol}{8.314JK^{-1}mol^{-1} }.(\frac{1}{505} -\frac{1}{525} )$

$In(\frac{K_{2} }{K_{1} } )$ = 19846.04×7.544× $10^{-5}$ = 1.497

$\frac{K_{2} }{K_{1} }$ = $e^{1.497}$ = 4.469

6 0

3 years ago

The freezing point of benzene C6H6 is 5.50°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in benzene is DDT . H

Vlad1618 [11]

Answer: 0.028 grams

Explanation:

Depression in freezing point :

Formula used for lowering in freezing point is,

$\Delta T_f=k_f\times m$

or,

$\Delta T_f=k_f\times \frac{\text{ Mass of solute in g}\times 1000}{\text {Molar mass of solute}\times \text{ Mass of solvent in g}}$

where,

$\Delta T_f$ = change in freezing point

$k_f$ = freezing point constant (for benzene} = $5.12^0Ckg/mol$

m = molality

Putting in the values we get:

$0.400^0C=5.12\times \frac{\text{ Mass of solute in g}\times 1000}{354.5\times 209.0}$

${\text{ Mass of solute in g}}=0.028g$

0.028 grams of DDT (solute) must be dissolved in 209.0 grams of benzene to reduce the freezing point by 0.400°C.

4 0

3 years ago