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3 years ago

How do photosynthesis and cellular respiration compare to each other

Chemistry

1 answer:

musickatia [10]3 years ago

4 0

Answer:

Photosynthesis takes place in the chloroplasts of cells. ... While photosynthesis requires energy and produces food, cellular respiration breaks down food and releases energy. Plants perform both photosynthesis and respiration, while animals can only perform respiration.

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REALLY NEED HELP!!! Chemistry

kicyunya [14]

Answer:

C.) 2

Explanation:

The pH equation is:

pH = -log[H⁺]

In this equation, [H⁺] is the molarity of the acid. In this case, the acid is HCl. Molarity can be found using the equation:

Molarity (M) = moles / volume (L)

Since you were given moles and volume, you can find the molarity of HCl.

Molarity = moles / volume

Molarity = 0.01 moles / 1.00 L

Molarity = 0.01 M

Now, you can plug the molarity of the acid into the pH equation.

pH = -log[H⁺]

pH = -log[0.01]

pH = 2

3 0

2 years ago

How many grams are in 3.14 x 1015 molecules of CO?

Vesnalui [34]

Answer:

Explanation:

Not Many

1 mol of CO has a mass of

C = 12

O = 16

1 mol = 28 grams.

1 mol of molecules = 6.02 * 10^23

x mol of molecules = 3.14 * 10^15 Cross multiply

6.02*10^23 x = 1 * 3.14 * 10^15 Divide by 6.02*10^23

x = 3.14*10^15 / 6.02*10^23

x = 0.000000005 mols

x = 5*10^-9

1 mol of CO has a mass of 28

5*10^-9 mol of CO has a mass of x Cross Multiply

x = 5 * 10^-9 * 28

x = 1.46 * 10^-7 grams

Answer: there are 1.46 * 10-7 grams of CO if only 3.14 * 10^15 molecules are in the sample

5 0

2 years ago

Calcium hydroxide is a strong base but is not very soluble ( Ksp = 5.02 X 10-6 ). What is the pH of a saturated solution of Ca(O

Misha Larkins [42]

Answer : The pH of a saturated solution is, 12.33

Explanation : Given,

$K_{sp}$ = $5.02\times 10^{-6}$

First we have to calculate the solubility of $OH^-$ ion.

The balanced equilibrium reaction will be:

$Ca(OH)_2\rightleftharpoons Ca^{2+}+2OH^-$

Let the solubility will be, 's'.

The concentration of $Ca^{2+}$ ion = s

The concentration of $OH^-$ ion = 2s

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ca^{2+}][OH^-]^2$

Let the solubility will be, 's'

$K_{sp}=(s)\times (2s)^2$

$K_{sp}=(4s)^3$

Now put the value of $K_[sp}$ in this expression, we get the solubility.

$5.02\times 10^{-6}=(4s)^3$

$s=1.079\times 10^{-2}M$

The concentration of $Ca^{2+}$ ion = s = $1.079\times 10^{-2}M$

The concentration of $OH^-$ ion = 2s = $2\times (1.079\times 10^{-2}M)=2.158\times 10^{-2}M$

First we have to calculate the pOH.

$pOH=-\log [OH^-]$

$pOH=-\log (2.158\times 10^{-2})$

$pOH=1.67$

Now we have to calculate the pH.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-1.67=12.33$

Therefore, the pH of a saturated solution is, 12.33

7 0

3 years ago

Determine the boiling point of a 3.70 m solution of phenol in benzene. Benzene has a boiling point of 80.1°C and a boiling point

xeze [42]

Answer: The boiling point of a 3.70 m solution of phenol in benzene is $89.5^0C$

Explanation:

Elevation in boiling point:

$\Delta T_b=i\times k_b\times m$

where,

$\Delta T_b$ = change in boiling point

i= vant hoff factor = 1 (for benzene which is a non electrolyte )

$k_b$ = boiling point constant = $2.53^0C/kgmol$

m = molality = 3.70

$T_{solution}-T_{solvent}=i\times k_b\times m$

$T_{solution}-80.1^0C=1\times 2.53\times 3.70$

$T_{solution}=89.5^0C$

Thus the boiling point of a 3.70 m solution of phenol in benzene is $89.5^0C$

8 0

2 years ago

The unheated Gas in the above system has a volume of 20.0 L at a temperature of 25.0 C and a pressure of 1.00 atm. The gas is he

kipiarov [429]

Answer:

1.25 atm.

Explanation:

Step 1:

Data obtained from the question. This includes the following:

Initial volume (V1) = 20L

Initial temperature (T1) = 25°C

Initial pressure = 1 atm

Final temperature (T2) = 100°C

Final volume (V2) = constant i.e remain the same

Final pressure (P2) =?

Step 2:

Conversion of celsius temperature to Kelvin temperature. This is illustrated below:

Temperature (Kelvin) = temperature (celsius) + 273

Initial temperature (T1) = 25°C = 25°C + 273 = 298K

Final temperature (T2) = 100°C = 100°C + 273 = 373K

Step 3:

Determination of the final pressure of the gas. This is illustrated below:

Since the volume is constant, the following equation, P1/T1 = P2/T2 will be used to obtain the final pressure of gas as follow:

P1/T1 = P2/T2

Initial temperature (T1) = 298k

Initial pressure = 1 atm

Final temperature (T2) = 373K

Final pressure (P2) =?

P1/T1 = P2/T2

1/298 = P2 /373

Cross multiply to express in linear form

298 x P2 = 1 x 373

Divide both side by 298

P2 = 373/298

P2 = 1.25 atm.

Therefore, the pressure of the heated gas is 1.25 atm.

3 0

3 years ago