Name (3) ways an animal can use energy from food?

We're given the [OH⁻] as 8.34 × 10⁻¹² M. Using the formula pOH = -log[OH⁻], the pOH of this solution would be -log(8.34 × 10⁻¹²) ≈ 11.08.

The pOH is, for lack of a better term, the "opposite" of pH: A pOH of 7 is neutral; a pOH less than 7 is basic; and a pOH greater than 7 is acidic.

This follows from the relation, pH + pOH = 14. In this case, with a pOH of 11.08, our pH would be 14 - 11.08 = 2.92, which is acidic (pH < 7).

Thus, the correct answer choice is B.

Answer:

1

because a dwarf star will seemlarge because of the in ability of any human being to see d sun

Answer: I think that is 55

Explanation: I hope I help

Answer: The concentration of [tex]H_2SO_4[/tex] is 0.234 M

Explanation:

According to the neutralization law,

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1[/tex] = basicity [tex]H_2SO_4[/tex] = 2

[tex]M_1[/tex] = molarity of [tex]H_2SO_4[/tex] solution = ?

[tex]V_1[/tex] = volume of  [tex]H_2SO_4[/tex] solution = 50.0 ml

[tex]n_2[/tex] = acidity of [tex]NaOH[/tex] = 1

[tex]M_1[/tex] = molarity of [tex]NaOH[/tex] solution = 0.375 M

[tex]V_1[/tex] = volume of  [tex]NaOH[/tex] solution = 62.5 ml

Putting in the values we get:

[tex]2\times M_1\times 50.0=1\times 0.375\times 62.5[/tex]

[tex]M_1=0.234M[/tex]

Therefore concentration of [tex]H_2SO_4[/tex] is 0.234 M

Answer:

0.20 g/L

Explanation:

Step 1: Calculate the molar mass of the gas (M)

At T = 350 °C = 623 K and P = 1 atm (we will assume this data), the density (ρ) of the gas is 0.087 g/L. We can calculate the molar mass of the gas using the following expression.

ρ = P × M/R × T

M = ρ × R × T/P

M = 0.087 g/L × (0.0821 atm.L/mol.K) × 623 K/1 atm = 4.5 g/mol

Step 2: Calculate the density of the gas at STP

At standard temperature (T = 273.15 K) and standard pressure (P = 1 atm), the density of the gas is:

ρ = P × M/R × T

ρ = 1 atm × 4.5 g/mol /(0.0821 atm.L/mol.K) × 273.15 K = 0.20 g/L

Answer:

The question wants you to determine the energy that the incoming photon must have in order to allow the electron that absorbs it to jump from

n

i

=

2

to

n

f

=

6

.

A good starting point here will be to calculate the energy of the photon emitted when the electron falls from

n

i

=

6

to

n

f

=

2

by using the Rydberg equation.

1

λ

=

R

⋅

(

1

n

2

f

−

1

n

2

i

)

Here

λ

si the wavelength of the emittted photon

R

is the Rydberg constant, equal to

1.097

⋅

10

7

m

−

1

Plug in your values to find

1

λ

=

1.097

⋅

10

7

.

m

−

1

⋅

(

1

2

2

−

1

6

2

)

1

λ

=

2.4378

⋅

10

6

.

m

−

1

This means that you have

λ

=

4.10

⋅

10

−

7

.

m

So, you know that when an electron falls from

n

i

=

6

to

n

f

=

2

, a photon of wavelength

410 nm

is emitted. This implies that in order for the electron to jump from

n

i

=

2

to

n

f

=

6

, it must absorb a photon of the same wavelength.

Answer:

a. pH = 2.04

b. pH = 3.85

c. pH = 8.06

d. pH = 11.56

Explanation:

The nitrous acid is a weak acid (Ka = 5.6x10⁻⁴) that reacts with NaOH as follows:

HNO₂ + NaOH → NaNO₂(aq) + H₂O(l)

a. At the beginning there is just a solution of 0.12M HNO₂. As Ka is:

Ka = [H⁺] [NO₂⁻] / [HNO₂]

Where [H⁺] and [NO₂⁻] ions comes from the same equilibrium ([H⁺] = [NO₂⁻] = X):

5.6x10⁻⁴ = X² / 0.15M

8.4x10⁻⁵ = X²

X = [H⁺] = 9.165x10⁻³M

As pH = -log [H⁺]

b. At this point we have HNO₂ and NaNO₂ (The weak acid and the conjugate base), a buffer. The pH of a buffer is obtained using H-H equation:

pH = pKa + log [NaNO₂] / [HNO₂]

Where pH is the pH of the buffer,

pKa is -log Ka = 3.25

And [NaNO₂] [HNO₂] could be taken as the moles of each compound.

The initial moles of HNO₂ are:

0.100L * (0.15mol / L) = 0.015moles

The moles of base added are:

0.0800L * (0.15mol / L) = 0.012moles

The moles of base added = Moles of NaNO₂ produced = 0.012moles.

And the moles of HNO₂ that remains are:

0.015moles - 0.012moles = 0.003moles

Replacing in H-H equation:

pH = 3.25 + log [0.012moles] / [0.003moles]

c. At equivalence point all HNO2 reacts producing NaNO₂. The volume added of NaOH must be 100mL. That means the concentration of the NaNO₂ is:

0.15M / 2 = 0.075M

The NaNO₂ is in equilibrium with water as follows:

NaNO₂(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq) + Na⁺

The equilibrium constant, kb, is:

Kb = Kw/Ka = 1x10⁻¹⁴ / 5.6x10⁻⁴ = 1.79x10⁻¹¹ = [OH⁻] [HNO₂] / [NaNO₂]

Where [OH⁻] = [HNO₂] = x

[NaNO₂] = 0.075M

1.79x10⁻¹¹ = [X] [X] / [0.075M]

1.34x10⁻¹² = X²

X = 1.16x10⁻⁶M = [OH⁻]

pOH = -log [OH-] = 5.94

pH = 14-pOH

d. At this point, 5mL of NaOH are added in excess, the moles are:

5mL = 5x10⁻³L * (0.15mol / L) =7.5x10⁻⁴moles NaOH

In 100mL + 105mL = 205mL = 0.205L. [NaOH] = 7.5x10⁻⁴moles NaOH / 0.205L =

3.66x10⁻³M = [OH⁻]

pOH = 2.44

pH = 14 - pOH

Answer:

FeCl₃ + 3 NaOH ⇒ 3 NaCl + Fe(OH)₃

Explanation:

Let's consider the following unbalanced equation. This is a double displacement reaction.

FeCl₃ + NaOH ⇒ NaCl + Fe(OH)₃

We will start balancing Cl atoms by multiplying NaCl by 3.

FeCl₃ + NaOH ⇒ 3 NaCl + Fe(OH)₃

Then, we will get the balanced equation by multiplying NaOH by 3.

FeCl₃ + 3 NaOH ⇒ 3 NaCl + Fe(OH)₃

Answer:

Show a picture

Explanation:

Answer:

[tex]30^{\circ}\text{C}[/tex]

Explanation:

To know the temperature at which KCl dissolves in water we need to refer to the general solubility curves.

In the case of [tex]KCl[/tex], [tex]35\ \text{g}[/tex] of it will dissolve in [tex]100\ \text{g}[/tex] of water at a minimum temperature of [tex]30^{\circ}\text{C}[/tex].

So, the the minimum temperature needed to dissolve 35 grams of KCl in 100 grams of water is [tex]30^{\circ}\text{C}[/tex].

Answer:

dfgh

Explanation:

Answer:

The law of conservation of energy states that the total energy is constant in any process. Energy may change in form or be transferred from one system to another, but the total remains the same.

Could you please help me with the two most recent questions of mine on my page? I will give u brainliest and 20 points! :))) X

Explanation:

B) 3CuCl2+2AL..….2AlCl3+3Cu

hope it helps.

Balanced Equation:

3CuCl₂ + 2Al = 2AlCl₃ + 3Cu

Answer:

deez cutz

Explanation:

did i get it right

Answer:

the reducing agent is Bromine

The oxidation state of an element is calculated by subtracting and the total sum of oxidation states of all the individual atom (excluding the one that has to be calculated) from total charge on the molecule. Bromine is the reducing agent in the following reaction.

Oxidation state of an element is a number that is assigned to an element in a molecule that represents the number of electron gained or lost during the formation of that molecule or compound.

The ionic equation is given as

2 Br⁻(aq) + H[tex]_2[/tex]O[tex]_2[/tex](aq) + 2 H⁺(aq) → Br[tex]_2[/tex](aq) + 2 H[tex]_2[/tex]O(l)

The oxidation state of bromine on reactant side is -1 while on product side it is 0 so, oxidation state of bromine has increased by 1 so,  bromine is oxidized. If it is oxidized that means it must have reduced someone. So, bromine is acting as a reducing agent.

Therefore, bromine is the reducing agent in the given reaction.

To learn more about Oxidation state, here:

https://brainly.com/question/11313964

#SPJ2

Answer:

3

Explanation: good luck dude I believe in u

Answer:

1. 24.45 moles

2. 437.90 grams

Bonus :

0.75 grams

Explanation:

Hope it was helpful ;)

Answer:

Explanation:

450 grams of Na2SO4? 450 Na soullnol Na2SO4 16.02x1023 molec, Na₂SO4 = 119x1024. | 142.05g Na25041 Imol Naz

Answer:

shhsss×<×>×××<××××

Explanation:

4×738×8<#329×

The number of moles of hydrogen to be 0.468.

The balanced chemical equation :

2H₂ + O₂ → 2H₂O

This implies that 2 moles of hydrogen and one mole of oxygen are required to produce 2 moles of water.

or

Thus molar ratio of ( H₂ : O₂)= 2 : 1.  

So, it requires double the moles of oxygen to get the required moles of water.

Number of moles of oxygen= 0.234 mol

So following the molar ratio concept:

We will get:

The number of moles of hydrogen to be 0.468.

Learn more:

brainly.com/question/21911991

Answer:

[tex]from \: the \: equation \\ 1 \: moles \: of \: propane \: produce \: 4 \: moles \: of \: water \\ 2.50 \:moles \: of \: propane \: will \: produce \: ( \frac{(2.50 \times 4)}{1} ) \: moles \\ = 10 \: moles \: of \: water[/tex]

Answer:

4.92 grams of sodium phosphate (Na₃PO₄) are required to make 125 milliliters of a 0.240 M.

Explanation:

Molarity is a measure of concentration that indicates the number of moles of solute that are dissolved in a given volume.

The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:

[tex]Molarity=\frac{number of moles}{volume}[/tex]

Molarity is expressed in units [tex]\frac{moles}{liter}[/tex].

In this case:

Replacing in the definition of molarity:

[tex]0.240 M=\frac{number of moles}{0.125 L}[/tex]

Solving:

number of moles= 0.240 M*0.125 L

number of moles= 0.03 moles

Being the molar mass of sodium phosphate 164 g/mole, that is, the mass of one mole of the compound, you can calculate the mass of 0.03 moles using the following rule of three: if 1 mole of the compound has 164 grams, 0.03 moles contains how much mass?

[tex]mass=\frac{0.03 moles*164 grams}{1 mole}[/tex]

mass= 4.92 grams

4.92 grams of sodium phosphate (Na₃PO₄) are required to make 125 milliliters of a 0.240 M.

Answer:

Energy released =  18.985 J

Explanation:

The  exponential decay of radioactive substance is given by -

N(t) = N₀ [tex]e^{-kt}[/tex]

where

N₀ = initial quantity

k = decay constant

Half life, [tex]t_{1/2} = \frac{ln 2}{k}[/tex]

⇒[tex]k = \frac{ln 2}{t_{1/2} }[/tex]

Given,

N₀ = 12.5 + 3 = 15.5 × 10⁻⁶ gm

[tex]t_{1/2}[/tex] = 32.6 + 18 = 50.6 × 10⁶ years

So,

[tex]k = \frac{ln 2}{50.6 * 10^{6} }[/tex] = 1.361 × 10⁻⁸ year⁻¹

Now,

N(1) = 15.5 × 10⁻⁶  [tex]e^{-1.361*10^{-8} *1}[/tex]

      = 15.49999978904

Now,

Substance decayed = N₀ - N(t)

                                 = 15.5 × 10⁻⁶ - 15.49999978904 × 10⁻⁶

                                 = 21.095 × 10⁻¹⁷ kg

⇒Δm = 21.095 × 10⁻¹⁷ kg

So,

Energy released = Δmc²

                           = 21.095 × 10⁻¹⁷ × 3 ×10⁸ × 3 × 10⁸

                          = 189.855 ×10⁻¹

                          = 18.985 J

⇒Energy released =  18.985 J