Can someone help me with this

Answer:

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Explanation:

4×738×8<#329×

Answer:

it is 0.50=0.5 M

Explanation:

Molarity definition is measure the concentration of solution.

and the rule is

molarity(M)= moles of solute/volume of solution in liters

Ex: 2.1 M MgCl2= 2.1 mole MgCl2/ L MgCl2 solution

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

Answer:

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Explanation:

Answer:

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Explanation:

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:

Water can be used for direct and indirect purposes. Direct purposes include bathing, drinking, and cooking, while examples of indirect purposes are the use of water in processing wood to make paper and in producing steel for automobiles. The bulk of the world's water use is for agriculture, industry, and electricity.

Explanation:

Answer:

water is used for drinking, bathing, cooking, agriculture etc

Answer:

133.34 g/moles

Explanation:

but to make life easy it could be 133.4 g/ moles as well

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:

1

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

Answer:

Heat absorbed = mS△T

51 = ( 8.2/1000)Kg ×S (11.5)

51×1000= 8.2 (11.5)S

S = 51000/94.3

S = 540.8 J/kg K

Answer:

1. 24.45 moles

2. 437.90 grams

Bonus :

0.75 grams

Explanation:

Hope it was helpful ;)

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:

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Explanation:

Answer:

60 pounds of CO₂ are emited into the air

Explanation:

The combustion of gasoline occurs as follows:

C₈H₁₈(l) + 25/2O₂(g) ⇄ 8CO₂(g) + 9H₂O

Where 1 mole of gasoline produce 8 moles of CO₂

To solve this question we must find the moles of gasoline in 3.1 gallons. 8 times these moles are the moles of CO₂ produced. With the moles of CO₂ we can find its pounds as follows:

Pounds gasoline:

3.1 gallons * (6.3 pounds / gallon) = 19.53 pounds

Grams gasoline:

19.53 pounds * (453.592g / pound) = 8859g

Moles gasoline -Molar mass C8H18: 114.23g/mol-

8859g * (1mol / 114.23g) = 77.55 moles gasoline

Moles CO₂:

77.55 moles gasoline * (8 mol CO₂ / mol Gasoline) = 620.4 moles CO₂

Mass CO₂ - Molar mass: 44.01g/mol-

620.4 moles CO₂ * (44.01g / mol) = 27304g CO₂

Pounds CO₂:

27304g CO₂ * (1lb / 453.592g) =

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: I think that is 55

Explanation: I hope I help

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:

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

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:

The answer would be sand.

Explanation:

Sand is weathered down tiny pieces of rock. Sand flows down rivers to shores and slowly builds up to form beaches.

Answer:

The Answer is gonna be B sand

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:

[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].