[H+] for a solution is 4.59 x 10-6 M.
This solution is
A. acidic
B. basic
C. neutral

Answer:

ΔG = -8.812 kJ/mol

Explanation:

To obtain the free energy of a reaction you can use the expression:

ΔG = ΔG° + RT ln Q

Where:

ΔG° is Standard Gibbs Free energy: -16.7kJ/mol = -16700J/mol

R is gas constant: 8.314472 J/molK

T is absolute temperature (37°C + 273.15 = 310.15K)

And Q is reaction quotient: 21.3

Replacing in the formula:

ΔG = ΔG° + RT ln Q

ΔG = -16700J/mol + 8.314472J/molK*310.15K ln 21.3

ΔG = -8812.4J/mol

Answer:

6.15.3 k

Explanation:

From the question we can see that

q = 0,  Δu = w

Then,

[tex]T_f = \frac{C_{V,m}+RP_{ext}P_i}{C_{V,m}+RP_{ext}P_f} T_i[/tex]

putting values wet

=[tex]\frac{2.5\times 8.314+8.314\left(10^5\right)\left(3.20\times 10^5\right)}{2.5\times 8.314+\left(8.314\right)\left(10^5\right)\left(10^5\right)}\times \:293[/tex]

T_f = 615.3 K

Answer:

The pressure in mmHg is 1253 (option C)

Explanation:

Two quantities are directly proportional if when multiplying or dividing one of them by a number, the other is multiplied or divided by that number. In other words, the magnitudes are directly proportional when one magnitude increases and so does the other in the same proportion; or when one magnitude decreases and so does the other in the same proportion.

The rule of three or is a way of solving proportionality problems between three known values ​​and an unknown value, which can be applied to directly proportional quantities as follows:

a ⇒ b

c ⇒ x

So [tex]x=\frac{c*b}{a}[/tex]

where a, b and c are data and x is the unknown value to be calculated.

In this case, knowing that 1 Torr = 1 mmHg, the rule of three can be applied as follows: if 1 torr is equal to 1 mmHg, 1253 torr is equal to how many mmHg?

[tex]pressure=\frac{1253 torr*1 mmHg}{1 torr}[/tex]

pressure= 1253 mmHg

The pressure in mmHg is 1253 (option C)

Answer:

The answer is

Explanation:

The density of a substance can be found by using the formula

[tex]density = \frac{mass}{volume} [/tex]

From the question

mass of box is 345 g

volume = 1125 cm³

The density of the object is

[tex]density = \frac{345}{1125} \\ = 0.3066666...[/tex]

We have the final answer as

Hope this helps you

1kg = 100000

0.459kg = 0.459*100000

= 45900cg

→ 0.459kg = 45900cg

According to unit conversion and as 1 kg=100,000 cg , there are 45900 cg in 0.459 kg.

Unit conversion is defined as a multi-step process which involves multiplication or a division operation by a numerical factor.The process  of unit conversion requires selection of appropriate number of significant figures and the rounding off procedure.

It involves a conversion factor  which is an expression  for expressing the relationship between the two units.A conversion ratio always has value which equals to one which indicates that numerator and denominator have values which are expressed in different units.

It is an easy process which involves unit conversion between the different conversion systems. While unit conversion care needs to be taken regarding accuracy and precision.

Learn more about unit conversion,here:

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Answer:

C

Explanation:

Compounds are substances that are made of elements linked by chemical bonds. "The simplest formula for a compound made from element X that is 21.0% nitrogen by mass is X₂N₃." Thus, option C is correct.

The empirical formula has been the representation of the ratio of the whole number of atoms involved to make a molecule or compound. It has been calculated by calculating the moles through mass and, molar mass.

The elements involved are X (unknown), and N (known).

Given,

Mass percentage of nitrogen = 21.0% = 21 gms

Mass percentage of X = 100.0 % - 21.0 % = 79.0% = 79 gms

Molar mass of X = 79.0 g /mol

Molar mass of Nitrogen = 14 g/mol

Moles of X is calculated as:

Moles X = 79 ÷ 79

= 1 mol

Moles of N are calculated as:

Moles N = 21 ÷ 14

= 1.5 mol

Now, the moles are divided by the smallest mole as:

X = 1 ÷ 1 = 1

N = 1.5 ÷ 1 = 1.5

The ratios are 1:1.5 or 2:3. 2X and 3N.

The empirical formula for the compound will be X₂N₃.

Therefore, the simplest formula for the compound will be X₂N₃.

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Answer:

The value is [tex] A = 679.5 \ ppm[/tex]

Explanation:

From the question we are told that

The half life of [tex]^{64}Cu[/tex] is [tex] t_h = 12.7 \ hr [/tex]

   The  initial concentration is   [tex] A_o  = 845 \  ppm [/tex]

    The time duration is   [tex] t =  4 \  hr  [/tex]

Generally the rate constant is mathematically represented as

       [tex] k =  \frac{0.693}{t_h} [/tex]

        [tex] k =  \frac{0.693}{12.7} [/tex]

       [tex] k = 0.0545 \  hr^{-1} [/tex]

This rate constant is also mathematically represented as

           [tex] k =  \frac{1}{t} *  ln (\frac{A_o}{A}) [/tex]

Here  A is the remaining concentration after t

So

             [tex]  0.0545 =  \frac{1}{4} *  ln (\frac{845}{A}) [/tex]

             [tex] 0.218 =  ln (\frac{845}{A}) [/tex]

             [tex] e^{0.218} =   \frac{845}{A} [/tex]

              [tex] 1.2436 =   \frac{845}{A} [/tex]

             [tex] A =   \frac{845}{1.2436} [/tex]

             [tex] A =   679.5 \  ppm[/tex]

Answer:

True

Explanation:

You must first observe your data then form a hypothesis.

Answer:

The most likely to occur when jagged edges of rock plates grind past each other is the presence of a high degree of frictional force.

This may cause the rocks to be broken down into smaller particles.

It also implies that the energy necessary for further disintegration and movement of rocks is stored up.

Answer:

(a) [tex]V_B=11.68L[/tex]

(b) [tex]x_{He}=0.533[/tex]

Explanation:

Hello,

In this case, since the both gases behave ideally, with the given information we can compute the moles of He in A:

[tex]n_A=\frac{0.082\frac{atm*L}{mol*K}*298K}{1.974 atm*6.00L}=2.063mol[/tex]

Thus, since the final pressure is 3.60 bar, we can write:

[tex]P=x_{Ar}P_A+x_{He}P_B\\\\P=\frac{n_{Ar}}{n_{Ar}+n_{He}} P_A+\frac{n_{He}}{n_{Ar}+n_{He}} P_B\\\\3.60bar=\frac{2.063mol}{2.063mol+n_{He}} *2.00bar+\frac{n_{He}}{2.063mol+n_{He}} *5.00bar[/tex]

The moles of helium could be computed via solver as:

[tex]n_{He}=2.358mol[/tex]

Or algebraically:

[tex]3.60bar=\frac{1}{2.063mol+n_{He}} *(4.0126+5.00*n_{He})\\\\7.314+3.60n_{He}=4.013+5.00*n_{He}\\\\7.314-4.013=5.00*n_{He}-3.60n_{He}\\\\n_{He}=\frac{3.3}{1.4}=2.358mol[/tex]

In such a way, the volume of the compartment B is:

[tex]V_B=\frac{n_{He}RT}{P_B}=\frac{2.358mol*0.082\frac{atm*L}{mol*K}*298.15K}{4.935atm}\\ \\V_B=11.68L[/tex]

Finally, he mole fraction of He is:

[tex]x_{He}=\frac{2.358}{2.358+2.063}\\ \\x_{He}=0.533[/tex]

Regards.

Answer:

10.4 qt

Explanation:

Step 1: Given data

Volume of water in the beaker (V): 9.80 L

Step 2: Convert the volume of water in the beaker to US quarts (qt)

In order to convert one unit into another, we need a conversion factor. In this case, the appropriate conversion factor is 1 L = 1.06 qt. The volume of water in the beaker, in US quarts, is:

9.80 L × (1.06 qt/1 L) = 10.4 qt

Answer:

The correct approach will be "Increased gravitational attraction".

Explanation:

Answer:

D = 4x10^-11

Explanation:

an increase in temperature would cause a resultant increase in diffusivity. as temperature rises, thermal energy of atoms would also rise and this would cause them to go faster.

4x10^-11cm²/s satisfies this condition because there is a temperature increase from 1300 to 1600.

DAI in Si = 4x10^-13cm²/sec at 1300

DAI in Si at 1600

D increases

temperature also increases

Answer:

Formic acid and Acetic acid is the best buffer at pH 3.7.

Explanation:

Given that,

The Ka values for several weak acids are given,

[tex]K_{a}\ of\ MES=7.9\times10^{-7}[/tex]

[tex]K_{a}\ of\ HEPES = 3.2\times10^{-3}[/tex]

[tex]K_{a}\ of\ Tris=6.3\times10^{-9}[/tex]

[tex]K_{a}\ of\ formic\ acid = 1.8\times10^{-4}[/tex]

[tex]K_{a}\ of\ Acetic\ acid = 1.8\times10^{-5}[/tex]

We need to calculate the pH of the weak acids with their [tex]pK_{a}[/tex] values

Using formula of [tex]pK_{a}[/tex]

For MES,

[tex]pK_{a}=-log K_{a}[/tex]

Put the value into the formula

[tex]pK_{a}=-log(7.9\times10^{-7})[/tex]

[tex]pK_{a}=7.0-log7.9[/tex]

[tex]pK_{a}=6.1[/tex]

pH range for best buffer,

[tex]pH=pK_{a}\pm 1[/tex]

Put the value into the formula

[tex]pH=6.1\pm 1[/tex]

[tex]pH=7.1, 5.1[/tex]

The pH value of the solution between 7.1 to 5.1.

This is not best buffer.

For HEPES,

[tex]pK_{a}=-log K_{a}[/tex]

Put the value into the formula

[tex]pK_{a}=-log(3.2\times10^{-3})[/tex]

[tex]pK_{a}=3.0-log3.2[/tex]

[tex]pK_{a}=2.5[/tex]

pH range for best buffer,

[tex]pH=pK_{a}\pm 1[/tex]

Put the value into the formula

[tex]pH=2.5\pm 1[/tex]

[tex]pH=3.5, 1.5[/tex]

The pH value of the solution between 3.5 to 1.5.

This is not best buffer.

For Tris,

[tex]pK_{a}=-log K_{a}[/tex]

Put the value into the formula

[tex]pK_{a}=-log(6.3\times10^{-9})[/tex]

[tex]pK_{a}=9.0-log6.3[/tex]

[tex]pK_{a}=8.2[/tex]

pH range for best buffer,

[tex]pH=pK_{a}\pm 1[/tex]

Put the value into the formula

[tex]pH=8.2\pm 1[/tex]

[tex]pH=9.2, 7.2[/tex]

The pH value of the solution between 9.2 to 7.2.

This is not best buffer.

For formic acid,

[tex]pK_{a}=-log K_{a}[/tex]

Put the value into the formula

[tex]pK_{a}=-log(1.8\times10^{-4})[/tex]

[tex]pK_{a}=4.0-log1.8[/tex]

[tex]pK_{a}=3.7[/tex]

pH range for best buffer,

[tex]pH=pK_{a}\pm 1[/tex]

Put the value into the formula

[tex]pH=3.7\pm 1[/tex]

[tex]pH=4.7, 2.7[/tex]

The pH value of the solution between 4.7 to 2.7.

This is best buffer.

For acetic acid,

[tex]pK_{a}=-log K_{a}[/tex]

Put the value into the formula

[tex]pK_{a}=-log(1.8\times10^{-5})[/tex]

[tex]pK_{a}=5.0-log1.8[/tex]

[tex]pK_{a}=4.7[/tex]

pH range for best buffer,

[tex]pH=pK_{a}\pm 1[/tex]

Put the value into the formula

[tex]pH=4.7\pm 1[/tex]

[tex]pH=5.7, 3.7[/tex]

The pH value of the solution between 5.7 to 3.7.

This is  best buffer

Hence, Formic acid and Acetic acid is the best buffer at pH 3.7.

Answer:

A = 0.334

B = -26

C = 1.666

D = -26

Explanation:

just took quiz

Answer:

A = ✔ 0.334

B = ✔ -26

C = ✔ 1.666

D = ✔ -26

Explanation:

edge 2023

Answer:

Characteristics. They are all relatively inert, corrosion-resistant metals. Copper and gold are colored. These elements have low electrical resistivity so they are used for wiring.

Explanation:

Hope this help u

️️

Answer:

When molecular hydrogen (H2) and oxygen (O2) are combined and allowed to react together, energy is released and the molecules of hydrogen and oxygen can combine to form either water or hydrogen peroxide. ... For both of the reactions shown, the hydrogen molecules are oxidized and the oxygen atoms are reduced.

Explanation: