Calculate the molality of a solution containing 141.5 g of glycine (NH2CH2COOH) dissolved in 4.456 kg of H2O

Answer:

[OH-] = 1.0 x 10-10 M

Explanation:

The acidity of a solution can be determined directly from the concentration of the hydrogen ions and indirectly from the concentrations of the hydroxide ions.

Generally, for a neutral solution we have;

[H3O+] = [OH-] = 1.0 x 10-7 M

For an acidic solution;

[H3O+] > 1.0 x 10-7 M

[OH-] < 1.0 x 10-7 M

Comparing the options the correct option is;

[OH-] = 1.0 x 10-10 M

Explanation:

Since pressure remained constant, we can eliminate P from the equation

[tex] \frac{pv }{t} = \frac{pv}{t} [/tex]

Doing some algebra and converting temperature to Kevin by adding 273, you should obtain the same result.

The statement that "substance A is rice, and substance B is sugar" is correct.

WHAT IS A MIXTURE?

According to this question, a mixture contains two substances A and B. Water is added to this mixture for easy separation.

Therefore, in the mixture of two substances A and B, substance A is rice, and substance B is sugar.

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Answer:

96 mmHg

[tex]h=96mmHg[/tex]

Explanation:

From this question,manometer end is closedw, So we can deduced that the height of the column will not be affected by the atmospheric pressure .

The difference of height of the mercury level is given as,

h=9.60cm

h=9.60(10mm/1cm)

[tex]h=96mm[/tex]

But it is obvious that in this closed end manometer.the pressure of the gas is equal to the height

P(gas)=h

P(gas)=96mmHg

This pressure is as a result of the presence of gas.

Therefore, the pressure of the argon gas in the container is 96mmHg.

The pressure of the argon in the container was 96mmHg.

We were told that the manometer has closed ends which means that the

height will not be affected by atmospheric pressure.

The height which is the difference in mercury level is

h=9.60cm

We can convert it to millimeter by multiplying it by 10

h=9.60 × 10 = 96mm

The pressure of the closed end manometer will be equal to the height

P(gas)=h

P(gas)=96mmHg

The pressure of the argon gas in the container is 96mmHg.

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The force exerted on the container by the particles of the matter is called pressure. The tank containing oxygen gas at 2.551 atm will have a pressure of 1939 mm Hg.

Pressure is the property used to estimate the force experienced by the system due to the liquid or the gas held in it. The pressure of the gas can be calculated by the ideal gas and force and area.

The pressure is created due to the collision of the particles of the gases and liquids on the wall perpendicularly. It is estimated in Pascal (Pa) as the standard unit along with atm and mmHg.

It is known that 1 atm = 760 mm Hg

Given,

The pressure of oxygen gas = 2.551 atm

Using the conversion factor the pressure from atm to mm Hg is calculated as,

1 atm = 760 mm Hg

2.551 atm = (2.551 atm × 760 mm Hg) ÷ 1 atm

= 1938.76 mmHg

Therefore, 1939 mm Hg is the pressure of the oxygen gas.

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Answer:

C;  ΔSuniv>0

Explanation:

In this question, we want to select which of the options must be true.

What we should understand is that for a process to be spontaneous, the change in entropy must be greater than 0 i.e the change in entropy must be positive.

Looking at the options we have; option C is the correct answer.

Option B looks correct but it is wrong. This is because if change in universal entropy is greater than zero, then change in Gibbs free energy must be less than zero for spontaneity to occur

It can be deduced that for a spontaneous process, B. ΔG>0.

It should be noted that a spontaneous process simply means a process that occurs without input of matter or electrical energy.

In this case, for a spontaneous process, it's true that ΔG>0, it should be noted that a spontaneous process related to the second law in thermodynamics. This is characterized by an increase in entropy.

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Answer:

12.5g

Explanation:

Half life = 2.4 Minutes.

The half life of a compound is the time it takes to decay to half of it's original concentration or mass.

Time lapsed= 7.2 minutes. This is equivalent to 3 half lives ( 3 * 2.4)

Initial mass = 100g

First half life;

100g --> 50g

Second half life;

50g --> 25g

Third half life;

25g --> 12.5g

The amount of Zn-71 that remains after 7.2 mins has elapsed is 12.5 g

We'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:

Half-life (t½) = 2.4 mins

Time (t) = 7.2 mins

[tex]n = \frac{t}{t_{1/2}} \\\\n = \frac{7.2}{2.4} \\\\[/tex]

Thus, 3 half-lives has elapsed.

Original amount (N₀) = 100 g

Number of half-lives (n) = 3

[tex]N = \frac{N_{0}}{2^{n}} \\\\N = \frac{100}{2^{3 }}\\\\N = \frac{100}{8}\\\\[/tex]

Thus, the amount of Zn-71 that remains after 7.2 mins is 12.5 g

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Answer:

B. NH3 < PH3< CH4

Explanation:

Hello,

In this case, taking into account that the boiling point of ammonia, methane and phosphorous trihydrate are -33.34 °C , -161.5 °C  and -87.7 °C , clearly, methane has the lowest boiling point (most negative) and ammonia the greatest boiling point (least negative), therefore, ranking is:

B. NH3 < PH3< CH4

Best regards.

answer should be 1.4 cm³

1 L = 10 and so

dL = 100 and then

cL = 1,000

mL = 0.001 m³

1 m³ = 1,000

dm³ = 1,000,000

cm³ = 1,000,000,000

mm³ = 1,000 L

So, 1 mL = 1 cm³ = 0.001 L = 0.1 cL

1,400 mL = 1,400 cm³ = 1.4 L = 140 cL

Answer:

1.4 cm^3

Explanation:

Answer:

The compound contains a o molecular orbital formed by the overlap of one carbon sp2 hybrid orbital and one hydrogen sp3 hybrid orbital.

Explanation:

Molecular orbital is function which describes wave like behavior of an electron in a molecule. The molecular orbital theory describes the electronic structure of molecule using quantum mechanics. Electrons are not assigned to individual bonds between atoms. The compound contains sp2 hybrid orbial which is polar C - CI bond.

Answer:

C. Gas

Explanation:

Answer:

Covalent bonding and non-covalent bonding

Answer:

The median would be 6700

Explanation:

Arrange data values from lowest to highest value

The median is the data value in the middle of the set

.

Ordering a data set x1 ≤ x2 ≤ x3 ≤ ... ≤ xn from lowest to highest value, the median x˜ is the data point separating the upper half of the data values from the lower half.

If the size of the data set n is odd the median is the value at position p where

Formula for the median

p=n+12

x˜=xp

If n is even the median is the average of the values at positions p and p + 1 where

p=n2

x˜=xp+xp+12

If there are 2 data values in the middle the median is the mean of those 2 values.

Answer:

C) 0.027

Explanation:

In this case we can start with the reaction between [tex]NaOH[/tex] and [tex]H_2SeO_4[/tex], so:

[tex]H_2SeO_4~+~NaOH~->~Na_2SeO_4~+~H_2O[/tex]

We have an acid ([tex]H_2SeO_4[/tex]) and a base ([tex]NaOH[/tex]), therefore we will have an acid-base reaction in which a salt is produced ([tex]Na_2SeO_4[/tex]) and water ([tex]H_2O[/tex]).

Now we can balance the reaction:

[tex]H_2SeO_4~+~2NaOH~->~Na_2SeO_4~+~2H_2O[/tex]

If we have the volume (45 mL= 0.045 L) and the concentration (0.3 M) of the acid we can calculate the moles using the molarity equation:

[tex]M=\frac{mol}{L}[/tex]

[tex]0.3~M~=~\frac{mol}{0.045~L}[/tex]

[tex]mol=0.3~M*0.045~L=0.0135~mol~of~H_2SeO_4[/tex]

In the balanced reaction, we have a 2:1 molar ratio between the acid and the base (for each mol of [tex]H_2SeO_4[/tex] 2 moles of [tex]NaOH[/tex] are consumed), with this in mind we can calculate the moles of NaOH:

[tex]0.0135~mol~of~H_2SeO_4\frac{2~mol~NaOH}{1~mol~of~H_2SeO_4}=0.027~mol~NaOH[/tex]

I hope it helps!

Answer:

0.8 part of the product is water

Explanation:

Volume (or parts) of water = 4

Volume (or parts) of lemonade = 1

Total volume = 4 + 1 = 5

Fraction of water = Volume of water / Total volume = 4 / 5 = 0.8

Answer:

The specific heat of the sample [tex]\mathbf{c_3 = 1011.056 \ J/kg.K}[/tex]

Explanation:

Given that:

mass of an unknown sample [tex]m_3[/tex] = 0.0850

temperature of the unknown sample [tex]t_{unknown}[/tex] = 100° C

initial temperature of the calorimeter can = 19° C

mass of copper [tex]m_1[/tex] = 0.150 kg

mass of water [tex]m_2[/tex]= 0.20 kg

the final temperature of the calorimeter can = 26.1° C

The objective is to compute the specific heat of the sample.

By applying the principle of conservation of energy

[tex]Q = mc \Delta T[/tex]

where;

[tex]Q_1 +Q_2 +Q_3 = 0[/tex]        

i.e

[tex]m_1 c_1 \Delta T_1 +m_2 c_2 \Delta T_2+m_3 c_3 \Delta T_3 =0[/tex]

the specific heat capacities of water and copper are 4.18 × 10³ J/kg.K and 0.39 × 10³ J/kg.K respectively

the specific heat of the sample [tex]c_3[/tex] can be computed by making [tex]c_3[/tex]  the subject of the above formula:

i.e

[tex]c_3 = \dfrac{m_1 c_1 \Delta T_1 +m_2 c_2 \Delta T_2}{m_3 c_3 \Delta T_3}[/tex]

[tex]c_3 = \dfrac{ 0.150 \times 0.39 \times 10^3 \times (26.1 -19) + 0.20 \times 4.18 \times 10^3 \times (26.1 -19) }{0.0850 \times (100-26.1 )}[/tex]

[tex]c_3 = \dfrac{ 0.150 \times 0.39 \times 10^3 \times (7.1) + 0.20 \times 4.18 \times 10^3 \times (7.1) }{0.0850 \times (73.9)}[/tex]

[tex]c_3 = \dfrac{415.35 + 5935.6 }{6.2815}[/tex]

[tex]c_3 = \dfrac{415.35 + 5935.6 }{6.2815}[/tex]

[tex]c_3 = \dfrac{6350.95}{6.2815}[/tex]

[tex]\mathbf{c_3 = 1011.056 \ J/kg.K}[/tex]

The specific heat of the sample [tex]\mathbf{c_3 = 1011.056 \ J/kg.K}[/tex]

Answer:

D) 2.01 x 10⁻⁴ .

Explanation:

pH = 2.11

[ H⁺ ] = [tex]10^{-2.11}[/tex]  

Let the acid be HA

It will ionise as follows .

                                        HA       ⇄       H⁺       +        A⁻

in equilibrium                 .30               [tex]10^{-2.11}[/tex]          [tex]10^{-2.11}[/tex]         

Acid ionisation constant Ka  =   [tex]\frac{(10^{-2.11})^2}{0.3}[/tex]

= 2 x 10⁻⁴                

Answer:

D) 2.01 x 10⁻⁴ is correct!

Explanation:

I got it in class!

Hope this Helps!! :))

Answer:

NSAIDs are steroidal anti-inflammatories, their action is on the phospholipase A2 enzyme, this enzyme is responsible for breaking down the phospholipids of the membrane to trigger an inflammatory response. This is how steroidal anti-inflammatory drugs inhibit ALL inflammatory pathways (not like NSAIDs that they only inhibit the COX pathway).

These corticosteroid drugs cannot exceed the systemic mineralocorticoid value 1 in the body, since this corticosteroid hormone is also released by the adrenal cortex.

The NSAIDs generate: sporadic peaks in blood glucose, hypertension, fluid retention, increase in body fat mass, possible suppression of the adrenal cortex over time, inhibiting endogenous synthesis of corticosteroids.

On the other hand, naproxen and ibuprofen are NSAIDs, that is, non-steroidal anti-inflammatory drugs that do not influence both routes of inflammation, but only COX, this enzyme is abbreviated as COX but is called cyclooxygenase, and is responsible for a single route of inflammation.

NSAIDs such as naproxen and ibuprofen can cause gastric disorders such as ulcers or gastritis if they are consumed in a very repetitive manner.

In addition, both drugs are anti-inflammatory, analgesic and antipyretic. Although its two main functions are the first two, it was shown to have an effect in lowering body temperature.

That they are anti-inflammatory means that they inhibit the path of inflammation and analgesics the path of pain.

Explanation:

Both types of drugs generate the same effect but by different mechanisms.

Some are steroids and others are not, although steroids are considered to have a greater risk of benefit that is why they are administered against more systematically compromised instances such as anaphylactic shock.

NSAIDs such as naproxen and ibuprofen are the most prescribed today, since they have few risks and very good benefits, meaning that their adverse effects are not lethal or highly relevant and have a good effect on symptoms.

Both must be administered with care when treating a diabetic patient since corticosteroids generate glycemic peaks or increase in blood glucose, and NSAIDs compete for plasma protein with oral hypoglycemic agents, thus generating that these are in higher free concentrations. high diffusing better through the tissues and increases the potency of the adverse effects of these.

Answer:

3-pentanol

Explanation:

In this case, we have alcohol as the main functional group (OH) with a molecular ion at 88. If the molecular ion is 88 the molar mass is also 88 g/mol therefore the formula for the unknown molecule is [tex]C_5H_1_2O[/tex].

Additionally, if the mass spectrum shows the molecular ion peak we can not have tertiary alcohols (tertiary alcohols often do not show M+ at  all). So, the structures only can be primary and secondary structures.

With this in mind, our options are:

-) 1-pentanol

-) 2-pentanol

-) 3-pentanol

Now we can analyze each structure:

-) 1-pentanol

The structure must explain all the fragments produced (73, 70, and 59). In this primary alcohol, we will have an alpha cleavage (the red bond would be broken). If this has to happen, we will have fragments at 31 and 57. These fragments dont fit with the reported ones, therefore this is not a possible structure (See figure 1).

-) 2-pentanol

On this structure, we will have also an alpha cleavage (red bond). In this rupture we will have fragments at 45 and 43, these m/z values dont fit with the reported ones, therefore this is not a possible structure (See figure 1).

-) 3-pentanol

In this structure, we have the "OH"  bonded to carbon three. So, we can analyze each fragment:

   -) m/z 59

This fragment, can be explained as an alpha cleavage. But, in this case we have two ruptures that can produce the same ion. The carbons on both sides of the C-OH bond.

   -) m/z 71

This fragment, can be explained as a loss of water (M-18) in which we have the production of a carbocation in the carbon where we previously have the C-OH bond.

   -) m/z 73

This fragment, can be explained as a beta cleavage. But, in this case, also we have two ruptures that can produce the same ion. The methyl groups on each end molecule.

See figure 2

I hope it helps!

Answer:

[tex]128~ATP[/tex]

Explanation:

The metabolic pathway by which energy can be obtained from a fatty acid is called "beta-oxidation". In this route, acetyl-Coa is produced by removing 2 carbons from the fatty acid for each acetyl-Coa produced. In other words, for each round, 1 acetyl Coa is produced and for each round 2 carbons are removed from the initial fatty acid. Therefore, the first step is to calculate the number of rounds that will take place for an 18-carbon fatty acid using the following equation:

[tex]Number~of~Rounds=\frac{n}{2}-1[/tex]

Where "n" is the number of carbons, in this case "18", so:

[tex]Number~of~Rounds=\frac{18}{2}-1~=~8[/tex]

We also have to calculate the amount of Acetyl-Coa produced:

[tex]Number~of~Acetyl-Coa=\frac{18}{2}~=~9[/tex]

Now, we have to keep in mind that in each round in the beta-oxidation we will have the production of 1 [tex]FADH_2[/tex] and 1 [tex]NADH[/tex]. So, if we have 8 rounds we will have 8 [tex]FADH_2[/tex] and 8 [tex]NADH[/tex].

Finally, for the total calculation of ATP. We have to remember the yield for each compound:

-) [tex]1~FADH_2~=~2~ATP[/tex]

-) [tex]1~NADH~=~3~ATP[/tex]

-) [tex]Acetyl~CoA~=~10~ATP[/tex]

Now we can do the total calculation:

[tex](8*2)~+~(8*3)~+~(9*10)=130~ATP[/tex]

We have to subtract  "2 ATP" molecules that correspond to the activation of the fatty acid, so:

[tex]130-2=128~ATP[/tex]

In total, we will have 128 ATP.

I hope it helps!

Answer:

See explanation

Explanation:

The dissolution of oxygen in water is exothermic. Hence, at higher temperature, less oxygen is dissolved in water.

If the chemist commits an error and records 27°C instead of 21°C, the dissolved oxygen concentration will be found to be lower than it should be at 27°C because dissolution of gases in water is exothermic.

Answer:

B. 21.07 x 10²³ molecules

Explanation:

Avogadro's Number: 6.022 x 10²³

Step 1: Set up equation

[tex]3.5 mols CO_2(\frac{6.022(10^{23}) moleculesCO_2}{1 mol CO_2})[/tex]

Step 2: Multiply and cancel out units

3.5(6.022 x 10²³) = 21.07 x 10²³ molecules CO₂

Step 3: Convert to proper scientific notation

≈ 2.11 x 10²³ molecules CO₂

Answer:

0

Explanation:

pure elements will always possess an oxidation number of 0, regardless of their charge.

Answer:

D.) 0

Explanation:

I got it correct on founders edtell

The advantage of a resource person would be that it will provide a hands-on activity that will allow the students to experience spacing between organs and on the body of the person.

It will also allow them to identify challenges when doing this and will engage them more in the activity and lesson.

Answer:A resource person add knowledge to the course

Explanation:

there will be no chemical reaction

We have a solution of NaOH and H₂CO₃

First, NaOH will dissociate into Na⁺ and OH⁻ ions

The Na⁺ ion will substitute one of the Hydrogen atoms on H₂CO₃ to form NaHCO₃

The H⁺ released from the substitution will bond with the OH⁻ ion to form a water molecule

If there were to be another NaOH molecule, a similar substitution will take place, substituting the second hydrogen from H₂CO₃ as well to form Na₂CO₃

Answer:

Explanation:

Given that:

The argon atoms are excited into an excited state before emitting the 488.0 nm laser.

the energy of the first ionization energy of argon is 1520 kJ mol-1.

SInce 1 eV = 96.49 kJ/mol

Therefore, the energy of the first ionization energy of argon in eV is = ( 1520/ 96.49) eV

=  15.75 eV

To find where  the energy level of the excited state lies below the vacuum energy level, let's first determine, the energy liberated by using planck expression.

[tex]E = \dfrac{hc}{\lambda}[/tex]

[tex]E = \dfrac{6.6 \times 10^{-34} \times 3 \times 10^8}{488 \times 10^{-9}}[/tex]

[tex]E = \dfrac{1.98 \times 10^{-25}}{488 \times 10^{-9}}[/tex]

[tex]E = \dfrac{1.98 \times 10^{-25}}{488 \times 10^{-9}}[/tex]

[tex]E =4.057 \times 10^{-19} \ J[/tex]

Converting Joules (J) to eV ; we get,

[tex]E =\dfrac{4.057 \times 10^{-19}}{1.6 \times 10^{-19}}[/tex]

E = 2.53 eV

The energy levels of the first exited state = -13.223 eV

Answer:

[tex]0.033g[/tex]

Explanation:

Hello,

In this case, since 11 mg per kilogram of body weight has the given lethality, the mg that turn out lethal for a chicken weighting 3 kg is computed by using a rule of three:

[tex]11mg\longrightarrow 1kg\\\\?\ \ \ \ \ \ \longrightarrow 3kg[/tex]

Thus, we obtain:

[tex]?=\frac{3kg*11mg}{1kg}\\ \\?=33mg[/tex]

That in grams is:

[tex]=33mg*\frac{1g}{1000mg} \\\\=0.033g[/tex]

Regards.