On a hot summer day, the density of air at atmospheric pressure at 35.5°C is 1.1970 kg/m3. (a) What is the number of moles contained in 1.00 m3 of an ideal gas at this temperature and

Answers

Answer 1

Complete question:

On a hot summer day, the density of air at atmospheric pressure at 35.5°C is 1.1970 kg/m3. (a) What is the number of moles contained in 1.00 m3 of an ideal gas at this temperature and pressure.

Answer:

The  number of moles contained by an ideal gas at this temperature and pressure is 41.32 moles.

Explanation:

Given;

density of dry air, ρ = 1.1970 kg/m³

temperature of the air, T = 35.5°C  = 273 + 35.5 = 308.5 K

air volume, V = 1 m³

Apply ideal gas law for dry to calculate the air pressure;

[tex]P = \rho R_dT[/tex]

where;

P is the air pressure

ρ is the air density

Rd is gas constant for dry air = 287 J/kg/K

P = 1.197 x 287 x 308.5 = 105,981.78 Pa

(a) Now, determine the number of moles contained by an ideal gas at this temperature and pressure, by applying ideal gas law;

PV = nRT

where;

P is the pressure of the gas (Pa)

V is the volume of the gas (m³)

n is number of gas moles

R is gas constant = 8.314 m³.Pa / mol.K

T is temperature (K)

n = (PV) / (RT)

n = (105,981.78 x 1) / (8.314 x 308.5)

n = 41.32 moles

Therefore, the  number of moles contained by an ideal gas at this temperature and pressure is 41.32 moles.

Answer 2

The number of moles of an ideal gas at this temperature and pressure is 41.5 moles.

Given that;

Density of dry air = 1.1970 kg/m3

Pressure of dry air = ?

Temperature of dry air = 35.5°C + 273 = 308.5 K

Hence;

P = Density × gas constant of dry air × Temperature

P = 1.1970 kg/m3 × 287.1 J/Kg/K × 308.5 K

P = 106019 Pa or 1.05 atm

Using the ideal gas equation;

PV = nRT

n = PV/RT

n = 1.05 atm × 1000 L/0.082 atmL/K.mol × 308.5 K

n = 41.5 moles

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Related Questions

Pre-Lab Study Questions / 9
1. Where are the valence electrons in an atom?
valen
2. How are positive and negative ions fomed?
3. Why are electrons shared in molecular compounds?​

Answers

Answer:

See detailed answer with explanation below.

Explanation:

Valence electrons are electrons found on the outermost shell of an atom. They are the electrons in an atom that participate in chemical combination. Recall that the outermost shell of an atom is also referred to as its valence shell. Let us consider an example; if we look at the atom, sodium-11, its electronic configuration is 2,8,1. The last one electron is the valence electron of sodium which is found in its outermost or valence shell.

Positive ions are formed when electrons are lost from the valence shell of an atom. For instance, if the outermost electron in sodium is lost, we now form the sodium ion Na^+ which is a positive ion. Positive ions possess less number of electrons compared to their corresponding atoms.

Negative ions are formed when one or more electrons is added to the valence shell of an atom. A negative ion possesses more electrons than its corresponding atom. For example, chlorine(Cl) contains 17 electrons but the chloride ion (Cl^-) contains 18 electrons.

In molecular compounds, a bond is formed when two electrons are shared between the bonding atoms. Each bonding atom may contribute one of the shared electrons (ordinary covalent bond) or one of the bonding atoms may provide the both shared electrons (coordinate covalent bond). The shared pair may be located at an equidistant position to the nucleus of both atoms. Similarly, the electron may be drawn closer to the nucleus of one atom than the other (polar covalent bond) depending on the electro negativity of the two bonding atoms.

The electrons are shared in order to complete the octet of each atom by so doing, the both bonding atoms now obey the octet rule. For example, two chlorine atoms may come together to form a covalent bond in which each chlorine atom has an octet of electrons on its outermost shell.

Enter an abbreviated electron configuration for magnesium: Express your answer in complete form, in order of increasing energy. For example, [He]2s22p2 would be entered as [He]2s^22p^2.

Answers

Answer:

[Ne]3s²

Explanation:

Mg

1s2 2s2 2p6 3s2   or [Ne]3s²

Abbreviated electronic configuration of magnesium is [Ne]3 s² and in complete form it is 1 s² 2 s² 2 p⁶ 3 s².

What is electronic configuration?

Electronic configuration is defined as the distribution of electrons which are present in an atom or molecule in atomic or molecular orbitals.It describes how each electron moves independently in an orbital.

Knowledge of electronic configuration is necessary for understanding the structure of periodic table.It helps in understanding the chemical properties of elements.

Elements undergo chemical reactions in order to achieve stability. Main group elements obey the octet rule in their electronic configuration while the transition elements follow the 18 electron rule. Noble elements have valence shell complete in ground state and hence are said to be stable.

Learn more about electronic configuration,here:

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Electrophilic substitution on 3-phenylpropenenitrile occurs at the meta position. Draw resonance structures to show how the ring is electron-poor at the ortho and para positions.

Answers

Answer:

See figure 1

Explanation:

In this question, we have to remember that a poor electron carbon is a carbon in which we have a positive charge, a carbocation. Therefore we have to start with the production of the carbocation. First, a double bond from the benzene is moved to the carbon in the top to produce a new double bond generating a positive charge in a carbon with ortho position (electron-poor). Then we can move another double bond inside the ring to produce a positive charge in the para carbon. Finally, we can move the last double bond to produce again another positive charge in the second ortho carbon.

See figure 1.

I hope it helps!

AMMONIUM CARBONATE
5. How many grams of nitrogen (N) are in a mass of ammonium carbonate that contains
1.23x10^23 carbon atoms?​

Answers

Answer:

Zero

Explanation:

Hello,

The question require us to calculate the mass of nitrogen present in aluminium carbonate.

This can easily be calculated using Avogadro's number as a constant with some minor calculations but however in this case, we can't because there's no single atom of nitrogen present in aluminium carbonate hence we can't calculate the mass of nitrogen present in it.

Chemical formula of aluminium carbonate = Al₂(CO₃)₃.

From the above chemical formula, we can see that there's no single atom of nitrogen present in the formula hence the mass of nitrogen present in aluminium carbonate that contains 1.23×10²³ carbon atoms is zero.

Use the link Standard Reduction Potentials. Write net equations for the spontaneous redox reactions that occur during the following or none (all lower case) if there is no extensive reaction. Use the lowest possible coefficients. A) Iron metal is dipped into a Ni2+ solution.B) Copper metal is added to hydrochloric acid.C) A silver wire is added to nitric acid (HNO3).D) Magnesium metal is added to waterE) Chromium metal is added to a solution of Mg2+

Answers

Answer:

a) Fe(s) + Ni^2+(aq) ----> Fe^2+(aq) + Ni(s)

b) no reaction

c) no reaction

d) 2Mg(s) + 2H2O(l)-----> 2Mg^2+(aq) + O2(g) +4H^+(aq)

e) no reaction

Explanation:

It is important to say here that the ability of a particular chemical specie to displace another chemical specie is dependent on the relative standard reduction potentials of the species involved.

All the reactions stated above are redox reactions. Let us take reaction E as an example. Mg^2+ has a reduction potential of -2.37 V while Cr^3+ has a reduction potential of -0.74V. Since the reduction potential of magnesium is more negative than that of chromium, there is no reaction when a piece of chromium metal is dipped into a solution of Mg^2+.

Similarly, though metals displace hydrogen gas from dilute acids, metals that are less than hydrogen in the reactivity series cannot do that. This explains why there is no reaction when copper and silver are dipped into dilute acid solutions.

Reaction occurs when iron is dipped into a nickel solution because the reduction potential of Fe^2+ is far more negative than that of Ni^2+.

Nitrosyl bromide, NOBr, is formed from NO and Br2.
2 NO(g) + Br2(g) 2 NOBr(g)
Experiments show the reaction is second order in NO and first order in Br2.
How does the reaction rate change if the concentration of Br2 is changed from 0.0022 mol/L to 0.0066 mol/L?
a) the rate will double
b) the rate will remain unchanged
c) the rate will triple

Answers

Answer:

C ) The rate will triple

Calculate ΔG (in kJ) for the following reaction at 1.0 atm for C2H6, 0.5 atm for O2, and 2.0 atm for CO2, and 25 oC: C2H6 (g) + O2 (g) ---> CO2 (g) + H2O (l) (unbalanced) ΔGfo C2H6 (g) = - 32.89 kJ/mol; ΔGfo CO2 (g) = - 394.4 kJ/mol; ΔGfo H2O (l) = - 237.13 kJ/mol

Answers

Answer:

Explanation:

Calculate ΔG (in kJ) for the following reaction at 1.0 atm for C2H6,

0.5 atm for O2, and

2.0 atm for CO2, and

25 oC:

C2H6 (g) + O2 (g) ---> CO2 (g) + H2O (l) (unbalanced)

ΔGfo C2H6 (g) = - 32.89 kJ/mol;

ΔGfo CO2 (g) = - 394.4 kJ/mol;

ΔGfo H2O (l) = - 237.13 kJ/mol

The balance equation of this reaction is

[tex]2C_2H_6 (g) + 7O_2 (g) ---> 4CO_2 (g) + 6H_2O (l)[/tex]

[tex]\Delta G_{rxn}=\sum G^o_f(product)-\sum G^o_f(reactant)[/tex]

[tex]=4G^o_f(CO_2)+6G^o_f(H_2O)-7G^o_f(O_2)-2G^o_f(C_2H_6)\\\\[/tex]

[tex][4(-394.4)+6(-237.13)-7(0)-2(-32.89)]kJ/mol\\\\=-1577.6-1422.78+65.78\\\\=-3000.38+65.78\\\\=-2934.6kJ/mol[/tex]

Nitrogen monoxide is produced by combustion in an automobile engine. For the following reaction, 4.36 grams of nitrogen monoxide are mixed with excess oxygen gas . The reaction yields 5.46 grams of nitrogen dioxide . nitrogen monoxide ( g ) oxygen ( g ) nitrogen dioxide ( g ) What is the theoretical yield of nitrogen dioxide

Answers

Answer:

[tex]6.68~g~NO_2[/tex]

Explanation:

We have to start with the combustion reaction:

[tex]NO~+~O_2~->~NO_2[/tex]

Then we can balance the reaction:

[tex]2NO~+~O_2~->~2NO_2[/tex]

If we want to find the theoretical yield, we have to calculate the amount of [tex]NO_2[/tex]. To do this, we have to first convert the 4.36 g of [tex]NO[/tex] to moles [tex]NO[/tex] (using the molar mass 30 g/mol), then we have to convert from moles of [tex]NO[/tex] to moles of [tex]NO_2[/tex] (using the molar ratio) finally, we have to convert from moles of [tex]NO_2[/tex] to grams of [tex]NO_2[/tex] (using the molas mass 46 g/mol), so:

[tex]4.36~g~NO\frac{1~mol~NO}{4.36~g~NO}\frac{2~mol~NO_2}{2~mol~NO}\frac{46~g~NO_2}{1~mol~NO_2}=6.68~g~NO_2[/tex]

I hope it helps!

The reason for the dramatic decline in the number of measles cases from the 1960s to 2010 in the United States was because the vaccine

Answers

Answer:

It was because the vaccine generated actively acquired immunity, that is, inoculation of a portion of the measles virus so that the body forms the antibodies for a second contact and thus can destroy it without triggering the pathology.

Explanation:

Vaccines are methods of active acquired immunity since the antibody is not passively inoculated, it is manufactured by the body with a physiological process once part of the virus is inoculated.

The measles virus most of all affected the lives of infants or newborn children with severe rashes and high fevers that led to death.

Classify the substances as atomic elements, molecular elements, molecular compounds, or ionic compounds. a. Agb. Cdc. MgCl2d. F2f. HIg. NO2h. NaCli. Cl2

Answers

Answer:

Explanation:

Hello,

We'll be doing some classification of some chemical substances based on molecules, elemental state or ionic or electrovalent properties.

A) Ag = atomic element : silver (Ag) in its elemental state is an atomic element.

B) Cd = atomic element : Cadmium (Cd) is an element of the periodic table and belongs to transition metal.

C) MgCl = ionic compounds: this is a compound formed between magnesium (Mg) and chlorine (Cl) to give MgCl. This compound has ionic or electrovalent properties since electron transfer occurred between the cation (Mg) and anion (Cl).

D) F₂ = moleculer element : Fluorine F₂ is moleculer element since two elements of fluorine combine together to form a molecule.

E) HI = molecular compound : this is a compound formed from the reaction between hydrogen and iodine. It's a molecular compound because they are two different elements combining together to form a compound.

F) NO₂ = molecular compound

G) NaCl = ionic compound

H) Cl₂ = molecular element

Enter your answer in the provided box. Before arc welding was developed, a displacement reaction involving aluminum and iron(III) oxide was commonly used to produce molten iron (the thermite process). This reaction was used, for example, to connect sections of iron railroad track. Calculate the mass of molten iron produced when 2.88 kg of aluminum reacts with 24.4 mol of iron(III) oxide.

Answers

Answer:

2.7255 kg Fe

Explanation:

Based on the reaction of the thermite process:

2 Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2 Fe(l)

2.88kg of Al (Molar mass: 26.98g/mol) are:

2880g ₓ (1mol / 26.98g) = 106.7 moles Al

For a complete reaction of these moles of Al are necessaries:

106.7 moles Al ₓ ( 1 mol Fe₂O₃ / 2 moles Al) = 53.35 moles Fe₂O₃

As you have just 24.4 moles of Fe₂O₃, Fe₂O₃ is limiting reactant.

1 mole of Fe₂O₃ produce 2 moles of Fe.

Thus, moles of Fe produced are 24.4×2 = 48.8 moles of Fe.

As molar mass of Fe is 55.85g/mol, mass of Fe is:

48.8 moles Fe ×(55.85g / mol) = 2725.5g of Fe =

2.7255 kg Fe

Cl2 + F2 → ClF3, 5. How many moles of Cl2 are needed to react with 3.44 moles of F2? 6. How many grams of ClF3 form when 0.204 moles of F2 react with excess Cl2? 7. How many grams of ClF3 form from 130.0 grams of Cl2 when F2 is in excess? 8. How many grams of F2 are needed to react with 3.50 grams of Cl2?

Answers

Answer:

5) 1.147 moles Cl2

6) 12.57 grams ClF3

7)  339.10 grams ClF3

8) 5.63 grams F2

Explanation:

Step 1: Data given

Number of moles F2 = 3.44 moles

Molar mass F2 = 38.00 g/mol

Step 2: The balanced equation

Cl2 + 3F2 → 2ClF3

Step 3: Calculate moles F2

For 1 mol Cl2 we need 3 moles F2 to produce 2 moles ClF3

For 3.44 moles F2 we'll need 3.44/3 = 1.147 moles Cl2

Step 1: Data given

Number of moles F2 = 0.204 moles

Molar mass F2 = 38.00 g/mol

Molar mass ClF3 = 92.448 g/mol

Step 2: The balanced equation

Cl2 + 3F2 → 2ClF3

Step 3: Calculate moles ClF3

For 1 mol Cl2 we need 3 moles F2 to produce 2 moles ClF3

For 0.204 moles F2 we'll have 2/3 * 0.204 = 0.136 moles

Step 4: Calculate mass ClF3

Mass ClF3 = Moles ClF3 * molar mass ClF3

Mass ClF3 = 0.136 moles * 92.448 g/mol

Mass ClF3 = 12.57 grams ClF3

Step 1: Data given

Mass of Cl2 = 130.0 grams

Molar mass F2 = 38.00 g/mol

Molar mass ClF3 = 92.448 g/mol

Step 2: The balanced equation

Cl2 + 3F2 → 2ClF3

Step 3: Calculate moles Cl2

Moles Cl2 = mass Cl2 / molar mass Cl2

Moles Cl2 = 130.0 grams / 70.9 g/mol

Moles Cl2 = 1.834 moles

Step 4: Calculate moles

For 1 mol Cl2 we need 3 moles F2 to produce 2 moles ClF3

For 1.834 moles Cl2 e'll have 2*1.834 = 3.668 moles ClF3

Step 5: Calculate mass ClF3

Mass ClF3 = Moles ClF3 * molar mass ClF3

Mass ClF3 = 3.668 moles * 92.448 g/mol

Mass ClF3 = 339.10 grams ClF3

Step 1: Data given

Mass of Cl2 = 3.50 grams

Molar mass F2 = 38.00 g/mol

Molar mass ClF3 = 92.448 g/mol

Step 2: The balanced equation

Cl2 + 3F2 → 2ClF3

Step 3: Calculate moles Cl2

Moles Cl2 = Mass Cl2 / molar mass Cl2

Moles Cl2 = 3.50 grams / 70.9 g/mol

Moles Cl2 = 0.0494  moles

Step 4: Calculate moles F2

For 1 mol Cl2 we need 3 moles F2

For 0.0494 moles we need 3*0.0494 = 0.1482 moles

Step 5: Calculate mass F2

Mass F2 = moles F2 * molar mass F2

Mass F2 = 0.1482 moles * 38.00 g/mol

Mass F2 = 5.63 grams F2

What force is needed to accelerate a truck with mass 2,000 kg at a rate of 5.0
m/s22
O A. 1,000 N
B. 20,000 N
C. 2,000 N
O D, 10,000 14

Answers

Answer:

D

Explanation:

[tex]F=ma \\\\F=2000\cdot 5=10,000N[/tex]

Hope this helps!

Ga3+ and Br1- is what formula?

Answers

[tex]\text{GaBr}_3[/tex]

A certain lightbulb containing argon at 1.20 atm and 18°C is
heated to 85°C at constant volume. Calculate its final pressure
(in atm).

Answers

Answer:

certain lightbulb containing argon at 1.20 atm and 18 0 C is heated to 85 0 C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? P 1 T 1 P 2 T 2 ... Ideal Gas Equation 5.4 Charles' law: V T (at constant n and P ) ... Consider a case in which two gases, A and B , are in a container of volume V.

Explanation:

Among three bases, X−, Y−, and Z−, the strongest one is Y−, and the weakest one is Z−. Rank their conjugate acids, HX, HY, and HZ, in order of decreasing strength. Rank the acids from strongest to weakest. To rank items as equivalent, overlap them.

Answers

Answer: HZ > HX > HY in order of decreasing strengths.

Explanation: Generally, the rule is that the stronger the acid, the weaker its conjugate base and vice versa; same rule applies for bases and their conjugate acids.

So the weakest base Z- would have the strongest conjugate acid. Consequently, the strongest base Y- would have the weakest conjugate acid.

I hope this was MORE helpful as this is the correct answer.

The ranking of the conjugate acids in order of decreasing strength (i.e from strongest to weakest) is; HZ < HX < HY

First we must know that the stronger a base is, the weaker is it's conjugate acid and the weaker a base is, the stronger is it's conjugate acid.

Therefore, the order of decreasing strength of the conjugate acid is; HZ < HX < HY

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Consider the heating curve for water. A graph of the heating curve for water has time in minutes on the horizontal axis and Temperature in degrees Celsius on the vertical axis. A line runs through the following points: 0 minutes, negative 20 degrees; 0.5 minutes, 0 degrees; 2 minutes, 0 degrees; 4 minutes, 100 degrees; 8 minutes, 100 degrees; 9.5 minutes, 160 degrees. At what temperature does the solid start melting? –20°C 0°C 20°C 80°C

Answers

Answer:

0°C.

Explanation:

Hello,

In this case, given the heating curve of water on the attached document, we can notice that at 0 °C the solid starts melting, which means that the melting point is reached. Melting point is known as a physical change whereby a solid changes to liquid by the addition of heat as it allows the molecules to separate to each other.

Best regards.

Answer:

0 degrees celcius

Explanation:

I took the test

The first three excited states of the nucleus Au-199 (gold) are at 0.075 Mev, 0.320 Mev and 0.475 MeV. If all transitions between theses states and the ground state occurred, what energy/wavelength gamma rays would be observed?

Answers

Answer and Explanation:

The computation of the energy or wavelength gamma rays observed is shown below:

Since the energy of gamma rays is higher than 0.10 MeV.

Now We have to calculate transitions in between the given levels of energy that correspond to this energy.

As per the given question, we have the following information

Ground state = E where E < 0.075 MeV

For  Level 1 = 0.075 MeV

For Level 2 = 0.320 MeV

For Level 3 = 0.475 MeV

Now we have to take the below transitions:

1. [tex]3 \rightarrow 2[/tex]

Difference of energy is

= 0.475 - 0.320

= 0.155 MeV

This represents  a gamma radiation

2. [tex]3 \rightarrow 1[/tex]

Difference of energy is

= 0.475 - 0.075

= 0.4 MeV

This represents  a gamma radiation

3. [tex]3 \rightarrow ground[/tex]

Difference of energy is

= 0.475 - E > 0.155 MeV

This represents  a gamma radiation

4. [tex]2 \rightarrow 1[/tex]

Difference of energy is

= 0.320 - 0.075

= 0.245 MeV

This represents  a gamma radiation

5. [tex]2 \rightarrow ground[/tex]

Difference of energy is

= 0.320 - E > 0.245 MeV

This represents  a gamma radiation

6. [tex]1 \rightarrow Ground[/tex]

Difference of energy is

= 0.075 - E < 0.10 MeV

This represents not a gamma radiation

We can see that there are 5 transitions that contain gamma rays

1. Determine whether the following hydroxide ion concentrations ([OH−]) correspond to acidic, basic, or neutral solutions by estimating their corresponding hydronium ion concentrations ([H3O+] using the ion product constant of water (Kw).
Kw = [H3O+][OH−] = [1×10−7 M][1×10−7 M] = 1×10−14 M
Hydronium ion concentration [H3O+] Solution condition
Greater than 1×10−7 M Acidic
Equal to 1×10−7 M Neutral
Less than 1×10−7 M Basic
Drag the appropriate items to their respective bins.
1. [OH−] = 6×10−12 M
2. [OH−] = 9×10−9 M
3. [OH−] = 8×10−10 M
4. [OH−] = 7×10−13 M
5. [OH−] = 2×10−2 M
6. [OH−] = 9×10−4 M
7. [OH−] = 5×10−5 M
8. [OH−] = 1×10−7 M
A. Acidic
B. Neutral
C. Basic
2. A solution has [H3O+] = 5.2×10−5M . Use the ion product constant of water
Kw=[H3O+][OH−]
to find the [OH−] of the solution.
3. A solution has [OH−] = 2.7×10−2M . Use the ion product constant of water
Kw=[H3O+][OH−]
to find the [H3O+] of the solution.

Answers

Answer:

Question 1.

1. [OH−] = 6×10−12 M  is less than 1 * 10⁻⁷, therefore is acidic.

2. [OH−] = 9×10−9 M  is less than 1 * 10⁻⁷, therefore is acidic.

3. [OH−] = 8×10−10 M  is less than 1 * 10⁻⁷, therefore is acidic.

4. [OH−] = 7×10−13 M  is less than 1 * 10⁻⁷, therefore is acidic.

5. [OH−] = 2×10−2 M  is greater than 1 * 10⁻⁷, therefore is basic.

6. [OH−] = 9×10−4 M  is greater than 1 * 10⁻⁷, therefore is basic.

7. [OH−] = 5×10−5 M  is greater than 1 * 10⁻⁷, therefore is basic.

8. [OH−] = 1×10−7 M  is equal to 1 * 10⁻⁷, therefore is neutral

Question 2:

[OH⁻] = 1.92 * 10⁻⁸ M

Question 3:

[H₃O⁺] = 3.70 * 10⁻¹¹ M

Explanation:

The ion product constant of water  Kw = [H₃O⁺][OH⁻] = 1 * 10⁻¹⁴ M² is a constant which gives the product of the concentrations of hydronium and hydroxide ions of dissociated pure water. The concentrations of the two ions are both equal to 1 * 10⁻⁷ in pure water.

A solution that has [OH⁻] greater than 1 * 10⁻⁷ is basic while one with [OH⁻] less than 1 * 10⁻⁷ is acidic.

1. [OH−] = 6×10−12 M  is less than 1 * 10⁻⁷, therefore is acidic.

2. [OH−] = 9×10−9 M  is less than 1 * 10⁻⁷, therefore is acidic.

3. [OH−] = 8×10−10 M  is less than 1 * 10⁻⁷, therefore is acidic.

4. [OH−] = 7×10−13 M  is less than 1 * 10⁻⁷, therefore is acidic.

5. [OH−] = 2×10−2 M  is greater than 1 * 10⁻⁷, therefore is basic.

6. [OH−] = 9×10−4 M  is greater than 1 * 10⁻⁷, therefore is basic.

7. [OH−] = 5×10−5 M  is greater than 1 * 10⁻⁷, therefore is basic.

8. [OH−] = 1×10−7 M  is equal to 1 * 10⁻⁷, therefore is neutral

Question 2:

Kw = [H₃O⁺][OH⁻] = 1 * 10⁻¹⁴ M²

[H₃O⁺][OH⁻] = 1 * 10⁻¹⁴ M²

[OH⁻] = 1 * 10⁻¹⁴ M²/ [H₃O⁺]

[OH⁻] = 1 * 10⁻¹⁴ M²/5.2*10⁻⁵ M

[OH⁻] = 1.92 * 10⁻⁸ M

Question 3:

Kw = [H₃O⁺][OH⁻] = 1 * 10⁻¹⁴ M²

[H₃O⁺][OH⁻] = 1 * 10⁻¹⁴

[H₃O⁺] = 1 * 10⁻¹⁴ M²/ [OH⁻]

[H₃O⁺] = 1 * 10⁻¹⁴ M²/ 2.7 * 10⁻² M

[H₃O⁺] = 3.70 * 10⁻¹¹ M

The first solution will be an unbuffered sucrose solution. Calculate the amount of 0.1 M sucrose stock solution and the amount of dH2O needed to prepare 100 mL of unbuffered 20 mM sucrose.

Answers

Answer:

20mL of the stock solution and dilute to 100mL

Explanation:

20mM = 0.020M is the concentration of the diluted solution. As you want to prepare this solution from a 0.1M solution, dilution factor is:

0.1M / 0.020M = 5

That means you need to dilute the stock solution 5 times to obtain the 20mM sucrose solution.

As you want to prepare 100mL, you need to add:

100mL / 5 =

20mL of the stock solution and dilute to 100mL

an auto of an element has 17 protons in its nucleus.a) write the electronic configuration of the atom.b)to what period and group does the element belong​

Answers

Answer:

i hope it will help you

Explanation:

electronic configuration 1s²,2s,²2p^6,3s²3p^6,4s^1

as it has one electron in its valence shell so it is the member of group 1A(ALKALI METALS)  and the number of shells is 4 so it is in period 4

In the first step of glycolysis, the given two reactions are coupled. reaction 1:reaction 2:glucose+Pi⟶glucose-6-phosphate+H2OATP+H2O⟶ADP+PiΔG=+13.8 kJ/molΔG=−30.5 kJ/mol Answer the four questions about the first step of glycolysis. Is reaction 1 spontaneous or nonspontaneous?

Answers

Answer: Reaction 1 is non spontaneous.

Explanation:

According to Gibb's equation:

[tex]\Delta G=\Delta H-T\Delta S[/tex]

[tex]\Delta G[/tex] = Gibbs free energy  

[tex]\Delta H[/tex] = enthalpy change

[tex]\Delta S[/tex] = entropy change  

T = temperature in Kelvin

When [tex]\Delta G[/tex] = +ve, reaction is non spontaneous

[tex]\Delta G[/tex]= -ve, reaction is spontaneous

[tex]\Delta G[/tex]= 0, reaction is in equilibrium

For the given reaction 1: [tex]glucose+Pi\rightarrow glucose-6-phosphate+H_2O[/tex] [tex]\Delta G=+13.8kJ/mol[/tex]

As for the reaction 1 , the value of Gibbs free energy is positive and thus the reaction 1 is non spontaneous.

For each of the following pairs of elements
(1C and N2) (1Ar and Br2)
pick the atom with
a. more favorable (exothermic) electron affinity.
b. higher ionization energy.
c. larger size.
How do you even go about do this?

Answers

Explanation:

Electron affinity is defined as the energy released by the addition of an electron to any gaseous atom. Electron affinity of an atom depends on the electronic configuration.

a).The carbon has vacant p-orbital and nitrogen has half-filled configuration which is more stable. Therefore, one electron can be easily added to carbon whereas nitrogen having more stable configuration releases more amount of energy on adding one electron. Therefore, nitrogen has more electron affinity than carbon.

The bromine has vacant p-orbital whereas argon has filled orbital which is most stable. Therefore, one electron can be easily added to bromine whereas argon having  more stable configuration releases more amount of energy on adding one electron.Therefore, argon has more electron affinity than bromine.

Answer:

1. a. C; b. N; c. C; 2. a. Br; b. Ar; c. Br

Explanation:

Use your Periodic Table and follow the trends in atomic properties (Fig. 1).

Electron affinity increases from left to right and from bottom to top.

The elements with the most exothermic EA are at the upper right corner

Exceptions are the noble gases (group 18) and the pnictogens (group 18).

The elements of Group 18 have a complete octet and have no tendency to accept electrons.  

The elements of Group 15 have half-filled p subshells. They are more stable than the elements immediately preceding them, so they are less exothermic when adding an electron.  

Ionization energy increases from left to right and from bottom to top.  

The atoms with the highest IE are at the upper right corner.

Atomic size increases from right to left and from bottom to top.  

The largest atoms are on the lower-left corner.

1. C vs N

(a)   EA: C. N is a Group 15 element

(b)    IE: N. N is further to the right.

(c) Size: C. C is further to the left.

2. Ar vs Br

(a)   EA: Br. Ar is a noble gas.

(b)    IE: Ar. Ar is further to the right.

(c) Size: Br. Br is nearer to the bottom.

What is the law of conservation and what happens when two substances at different temperatures come into contact.

Answers

Answer:

- Both energy and matter cannot be neither created nor destroyed.

- An equilibrium temperature will be reached.

Explanation:

Hello,

In this case, the law of conservation is applied to both matter and energy, and it states that both energy and matter cannot be neither created nor destroyed. Specifically, in chemical reactions, it states that in closed systems, the mass of the reactants equals the mass of the products even when the number of moles change. Moreover, for energy, if two substances at different temperatures come into contact, the hot one will cool down and the cold one will heat up until an equilibrium temperature so the energy lost by the hot one is gained by the cold one, which accounts for the transformation of energy.

Best regards.

The lock-and-key model and the induced-fit model are two models of enzyme action explaining both the specificity and the catalytic activity of enzymes. Following are several statements concerning enzyme and substrate interaction. Indicate whether each statement is part of the lock-and-key model, the induced-fit model, or is common to both models.

a. Enzyme conformation changes when it binds the substrate so the active site fits the substrate.
b. Substrate binds to the enzyme at the active site, forming an enzyme-substrate complex.
c. Enzyme active site has a rigid structure complementary
d. Substrate binds to the enzyme through noncovalent interactions

Answers

Answer:

The lock-and-key model:

c. Enzyme active site has a rigid structure complementary

The induced-fit model:

a. Enzyme conformation changes when it binds the substrate so the active site fits the substrate.

Common to both The lock-and-key model and The induced-fit model:

b. Substrate binds to the enzyme at the active site, forming an enzyme-substrate complex.

d. Substrate binds to the enzyme through non-covalent interactions

Explanation:

Generally, the catalytic power of enzymes are due to transient covalent bonds formed between an enzyme's catalytic functional group and a substrate as well as non-covalent interactions between substrate and enzyme which lowers the activation energy of the reaction. This applies to both the lock-and-key model as well as induced-fit mode of enzyme catalysis.

The lock and key model of enzyme catalysis and specificity proposes that enzymes are structurally complementary to their substrates such that they fit like a lock and key. This complementary nature of the enzyme and its substrates ensures that only a substrate that is complementary to the enzyme's active site can bind to it for catalysis to proceed. this is known as the specificity of an enzyme to a particular substrate.

The induced-fit mode proposes that binding of substrate to the active site of an enzyme induces conformational changes in the enzyme which better positions various functional groups on the enzyme into the proper position to catalyse the reaction.

The amount of calcium in a 15.0-g sample was determined by converting the calcium to calcium oxalate, CaC2O4. The CaC2O4 weighed 40.3 g. What is the percent of calcium in the original sample

Answers

Answer:

128 gram of CaC2O4 contains 40 gram of Calcium

40.3 gram of CaC2O4 cotnains = 40*40.3/128 = 12.59 gram of Calcium

out of 15 gram 12.59 gram is Calcaim that means around 50% of orginal sample has Calcium

Consider each of the following molecules in turn: (a) dimethyl ether, (CH3)2O; (b) trimethylamine, (CH3)3N; (c) trimethylboron, (CH3)3B; and (d) carbon dioxide (CO2). Describe the hybridization state of the central atom (i.e., O, N, B, or C) of each molecule, tell what bond angles you would expect at the central atom, and state whether the molecule would have a dipole moment.

Answers

Answer:

Consider each of the following molecules in turn: (a) dimethyl ether, (CH3)2O; (b) trimethylamine, (CH3)3N; (c) trimethylboron, (CH3)3B; and (d) carbon dioxide (CO2). Describe the hybridization state of the central atom (i.e., O, N, B, or C) of each molecule, tell what bond angles you would expect at the central atom, and state whether the molecule would have a dipole moment.

Rank the following transitions in a hydrogen atom in order of increasing wavelength of electromagnetic radiation that could produce them. Answer this question without doing any calculations. Explain the order.

n=2 to n=4
n=6 to n=8
n=10 to n=12
n=14 to n=16

Answers

Answer:

n=2 to n=4 < n=6 to n=8 < n=10 to n=12 < n=14 to n=16

Explanation:

According to Neils Bohr, electrons in an atom are found in specified energy levels. Transitions are possible from one energy level to another when the electron receives sufficient energy usually in the form of a photon of electromagnetic radiation of appropriate frequency and wavelength. The energy of this photon corresponds to the energy difference between the two energy levels. Thus the higher the energy difference between energy levels, the greater the energy of the photon required to cause the transition and the shorter the wavelength of the photon.

High energy photons have a very short wavelength. It should be noted that as n increases, the energy of successive energy levels decreases and transitions between them now occurs at longer wavelengths. Hence, the highest energy and shortest wavelength of photons are required for transition involving lower values of n because such electrons are closer to the nucleus and are more tightly bound to it than electrons found at a greater distance from the nucleus.

Hence transition involving electrons at higher energy levels occur at a longer wavelength compared to transition involving electrons closer to the nucleus. This is the basis for the arrangement of wavelengths required to effect the various electronic transitions shown in the answer.

Write the complete balanced equation for the neutralization reaction that occurs when aqueous hydroiodic acid, HI, and sodium hydrogen carbonate, NaHCO3, are combined. Include physical states.

Answers

Answer:

[tex]HI_(_a_q_)~+~NaHCO_3_(_a_q_)~->~NaI_(_a_q_)~+~H_2O_(_l_)~+~CO_2_(_g_)[/tex]

Explanation:

In this case, we will have a neutralization reaction. We have a base ([tex]HI[/tex]) and a base ([tex]NaHCO_3[/tex]). Additionally, we have a strong acid and a strong base, therefore both will be soluble on water, so we will have an aqueous state for these compounds. If we will have a neutralization reaction, we will have as a salt as a product. With this in mind the reaction would be:

[tex]HI_(_a_q_)~+~NaHCO_3_(_a_q_)~->~NaI_(_a_q_)~+~H_2O_(_l_)~+~CO_2_(_g_)[/tex]

All the sodium salts are soluble in water, therefore we will have an aqueous state. Water is a liquid and carbon dioxide is a gas.

I hope it helps!

Given the following Fischer projection: Fischer projection for an entantiomer of 2-bromo-2,3-dihydroxypropanal with the bromine oriented horizontally to the left and the hydroxide group oriented horizontally to the right. draw the perspective formula of the molecule. Be sure to correctly indicate stereochemistry in your answer.

Answers

Answer:

Explanation:

Stereoisomers are two or more atoms that  have the same bonding order of atoms but there is a difference  spatial arrangement of  the  atoms in space.

A plane of symmetry divides a molecule into two equal halves.

A chiral stereoisomer are not superimposed on a mirror image , Hence they do not posses a plane of symmetry.

As a result to that. these non-superimposable mirror images are said to be Enantiomers.

However, a Fischer Projection emanates from a two - dimensional figure which is used for presenting a three - dimensional organic molecules.

From the given question;

Fischer projection for an enantiomer of 2-bromo-2,3-dihydroxypropanal with the bromine oriented horizontally to the left and the hydroxide group oriented horizontally to the right.

we can sketch the way the enantiomer of 2-bromo-2,3-dihydroxypropanal can  be seen like the one shown below:

              CH₂OH

                   |

                   |

                   |

Br -------------|----------------OH

                   |

                   |

                   |

                 CHO

The objective of this question is to drawn the perspective formula of the molecule.

So , from the attached file below; we can see the perspective formula of the molecule in a well structured 3-D format.

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