The solubility of calcium oxalate, CaC2O4, in pure water is 4.8 × 10‑5 moles per liter. Cal­culate the value of Ksp for silver carbonate from this data.

Answer:

NaN3 = 47.2 g

Explanation:

Given:

2 NaN3 ⇒ 2 Na  + 3 N2

Find:

Amount of NaN3

Computation:

N2 moles = Product of N2 / molar mass of N2

N2 moles =30.5/28

N2 moles = 1.0893

2NaN3 makes 3(N2 )

So,

NaN3 moles = (2/3) moles of N2  

NaN3 moles = ( 2/3) × 1.0893

NaN3 moles = = 0.7262

NaN3 mass = 0.7262 x 65

NaN3 = 47.2 g

Answer:

NaN3 = 47.2 g

Explanation:

Given:

2 NaN3 ⇒ 2 Na  + 3 N2

Find:

Amount of NaN3

Computation:

N2 moles = Product of N2 / molar mass of N2

N2 moles =30.5/28

N2 moles = 1.0893

2NaN3 makes 3(N2 )

So,

NaN3 moles = (2/3) moles of N2  

NaN3 moles = ( 2/3) × 1.0893

NaN3 moles = = 0.7262

NaN3 mass = 0.7262 x 65

NaN3 = 47.2 g

Explanation:

Answer:

+ small

Explanation:

The entropy is obtained from;

∆S= ∆H/T

Where;

∆S= entropy of the system

∆H= enthalpy if the system = +23.7 KJ

T= absolute temperature of the system = 390 K

∆S= 23.7 ×10^3/390 = 60.8 JK^-

There is a small positive change in entropy.

Answer:

The volume of the container is 59.112 L

Explanation:

Given that,

Number of moles of Oxygen, n = 3

Temperature of the gas, T = 300 K

Pressure of the gas, P = 1.25 atm

We need to find the volume of the container. For a gas, we know that,

PV = nRT

V is volume

R is gas constant, R =  0.0821 atm-L/mol-K

So,

[tex]V=\dfrac{nRT}{P}\\\\V=\dfrac{3\ mol\times 0.0821\ L-atm/mol-K \times 300\ K}{1.25\ atm}\\\\V=59.112\ L[/tex]

So, the volume of the container is 59.112 L

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:

[tex]91°C[/tex]

Explanation:

CHECK THE COMPLETE QUESTION BELOW;

Suppose that 0.95 g of water condenses on a 75.0 g block of iron that is initially at 22 °c. if the heat released during condensation is used only to warm the iron block, what is the final temperature (in °c) of the iron block? (assume a constant enthalpy of vaporization for water of 44.0 kj/mol.)

Heat capacity which is the amount of heat required to raise the temperature of an object or a substance by one degree

From the question, it was said that that 0.95 g of water condenses on the block thenwe know that Heat evolved during condensation is equal to the heat absorbed by iron block.

Then number of moles =given mass/ molecular mass

Molecular mass of water= 18 g/mol

Given mass= 0.95 g

( 0.95 g/18 g/mol)

= 0.053 moles

Then Heat evolved during condensation = moles of water x Latent heat of vaporization

Q= heat absorbed or released

H=enthalpy of vaporization for water

n= number of moles

Q=nΔH

Q = 0.053 moles x 44.0 kJ/mol

= 2.322 Kj

=2322J

We can now calculate Heat gained by Iron block

Q = mCΔT

m = mass of substance

c = specific heat capacity

=change in temperature

m = 75 g

c = 0.450 J/g/°C

If we substitute into the above formula we have

Q= 75 x 0.450 x ΔT

2322 = 75 x 0.450 x ΔT

ΔT = 68.8°C

Since we know the difference in temperature, we can calculate the final temperature

ΔT = T2 - T1

T1= Initial temperature = 22°C

T2= final temperature

ΔT= change in temperature

T2 = T1+ ΔT

= 68.8 + 22

= 90.8 °C

=91°C

Therefore, final temperature is [tex]91°C[/tex]

The final temperature of the iron block is 91∘C.

Given that;

Heat lost during condensation of the water = Heat gained by iron block

Mass of water(mw) = 0.95 g

Latent heat of vaporization =  Latent heat of condensation(L) = 44.0 kJ/mol

Mass of iron(mi) = 75.0 g

Initial temperature of iron(T1) =  22∘C

Final temperature of iron(T2) = ?

Heat capacity of iron(ci) =  0.449 J⋅g−1⋅∘C−1

So;

mwL = mici(T2 - T1)

Substituting values;

(0.95g/18g/mol) ×  44.0 × 10^3(J/mol) = 75.0(g) × 0.449 J⋅g−1⋅∘C−1 (T2 - 22∘C)

2322.2 = 33.7T2 - 741.4

2322.2 +  741.4 = 37.4T2

T2 = (2322.2 +  741.4)/ 33.7

T2 =91∘C

Missing parts;

Suppose that 0.95 g of water condenses on a 75.0 g block of iron that is initially at 22 °c. if the heat released during condensation is used only to warm the iron block, what is the final temperature (in °c) of the iron block? (assume a constant enthalpy of vaporization for water of 44.0 kj/mol.)

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Answer:

0.0002949 moles

Explanation:

Concentration = 2.02×10−2 M

Volume = 14.6 mL = 0.0146 L (Upon converting to litres)

Number of moles = ?

These variables are related by the fllowing equation;

Concentration = Number of moles / Volume

Number of moles = Concentration * Volume

Number of moles = 2.02×10−2 * 0.0146 = 0.0002949 moles

Answer:

The concentration in mol/L =  4.342 mol/L

Explanation:

Given that :

mass of sodium chloride = 25.4 grams

Volume of the volumetric flask = 100 mL

We all know that the molar mass of sodium chloride NaCl = 58.5 g/mol

and number of moles = mass/molar mass

The number of moles of sodium chloride = 25.4 g/58.5 g/mol

The number of moles of sodium chloride = 0.434188 mol

The concentration in mol/L = number of mol/ volume of the solution

The concentration in mol/L = 0.434188 mol/ 100 × 10⁻³ L

The concentration in mol/L =  4.34188 mol/L

The concentration in mol/L =  4.342 mol/L

Answer:

THE VOLUME OF THE OXYGEN GAS AFTER DEFLATION TILL A PRESSURE OF 735 TORR IS ATTAINED IS 7836.99 L

Explanation:

Using Boyle's law,

P1V1 = P2V2

P1 = 150 atm

V1 = 50 L

P2 = 735 Torr

V2 = unknown

We must first convert the pressures into the same SI unit for easy calculation

1torr = 1/760 atm

So converting 735 torr to atm; we have:

1 torr = 1/ 760 atm

735 torr = 735 * 1 / 760 atm

= 0.967 atm

In other words, P2 = 0.957 atm

So rearranging the formula by making V2 the subject of the equation, we have:

V2 = P1 V1 / P2

V2 = 150 * 50 / 0.957

V2 = 7836.99 L

The volume of the oxygen cylinder after deflation to a final pressure of 735 torr or 0.967 atm pressure is 7836.99 L.

Answer:

Explanation:

lewis structure can be defined as a process of how the valence shell electrons of a molecule is being arranged, the pattern of it arrangement and the relationship between the bonding atoms and the lone pairs present in the molecule.

In order to draw the Lewis structure for Arsenate ion  [tex]\mathsf{AsO \ _4^{3-}}[/tex], first thing is to count the valence electrons in the molecule. Once we determine the valence electrons, then we distribute them around the central atom. The Arsenate ion structure is tetrahedral in nature with a bond angle of 109.5° and it is sp³ hybridized.

Answer:

The temperature of the air at this given elevation will be 53.32425°C

Explanation:

We can calculate the final temperature through the combined gas law. Therefore we will need to know 1 ) The initial volume, 2 ) The initial temperature, 3 ) Initial Pressure, 4 ) Final Volume, 5 ) Final Pressure.

Initial Volume = 1.2 L ; Initial Temperature = 25°C = 298.15 K ; Initial pressure = 1.0 atm  ; Final Volume = 1.8 L ; Final pressure = 0.73 atm  

We have all the information we need. Now let us substitute into the following formula, and solve for the final temperature ( T[tex]_2[/tex] ),

P[tex]_1[/tex]V[tex]_1[/tex] / T[tex]_1[/tex] = P[tex]_2[/tex]V[tex]_2[/tex] / T[tex]_2[/tex],

T[tex]_2[/tex] = P[tex]_2[/tex]V[tex]_2[/tex]T[tex]_1[/tex] / P[tex]_1[/tex]V[tex]_1[/tex],

T[tex]_2[/tex] = 0.73 atm [tex]*[/tex] 1.8 L [tex]*[/tex] 298.15 K / 1 atm [tex]*[/tex] 1.2 L = ( 0.73 [tex]*[/tex] 1.8 [tex]*[/tex] 298.15 / 1 [tex]*[/tex] 1.2 ) K = 326.47425 K,

T[tex]_2[/tex] = 326.47425 K = 53.32425 C

Answer:

THE MASS OF NITROGEN GAS IN THIS CONDITIONS IS 0.0589 g

Explanation:

In an ideal condition

PV = nRT or PV = MRT/ MM where:

M = mass = unknown

MM =molar mass = 28 g/mol

P = pressure = 2 atm

V = volume = 25 mL = 0.025 L

R = gas constant = 0.082 L atm/mol K

T = temperature = 290 K

n = number of moles

The gas in the question is nitrogen gas

Molar mass of nitrogen gas = 14 * 2 = 28 g/mol

Then equating the variables and solving for M, we have

M = PV MM/ RT

M = 2 * 0.025 * 28 / 0.082 * 290

M = 1.4 / 23.78

M = 0.0589 g

The mass of the nitrogen gas at ideal conditions of 2 atm, 25 mL volume and 290 K temperature is 0.0589 g

Answer:

C

Explanation:

According to the mole ratio, using 2NH3 will give you 3H2. Which means in order to find the moles of H2 you would only need to divide 2 and multiply 3 to get the amount of moles of H2 produced.

Answer:

I think it's C

Explanation:

Please, tell me if I'm incorrect.

Answer:

Ca²⁺ and Cl⁻

Explanation:

In a chemical reaction, spectator ions are ions that are not involved in the reaction, that means are the same before and after the reaction.

In water, calcium hydroxide, Ca(OH)₂ is dissociated in Ca²⁺ and OH⁻. Also, hydrochloric acid, HCl, dissociates in H⁺ and Cl⁻. The reaction is:

Ca²⁺ + 2OH⁻ + 2H⁺ + 2Cl⁻ → 2H₂O + Ca²⁺ + 2Cl⁻

The ions that react are H⁺ and OH⁻ (Acid and base producing water)

And the ions that are not reacting, spectator ions, are:

Answer:

[tex]M_{acid}=0.141M[/tex]

Explanation:

Hello,

In this case, the reaction between sulfuric acid and hydroxide is:

[tex]H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O[/tex]

We can notice a 1:2 molar ratio between the acid and the base respectively, therefore, at the equivalence point we have:

[tex]2*n_{acid}=n_{base}[/tex]

And in terms of volumes and concentrations:

[tex]2*M_{acid}V_{acid}=M_{base}V_{base}[/tex]

So we compute the molarity of sulfuric acid as shown below:

[tex]M_{acid}=\frac{M_{base}V_{base}}{2*V_{acid}} =\frac{0.100M*28.15mL}{2*10.0mL}\\ \\M_{acid}=0.141M[/tex]

Best regards.

Answer:

B.3/5p

Explanation:

For this question, we have to remember "Dalton's Law of Partial Pressures". This law says that the pressure of the mixture would be equal to the sum of the partial pressure of each gas.

Additionally, we have a proportional relationship between moles and pressure. In other words, more moles indicate more pressure and vice-versa.

[tex]P_i=P_t_o_t_a_l*X_i[/tex]

Where:

[tex]P_i[/tex]=Partial pressure

[tex]P_t_o_t_a_l[/tex]=Total pressure

[tex]X_i[/tex]=mole fraction

With this in mind, we can work with the moles of each compound if we want to analyze the pressure. With the molar mass of each compound we can calculate the moles:

moles of hydrogen gas

The molar mass of hydrogen gas ([tex]H_2[/tex]) is 2 g/mol, so:

[tex]6g~H_2\frac{1~mol~H_2}{2~g~H_2}=~3~mol~H_2[/tex]

moles of oxygen gas

The molar mass of oxygen gas ([tex]O_2[/tex]) is 32 g/mol, so:

[tex]64g~H_2\frac{1~mol~H_2}{32~g~H_2}=~2~mol~O_2[/tex]

Now, total moles are:

Total moles = 2 + 3 = 5

With this value, we can write the partial pressure expression for each gas:

[tex]P_H_2=\frac{3}{5}*P_t_o_t_a_l[/tex]

[tex]P_O_2=\frac{2}{5}*P_t_o_t_a_l[/tex]

So, the answer would be 3/5P.

I hope it helps!

Answer:

ΔG°′ = 1.737 KJ/mol

Explanation:

The reaction involves the transfer of two electrons in the form of hydride ions from reduced coenzyme Q, CoQH₂ to fumarae to form succinate and oxidized coenzyme Q, CoQ.

The overall equation of reaction is as follows:

fumarate²⁻ + CoQH₂ ↽⇀ succinate²⁻ + CoQ ;    ΔE∘′=−0.009 V

Using the equation  for standard free energy change; ΔG°′ = −nFΔE°′

where n = 2; F = 96.5 KJ.V⁻¹.mol⁻¹; ΔE°′ = 0.009 V

ΔG°′ = - 2 * 96.5 KJ.V⁻¹.mol⁻¹ * 0.009 V

ΔG°′ = 1.737 KJ/mol

Answer:

ΔHrxn = -635.14kJ/mol

Explanation:

We can make algebraic operations of reactions until obtain the desire reaction and, ΔH of the reaction must be operated in the same way to obtain the ΔH of the desire reaction (Hess's law). Using the reactions:

(1)Ca(s) + 2 H+(aq) → Ca2+(aq) + H2(g) ΔH = 1925.9 kJ/mol

(2) 2H2(g) + O2 g) → 2 H2O(l) ΔH = −571.68 kJ/mole

(3) CaO(s) + 2 H+(aq) → Ca2+(aq) + H2O(l) ΔH = 2275.2 kJ/mole

Reaction (1) - (3) produce:

Ca(s) + H2O(l) → H2(g) + CaO(s)

ΔH = 1925.9kJ/mol - 2275.2kJ/mol = -349.3kJ/mol

Now this reaction + 1/2(2):

Ca(s) + ½ O2(g) → CaO(s)

ΔH = -349.3kJ/mol + 1/2 (-571.68kJ/mol)

Answer:

The gas is Methane at 340K

Answer:

Explanation:

Given:

Mass of solute = 30 gram

Volume of water = 500 ml

Density = 1.15g/ml

Find:

(i)ppm

(ii) %w/v

(iii) %w/w

Computation:

Water in gram = 500 ml × 1.15 g/ml

Water in gram = 575 gram

In ppm

Concentration of MgSO4 = [30 / 575] × 10⁶

Concentration of MgSO4 = 0.0521 × 10⁶ ppm

in % w/v

Concentration of MgSO4 = [30 / 500] × 100

Concentration of MgSO4 = 6% w/v

in % w/w

Concentration of MgSO4 = [30 / 575] × 100

Concentration of MgSO4 = 5.21% w/w

Answer:

d. None of the above

Explanation:

A buffer works when pH of the buffer is ± 1. Out of this range, the buffering capacity is very low.

Acetic acid, CH₃CO₂H, has a pKa of 4.74. That means its buffering capacity is between 3.74 and 5.74 of pH. Is not a good buffer to pH 7

Pyridine is a weak base with pKa of 5.52. Its buffering capacity is between 6.52 and 4.52. Is not a good buffer to pH 7

HCl is a strong acid. Just weak acids and bases can produce a buffer with its conjugate base. HCl can't produce a buffer.

Thus, right answer is:

Answer:

CH3CO2H

Explanation:

A weak acid, such as acetic acid, paired with its conjugate base (here, the acetate ion) will be an ideal system for creating an acidic buffer solution with a pH below 7.

Answer:

Percentage lithium by mass in Lithium carbonate sample = 19.0%

Explanation:

Atomic mass of lithium = 7.0 g; atomic mass of Chlorine = 35.5 g; atomic mass of carbon = 12.0 g; atomic mass of oxygen = 16.0 g

Molar mass of lithium chloride, LiCl = 7 + 35.5 = 42.5 g

Percentage by mass of lithium in LiCl = (7/42.5) * 100% = 16.4 % aproximately 16%

Molar mass of lithium carbonate, Li₂CO₃ = 7 * 2 + 12 + 16 * 3 =74.0 g

Percentage by mass of lithium in Li₂CO₃ = (14/74) * 100% = 18.9 % approximately 19%

Mass of Lithium carbonate sample = 2 * 42.5 = 85.0 g

mass of lithium in 85.0 g Li₂CO₃ = 19% * 85.0 g = 16.15 g

Percentage by mass of lithium in 85.0 g Li₂CO₃ = (16.15/85.0) * 100 % = 19.0%

Percentage lithium by mass in Lithium carbonate sample = 19.0%

Answer:

2.77 mol/kg

Explanation:

Molality is a sort of concentration that indicates the moles of solute in 1kg of solvent. In this case our solvent is water and, if we consider water's density as 1g/mL, we determine that the mass of solvent is 750 g.

We convert the mass to kg → 750 g . 1kg /1000g = 0.750 kg

Our solute is the NaOH → 83 g.

We convert the mass to moles → 83 g . 1mol /40g = 2.075 mol

Molality (mol/kg) = 2.075 mol / 0.75kg = 2.77 m

Answer:

The total photons required for this radiation = 5.1938 × 10²⁸ photons

Explanation:

Given that:

A microwave oven heats by radiating food with microwave radiation, which is absorbed by the food and converted to heat.

If the radiation wavelength is 12.5 cm,

density of water = 1g/cm³

volume of the container = 0.250 L = 250 cm³

density = mass/volume

mass of the water = density × volume

mass of the water =  1g/cm³  × 250 cm³

mass of the water = 250 g

specific heat capacity of water = 4.182 J/g°C

The change in temperature was from 20.0° C to 99° C

ΔT =( 99 -20.0)° C

ΔT = 79.0° C

The heat absorbed in the process is calculated by using the formula,

q = mcΔT

q = 250 g × 4.182 J/g°C ×  79.0° C

q = 82594.5 Joules

Recall that the radiation wavelength λ = 12.5 cm = 0.125 m

The amount of energy of one photon of the radiation wavelength is determined by using the formula:

E = hv  

since v = c/λ

E = hc/λ

where;

h = Planck's constant = 6.626 × 10⁻³⁴ J.s

c = velocity of light = 3.0 × 10⁸ m/s

∴

E = (6.626 × 10⁻³⁴ J.s × 3.0 × 10⁸ m/s)/ 0.125 m

E = 1.59024⁻²⁴ Joules

The total photons required for this radiation = total heat energy/energy of radiation

The total photons required for this radiation = 82594.5 Joules/1.59024⁻²⁴ Joules

The total photons required for this radiation = 5.1938 × 10²⁸ photons

Answer:

Methyl radical

Explanation:

A radical is any specie that contains an odd number of electrons. We must note that the greater the number of alkyl groups which are attached to a carbon atom that bears the odd electrons, the more the degree of delocalization of the odd electrons and consequently the more stable we expect the free radical to be.

Hence the order of free radical stability is; Methyl < Primary < Secondary < Tertiary. Hence, we can easily see that the methyl radical is the least stable free radical.

Answer: Methyl radical

Explanation:

Answer:

The 3rd one

Explanation:

ANSWER

I need great answers

EXPLANATION

please rate my answer as great

Answer:

The process will be spontaneous above 702 K.

Explanation:

Step 1: Given data

Step 2: Calculate the temperature range in which the process will be spontaneous

The reaction will be spontaneous when the standard Gibbs free energy (ΔG°) is negative. We can calculate ΔG° using the following expression.

ΔG° = ΔH° - T × ΔS°

When ΔG° < 0,

ΔH° - T × ΔS° < 0

ΔH° < T × ΔS°

T > ΔH°/ΔS°

T > (308,000 J/mol)/(439 J/mol.K)

T > 702 K

The process will be spontaneous above 702 K.

Answer:

AS WE KNOW THAT , when non-metallic elements gain electrons to form anions, SO sulphur is non metal and have the capacity to gain two electrons as lies in 6th group so it can gain electron and become sulphide ion(S-).

Explanation:

Answer:

Below

Explanation:

Let n be the quantity of matter in the Calcium Bromide

● n = m/ M

M is the atomic weight and m is the mass

M of CaBr2 is the sum of the atomic wieght of its components (2 Bromes atoms and 1 calcium atom)

M = 40.1 + 2×79.9

● 0.422 = m/ (40.1+2×79.9)

●0.422 = m/ 199.9

● m = 0.422 × 199.9

● m = 84.35 g wich is 88.4 g approximatively

88.4 g approximatively is  the mass in grams of 0.442 moles of calcium bromide, CaBr2 ,therefore option (d) is correct.

Mass is the amount of matter that a body possesses. Mass is usually measured in grams (g) or kilograms (kg) .

To calculate mass in grams of 0.442 moles of calcium bromide, CaBr2,

Let n be the quantity of matter in the Calcium Bromide

M is the atomic weight and m is the mass

M of CaBr2 is the sum of the atomic weight of its components

Mass of  Ca = 40.1 , Mass of Br = 79.9

M = 40.1 + 2×79.9

  0.422 = m/ (40.1+2×79.9)

  0.422 = m/ 199.9

  m = 0.422 × 199.9

  m = 84.35 g which is 88.4 g approximatively .

Thus ,88.4 g approximatively is  the mass in grams of 0.442 moles of calcium bromide, CaBr2 , hence option (d) is correct .

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