Questions posted in Chemistry

What are the applications of chromatography

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ANSWERED AT 18/11/2019 - 04:12 AM


QUESTION POSTED AT 18/11/2019 - 04:12 AM

What is the coefficient for the oxygen molecule in the chemical equation below? 2hgo → 2hg + ? o2?

This question is purely balancing the chemical equation where the number of atoms in reactants is equal to the number of atoms in products.
For the mentioned equation:
In the reactants, we have 2 atoms of Hg and 2 atoms of oxygen.
In the products, we should have 2 atoms of Hg and 2 atoms of oxygen.
Since the oxygen molecule is composed of 2 oxygen atoms, therefore the coefficient should be equal to 1 for the equation to be balanced.

ANSWERED AT 18/11/2019 - 04:02 AM


QUESTION POSTED AT 18/11/2019 - 04:02 AM

Given that 4 nh3 + 5 o2 → 4 no + 6 h2o, when 4.50 mol of h2o are formed, the amount of no formed is 3.00 mol.

We need to examine the balancing coefficients of this equation.
4 NH3 + 5 O2 → 4 NO + 6 H2O

From the above equation, it is clear that the formation of 4 moles of NO is accompanied by the formation of 6 moles of H2O.
This means that the ratio between NO:H2O is 2:3

Based on this, if 4.5 moles of H2O are formed then number of NO moles formed (given the symbol n) will be calculated as:
n = 4.5 x (2/3) = 3 moles (cross multiplication)

ANSWERED AT 18/11/2019 - 04:02 AM


QUESTION POSTED AT 18/11/2019 - 04:02 AM

How to tell if a reaction is exothermic or endothermic from an equation?

Hope this table might help!

ANSWERED AT 18/11/2019 - 04:02 AM


QUESTION POSTED AT 18/11/2019 - 04:02 AM

What is one of the products produced when al(no3)3 and cao react together? alo2 cano3 al2o3 ca(no3)3?

Answer: Option (c) is the correct answer.

Explanation:

When Al(NO_{3})_{3} reacts with CaO then it results in the formation of aluminium oxide and calcium nitrate.

The chemical reaction equation will be as follows.

2Al(NO_{3})_{3} + 3CaO \rightarrow Al_{2}O_{3} + 3Ca(NO_{3})_{2}

Thus, we can conclude that out of the given options Al_{2}O_{3} is one of the products produced when Al(NO_{3})_{3} and CaO react together.


ANSWERED AT 18/11/2019 - 04:02 AM


QUESTION POSTED AT 18/11/2019 - 04:02 AM

The feed to a batch process contains equimolar quantities of nitrogen and methane. write an expression for the kilograms of nitrogen in terms of the total moles n(mol) of this mixture.

1) number of moles of N2 = n/2

2) Number of moles of CH4 = n/2

3) Total number of moles of the mixture = n/2 + n/2 = n

4) Kg of N2

mass in grams = number of moles * molar mass

molar mass of N2 = 2 * 14.0 g/mol = 28 g/mol

=> mass of N2 in grams = (n/2) * 28 = 14n

mass of N2 in Kg = mass of N2 in grams * [1 kg / 1000g] = 14n/1000 kg = 0.014n kg

Answer: mass of N2 in kg = 0.014n kg

ANSWERED AT 18/11/2019 - 04:02 AM


QUESTION POSTED AT 18/11/2019 - 04:02 AM

Calculate the ph of a 0.060 m carbonic acid solution, h2co3(aq), that has the stepwise dissociation constants ka1 = 4.3 × 10-7 and ka2 = 5.6 × 10-11.

1) First dissociation

H2 CO3    =    H(+) + HCO3(-)          Ka1 = 4.3 * 10^ -7

0.06 - x              x          x

Ka1 = x^2 / (0.06 - x) = 4.3 * 10^ - 7

A low Ka => x << 0.06 => 0.06 -x ≈ 0.06

=> Ka1 ≈ x^2 / 0.06 => x^2 ≈ 0.06 * Ka1 = 0.06 * 4.3 * 10^-7

=> x ≈ √[ 2.58 * 10 ^ -8] = 1.606 * 10^ - 4 = 0.0001606

2) Second dissociation

          HCO3(-)    =   H(+) + CO3(2-)         Ka2 = 5.6 * 10^ - 11

0.0001606 - y            y             y

Ka2 ≈ y^2 / 0.0001606 => y = √[0.0001606 * 5.6* 10^ -11]

y = 9.48 * 10^ -8

3) [H+] = x + y = 1.607 * 10^ -4

4) pH = - log [H+] = 3.79

Answer: 3.79

ANSWERED AT 18/11/2019 - 04:01 AM


QUESTION POSTED AT 18/11/2019 - 04:01 AM

Lava is molten rock that cools: a)on the surface. b)underground

A because lava is above ground and magma is when it is in the volcano and not on the ground. Please rate brainliest and everyone thank ;)

ANSWERED AT 18/11/2019 - 03:59 AM


QUESTION POSTED AT 18/11/2019 - 03:59 AM

What is the [h3o+] at equilibrium of a 0.50 m weak acid (ha) solution if the ka of the acid is 4.6 × 10−4?

Answer : The concentration of H_3O^+ at equilibrium is, 0.015 M

Solution :

The balanced equilibrium reaction will be,

HA+H_2O\rightleftharpoons H_3O^++A^-

The expression for dissociation constant of weak aciod will be,

k_a=\frac{[H_3O^+]\times [A^-]}{[HA]}

where,

k_a = dissociation constant of weak acid

Let the concentration of H_3O^+ and A^- be 'x'

Now put all the given values in this expression, we get

4.6\times 10^{-4}=\frac{(x)\times (x)}{0.50}

x=0.015M

The concentration of H_3O^+ = A^- = x = 0.015 M

Therefore, the concentration of H_3O^+ at equilibrium is, 0.015 M

ANSWERED AT 18/11/2019 - 03:57 AM


QUESTION POSTED AT 18/11/2019 - 03:57 AM

What would be the formula of the precipitate that forms when pb(no3)2 (aq) and k2so4 (aq) are mixed? pbk2 pbso4 h2o none of the above k(no3)2?

The formula of the precipitate that will be formed is PbSO4.
When lead ll nitrate react with potassium sulphate, one of the product formed is lead sulphate. The reaction is a double decomposition precipitation reaction; PbSO4 is the precipitate formed. 

ANSWERED AT 18/11/2019 - 03:55 AM


QUESTION POSTED AT 18/11/2019 - 03:55 AM

If solutions of nh4cl(aq) and nh3(aq) are mixed, which ions in the resulting solution are spectator ions in any acid-base chemistry occurring in the solution?

If solutions of nh4cl(aq) and nh3(aq) are mixed, the ions in the resulting solution that would be the spectator ions in any acid-base chemistry occurring in the solution would be the chloride ions. The mixture is a common buffer of ammonia and ammonium chloride.  NH4Cl is a salt that is soluble in water dissociating into NH4+ and Cl-. The chloride ions are the spectator ions while NH4+ ions would be the conjugate acid for the buffer solution. Buffer solutions are solution which would resist change in pH when acid or a base is being added to the solution. The equilibrium reaction of the buffer solution in this problem would be

NH3 + H2O = OH– + NH4+

ANSWERED AT 18/11/2019 - 03:55 AM


QUESTION POSTED AT 18/11/2019 - 03:55 AM

What is present in the glucose molecule that is missing in the carbon dioxide molecule?

Hydrogen .carbon dioxide is CO2 and glucose is C6H12O6

ANSWERED AT 18/11/2019 - 03:53 AM


QUESTION POSTED AT 18/11/2019 - 03:53 AM

Aluminum reacts with chlorine gas to form aluminum chloride. 2al(s)+3cl2(g)→2alcl3(s) what minimum volume of chlorine gas (at 298 k and 225 mmhg) is required to completely react with 7.85 g of aluminum

The balanced chemical reaction is expressed as:

2Al(s)+3Cl2(g)→2AlCl3(s)

To determine the volume of chlorine gas needed given the mass of aluminum metal to be used, we need to calculate for the moles of chlorine needed and use a relation that relates moles and volume by assuming the gas to be an ideal gas. We use the equation PV =nRT. We calculate as follows:

7.85 g Al ( 1 mol / 26.98 g ) ( 3 mol Cl2 / 2 mol Al ) = 0.43643 mol Cl2

PV = nRT
V = nRT / P
V = 0.43643 (0.08205) (298) / (225/760)
V = 36.04 L chlorine gas

The minimum volume needed would be 36.04 L.

ANSWERED AT 18/11/2019 - 03:51 AM


QUESTION POSTED AT 18/11/2019 - 03:51 AM

A key step in the extraction of iron from its ore is feo(s) + co(g) fe(s) + co2(g) kp = 0.403 at 1000°c. this step occurs in the 700°c to 1200°c zone within a blast furnace. what are the equilibrium partial pressures of co(g) and co2(g) when 1.58 atm co(g) and excess feo(s) react in a sealed container at 1000°c?

The complete reaction of the problem, for better illustration, is

FeO(s) + CO(g) <--> Fe(s) + CO2(g)

The double-tailed arrow signifies that the reaction is in a dynamic chemical equilibrium. When the system is in equilibrium, the forward and the backward reaction rates have an equal ratio of Kp = 0.403 at 1000°C. The formula for Kp is

Kp = [partial pressure of products]/[partial pressure of reactants]

So, first, let's find the partial pressure of the compounds in the reaction.

                      FeO(s) + CO(g) <--> Fe(s) + CO2(g)
Initial                 x            1.58            0           0
Change           -1.58        -1.58       +1.58     +1.58
------------------------------------------------------------------
Equilbrium      x-1.58          0             1.58       1.58

Kp = [(1.58)(1.58)]/[(x-1.58)] = 0.403
x = 7.77 atm (this is the amount of excess FeO)

Therefore, the partial pressure of CO2 at equilibrium is 1.58 atm. There is no more CO because it has been consumed due to excess FeO.

ANSWERED AT 18/11/2019 - 03:45 AM


QUESTION POSTED AT 18/11/2019 - 03:45 AM

Write the net chemical equation for the production of aluminum from aluminum hydroxide and carbon. be sure your equation is balanced.

The net chemical equation for the production of aluminium from aluminium hydroxide and carbon is this:
4AL[OH]3 + 3C = 4AL + 6H2O + 3CO2.
Four molecules of aluminium hydroxide reacts with three molecules of carbon to produce four molecules of aluminium, six water molecules and three molecules of carbon dioxide. All the reactants and products are in solid state except water, which is given off as steam and carbon dioxide; these two are gases.

ANSWERED AT 18/11/2019 - 03:38 AM


QUESTION POSTED AT 18/11/2019 - 03:38 AM

When this reaction is coupled to the conversion of graphite to carbon dioxide, it becomes spontaneous. what is the chemical equation of this coupled process? show that the reaction is in equilibrium, include physical states, and represent graphite as c(s)?

The reaction of a graphite which is basically a carbon with oxygen which is a gas forms a carbon dioxide which is also a gas. 

The balance chemical reaction is shown below,
C + O2 --> CO2

This is already a balanced chemical equation because the number of elements in the left side of the equation is same that of the number on the right side.
C = 1 
O = 2

If we are to incorporate the physical states of the elements and compounds which are already given above, the equation would become,
C(s) + O2(g) --> CO2(g)

ANSWERED AT 18/11/2019 - 03:37 AM


QUESTION POSTED AT 18/11/2019 - 03:37 AM

How many grams of naoh are needed to make 750 ml of a 2.5 (w/v) solution?

Concentration: 2.5% w/v

Concentration w/v % = [mass of solute / volume of solution ] * 100

=> mass of solute = concentration w/v * volume of solution / 100

=> mass of solute = 2.5 * 750 ml / 100 = 18.75 grams

Also, using proportions you get to the same result:

2.5% w/v => 2.5 grams of NaOH / 100 ml of water

=> 2.5 grams of NaOH / 100 ml water = x grams of NaOH / 750 ml water

=> x grams of NaOH = 750 ml water * 2.5 grams of NaOH / 100 ml water

=> x = 18.75 grams of NaOH

Answer = 18.75 grams of NaOH

ANSWERED AT 18/11/2019 - 03:33 AM


QUESTION POSTED AT 18/11/2019 - 03:33 AM

Calculate the dipole moment for hf (bond length 0.917 å), assuming that the bond is completely ionic.

The dipole moment  =  [the charge magnitude at one of the ends (you can choose either one)] [the distance between charges]
It is usually calculated in SI units of Debye which represents the positive and negative charges separated by 0.2082 angestron

Thus,
dipole moment = 0.917/0.2082 = 4.4044 debye.

ANSWERED AT 18/11/2019 - 03:32 AM


QUESTION POSTED AT 18/11/2019 - 03:32 AM

The most probable mechanism of reactivity for the solvolysis of 2-chloro-norbornane is

The rate determining step for the reactivity for the solvolysis of 2-chloro-norbornane depends only on the decomposition of a single molecular species which is the 2-chloro-norbornane itself. For unimolecular reactions, the mechanism pathway being followed is that of an SN1 mechanism.

 

Answer:

SN1 mechanism

ANSWERED AT 18/11/2019 - 03:28 AM


QUESTION POSTED AT 18/11/2019 - 03:28 AM

Calculate the mass of o2 produced by the decomposition of kclo3 by heating in the presence of manganese dioxide

The balanced chemical reaction would be expressed as follows:

2KClO3 = 2KCl + 3O2

To determine the mass of oxygen gas that is produced from a given amount of potassium chlorate, we use the initial amount of the reactant and use the relations from the reaction of the substances involved. However, from the problem, the amount of reactant is not given. We assume 1 g of potassium chlorate.

1 g KClO3 ( 1 mol / 122.55 g ) ( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 0.3917 g of O2 produced.

If the given amount of the reactant is not 1 g, then you can change the value of the mass from the calculation and evaluate the new value of oxygen gas produced. 

ANSWERED AT 18/11/2019 - 03:24 AM


QUESTION POSTED AT 18/11/2019 - 03:24 AM

A compound contains 40.0% c, 6.71% h, and 53.29% o by mass. the molecular weight of the compound is 60.05 amu. the molecular formula of this compound is

To determine the molecular formula of the compound given, we first need to determine the empirical formula. To do that we assume to have 100 grams sample of the compound with the given composition. Then, we calculate for the number of moles of each element. We do as follows:
         mass       moles
C       40.0        3.33
H       6.71        6.64
O      53.29       3.33

Dividing the number of moles of each element with the smallest value, we will have the empirical formula:

CH2O

To determine the molecular formula, we multiply a value to the empirical formula. Then, calculate the molar mass and see whether it is equal to the one given (60.05 g/mol).

              molar mass 
CH2O      30.03
C2H4O2  60.06

Most likely, the molecular formula of the compound would be C2H4O2.

ANSWERED AT 18/11/2019 - 03:20 AM


QUESTION POSTED AT 18/11/2019 - 03:20 AM

How many moles of nacl are required to prepare 0.80 l of 6.4 m nacl? 0.13 mol nacl 5.1 mol nacl 7.2 mol nacl 8.0 mol nacl?

NaCl = n/V 

6.4M = n/0.80L 

n= 5.12mol

ANSWERED AT 18/11/2019 - 03:20 AM


QUESTION POSTED AT 18/11/2019 - 03:20 AM

A chemist determines by measurements that 0.030 moles of nitrogen gas participate in a chemical reaction. calculate the mass of nitrogen gas that participates.

Remember that: 
number of moles = mass/molar mass

First, we get the molar mass of the nitrogen gas molecule:
It is known the the nitrogen gas is composed of two nitrogen atoms, each with molar mass 14 gm (from the periodic table)
Therefore, molar mass of nitrogen gas = 14 x 2 = 28 gm

Second we calculate the mass of the precipitate:
we have number of moles = 0.03 moles (given)
and molar mass = 28 gm (calculated)
Using the equation mentioned before,
mass = number of moles x molar mass = 0.03 x 28 = 0.84 gm

ANSWERED AT 18/11/2019 - 03:19 AM


QUESTION POSTED AT 18/11/2019 - 03:19 AM

How many moles of libr are contained in 347 g of water in a 0.175 m libr solution?

Molarity = number of moles of solute/Kg of solvent

Molarity is given in the problem as 0.175 m
Mass of solvent = 347 gm = 0.347 Kg

Therefore,
0.175 = number of moles of LiBr / 0.347
number of moles of LiBr = 0.175 x 0.347 = 0.060725 moles

ANSWERED AT 18/11/2019 - 03:17 AM


QUESTION POSTED AT 18/11/2019 - 03:17 AM

A certain weak acid, ha, has a ka value of 8.4×10−7. calculate the percent ionization of ha in a 0.10 m solution

To determine the percent ionization of the acid given, we make use of the acid equilibrium constant (Ka) given. It is the ratio of the equilibrium concentrations of the dissociated ions and the acid. The dissociation reaction of the HF acid would be as follows:

HA = H+ + A-

The acid equilibrum constant would be expressed as follows:

Ka = [H+][A-] / [HA] = 8.4 x 10^-7

To determine the equilibrium concentrations we use the ICE table,
         HF             H+              A-
I      0.10           0                 0
C      -x              +x               +x
---------------------------------------------
E    0.10-x        x                   x 

8.4 x 10^-7= [H+][A-] / [HA] 
8.4 x 10^-7 = [x][x] / [0.10-x] 

Solving for x,

x = 0.0002894 = [H+] = [A-]

percent ionization = 0.0002894 / 0.10 x 100  = 0.289%

ANSWERED AT 18/11/2019 - 03:16 AM


QUESTION POSTED AT 18/11/2019 - 03:16 AM

Calculate the ph of a 0.20 m solution of kcn at 25.0 ∘c. express the ph numerically using two decimal places

The pH of the solution is basically the negative logarithm of the concentration of hydrogen ions or H+. In equation form, pH = -log[H+]. It could also be in terms of the concentration of hydroxide ions or OH- as pOH, where pOH = -log[OH-]. The sum of pH and pOH is 14. These are the important equations to know when it comes to equilibrium pH problems.

KCN is a basic salt coming from the reaction of a weak acid, HCN, and a strong base, KOH. In the hydrolysis of KCN, only the strong conjugate base (SCB) is involved. Since HCN is the weak acid, the SCB is CN-. The reaction would be

CN- + H2O ⇔ HCN + OH-

The important data is the equilibrium constant of acidity of the weak acid. Ka for HCN is 6.2×10^-10. Then, let's do the ICE(Initial-Change-Equilibrium) analysis.

          CN-    +    H2O    ⇔    HCN +    OH-

I      0.2 m             ∞                   0             0
C      -x                 ∞                   +x        +x
-----------------------------------------------------------
E      0.2-x                               +x         +x

The value x denotes the number of moles CN- reacted. There is no value for H2O because the solution is dilute such that H2O>>>CN-. Then, we apply the ratio:

 K_{H} = \frac{ K_{W} }{ K_{A} } = \frac{[HCN][OH-]}{[CN-]}

where K,H is the equilibrium constant of hydrolysis and Kw is equilibrium constant for water solvation which is equal to 1×10^-14. Therefore,

K_{H} = \frac{ 1x10^-14}{ 6.2x10^-10 } = \frac{[X][X]}{[0.2-X]}

x = 0.001788 m
Since the value of OH- is also x, then OH-=0.001788 m. Consequently,

pOH = -log(0.001788) = 2.75
pH = 14 - pOH = 14 - 2.75
pH = 11.25

ANSWERED AT 18/11/2019 - 03:14 AM


QUESTION POSTED AT 18/11/2019 - 03:14 AM

1.64 mol of nickel at 150.13ºC is placed in 1.00 L of water at 25.09ºC. The final temperature of the nickel-water mixture is 26.34ºC. What is the specific heat of nickel? a. 0.439 J K-1 mol-1 b. 0.554 J K-1 mol-1 c. 2.28 J K-1 mol-1 d. 25.7 J K-1 mol-1 e. 31.6 J K-1 mol-1

Heat gained or released by a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:

Heat = mC(T2-T1) 

When two objects are in contact, it should be that the heat lost is equal to what is gained by the other. From this, we can calculate the what is being asked in the problem. We do as follows:

Heat gained = Heat lost
m C (Tf-Ti) = - m C (Tf-Ti) 
Since what is aksed is the specific heat capacity in units of J/mol-K, we change m into n which is the number of moles of a substance.
1.64 C (150.13 - 26.34) = (1000 /18.02) (4.18 x 18.02 )(26.34 - 25.09)
C = 25.7 J / mol-K <----------OPTION D

ANSWERED AT 18/11/2019 - 03:12 AM


QUESTION POSTED AT 18/11/2019 - 03:12 AM

How many molecules of n2 gas can be present in a 2.5 l flask at 50c and 650 mmhg?

First step is to change temperature and pressure into SI units:
50 degree celcius = 50 + 273 = 323 degree kelvin
650 mmHg = 650/760 = 0.855 atm

Second step is to calculate the number of moles of N2 that can be found under the mentioned conditions using the rule:
PV = nRT where:
P is the pressure, V is the volume, n is the number of moles, R is the gas constant (= 0.08206) and T is the temperature.
Substituting in the equation we get:
0.855 x 2.5 = n x 0.08206 x 323
n = 0.0806 moles

Third step is to get the number of molecules in 0.0806 moles by multiplying the number of moles by Avogadro's number:
number of molecules = 0.0806 x 6.02 x 10^23 = 0.485212 x 10^23 molecule

ANSWERED AT 18/11/2019 - 03:12 AM


QUESTION POSTED AT 18/11/2019 - 03:12 AM

Sheila's measured glucose level one hour after a sugary drink varies according to the normal distribution with μμ = 120 mg/dl and σσ = 20 mg/dl. what is the level l such that there is probability only 0.15 that the mean glucose level of 5 test results falls above l?

Since the sample size is below 30, in this case we use the t statistic. The formula for t score is:

t = (x – u) / (σ / sqrt n)

where,

x = the level l = unknown

u = sample mean = 120 mg / dl

σ = standard deviation = 20 mg / dl

n = sample size or number of results = 5

Using the standard distribution tables for t, we can find the value of t given the probability (P = 0.15) and degrees of freedom (DOF).

t  = 1.036

Going back to the formula for t score:

1.036 = (x – 120) / (20 / sqrt 5)

x = 129.27 mg / dl = l

ANSWERED AT 18/11/2019 - 03:11 AM


QUESTION POSTED AT 18/11/2019 - 03:11 AM

How many orbitals are there in the n = 4 level of the h-atom?

The H-atom has fourth energy level with n=4.

This energy level can have values of orbital angular momentum l = 0, 1, 2, 3

Now each l can also have magnetic momentum ml from – l to l.

Therefore,

l = 0

ml = 0

l = 1

ml = -1, 0, 1

l = 2

ml = -2, -1, 0, 1, 2

l = 3

ml = -3, -2, -1, 0, 1, 2, 3

Now adding up all the number of ml’s will give us the total number of orbitals:

orbitals = 1 + 3 + 5 + 7

orbitals = 16

 

Alternatively, we can simply use the formula:

orbitals = n^2 = 4^2 = 16

ANSWERED AT 18/11/2019 - 03:11 AM


QUESTION POSTED AT 18/11/2019 - 03:11 AM