1. List the ketose carbon(s) that is(are) reduced with NaBH4.

Answer:

d

Explanation:

since it is much convenient since the email will not get lost and it's contents will not be forgotten

Answer:

a) E2

b) SN2

c) SN2

Explanation:

A substitution reaction involves replacement of an atom or group in a molecule by another atom or group. An elimination reaction is the loss of two atoms from the same molecule leading to the formation of a multiple bond in the molecule.

We must note that primary alkyl halides never undergo SN1/E1 reactions. However, the presence of a strong bulky base such as tert BuO- , E2 reactions predominate. In the presence of strong bases such as OH^- and good nucleophiles such as I^-, SN2 mechanism predominates.

Finding Nemo??? sorry I really need these points

Answer:

2-nonyne

explanation:

First consider the type of bonds, there are tripple bonds of carbon to carbon, in position 2 from the right.

hence it us alkyn.

There are 9 carbons.

Answer:

[tex]\boxed {\boxed {\sf 0.75 \ mol \ Li}}[/tex]

Explanation:

We are asked to convert 5.2 grams of lithium to moles of lithium.

To convert from grams to moles, we need the molar mass. This is the measurement of the mass in 1 mole of a substance. It can be found on the Periodic Table because it is the same value as the atomic mass, but the units are grams per mole instead of atomic mass units.

Look up the molar mass of lithium.

Create a ratio using the molar mass of lithium.

[tex]\frac { 6.94 \ g \ Li}{ 1 \ mol \ Li}[/tex]

Multiply by the value we are converting: 5.2 grams of lithium.

[tex]5.2 \ g \ Li *\frac { 6.94 \ g \ Li}{ 1 \ mol \ Li}[/tex]

Flip the ratio so the units of grams of lithium cancel.

[tex]5.2 \ g \ Li *\frac{ 1 \ mol \ Li} { 6.94 \ g \ Li}[/tex]

[tex]5.2 *\frac{ 1 \ mol \ Li} { 6.94 }[/tex]

[tex]\frac{5.2} { 6.94 } \ mol \ LI[/tex]

[tex]0.749279538905 \ mol \ Li[/tex]

The original measurement of grams (5.2) has 2 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place. The 9 in the thousandths place to the right tells us to round the 4 up to a 5.

[tex]0.75 \ mol \ Li[/tex]

5.2 grams of lithium is equal to 0.75 moles of lithium atoms.

We measured the Fe(II) reduction of one of the Co(III) complexes by water at a rate of about 0.545 kJ/mol (to three significant figures).

Calculating a Reaction's Activation Energy A reaction's rate is influenced by the temperature at which it is carried out. The molecules travel more quickly and clash more frequently as the temperature rises. Moreover, the molecules contain greater kinetic energy.

We can use the Arrhenius equation to calculate the reaction's activation energy:

k = A × exp(-Ea/RT)

When the activation energy Ea, the rate constant k, the gas constant R, and the temperature T in Kelvin are all present.

Finding the natural logarithm of the equation's two sides results in:

ln(k) = ln(A) - (Ea/RT)

This equation can be rearranged to take a linear form:

ln(k) = (-Ea/R) × (1/T) + ln(A)

y = mx + b, where (1/T) is x, (-Ea/R) is the slope, and ln(A) is the y-intercept, has the form of a linear equation.

We can get the slope of the line using the given data:

slope = (-Ea/R) = (ln(k2/k1)) / (1/T2 - 1/T1)

where the rate constants for temperatures T1 and T2, respectively, are k1 and k2.

substituting the specified values:

k1 = 0.054s⁻¹ at 293 K

k2 = 0.100s⁻¹ at 298 K

T1 = 293 K

T2 = 298 K

slope = (-Ea/R)

= (ln(0.100/0.054)) / (1/298 - 1/293)

= 65.5 kJ/mol

Therefore, the activation energy of the reaction is:

Ea = slope * R = 65.5 kJ/mol × 8.314 J/mol-K = 545 J/mol

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The mass of radon-222 that will remain after 15.2 days given that it was originally 6 g is 0.375 g (Option B)

This is the time taken for half a substance to decay.

We'll begin our calculation by calculating the number of half-lives that has elapsed after 15.2 days. This is illustrated below:

n = t / t½

n = 15.2 / 3.8

n = 4

Thus, 4 half-lives has elapsed.

The amount of radon-222 remaining can be obtained as illustrated below:

N = N₀ / 2ⁿ

N = 6 / 2⁴

N = 6 / 16

N = 0.375 g

Thus, the amount of radon-222 remaining after 15.2 days is 0.375 g

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Answer:

An atom is the smallest particle of an element that can take part in chemical reaction.

Explanation:

hope it will help u Amri

Answer:

A BEC ( Bose - Einstein condensate ) is a state of matter of a dilute gas of bosons cooled to temperatures very close to absolute zero is called BEC.

Examples - Superconductors and superfluids are the two examples of BEC.

Explanation:

Explanation:

here's the answer to your question

2.0 L

Answer:

Solution given:

no. of mole(n)=3.0mole

molarity(M)=1.5M

we have

Volume of a substance is equal to the ratio of its no of mole to its molarity.

By this we get a relation:

Volume=no.of mole/molarity

substituting value

Volume=3.0/1.5=2

The required volume is 2 litre.

Answer:

This is an acid-base reaction (neutralization): Ba(OH) 2 is a base, H 2SO 4 is an acid. This is a precipitation reaction: BaSO 4 is the formed precipitate.

Answer:

3.310308*10^26

Explanation:

nO=9nFe2(SO3)3=9*60.1=540.9 moles

number of atoms: 540.9*6.02*10^23

Water, mercury chloride and nitrogen oxide.

Water, mercury chloride and nitrogen oxide will present in the precipitate when we slowly add a solution of mercury(II) nitrate to a solution of aqueous hydrochloric acid having halide ions both in equal concentrations. The equation of this reaction is Hg2(NO3)2 + 4 HCl ----> 2 HgCl2 + 2 H2O + 2 NO so it is concluded that from this reaction we get precipitate of water, mercury chloride and nitrogen oxide.

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Answer:

% NiCl2 = 24.13%

% water = 78.57%

Explanation:

Mass percentage = mass of solute/mass of solution × 100

According to this question, a solution contains 159 g of NiCl2 in 500 g of water. Hence, mass of the solution is calculated as follows:

Mass of solution = 159g + 500g

Mass of solution = 659g

Therefore;

A) % Mass of NiCl2 in solution = mass of NiCl2/mass of solution × 100

% Mass of NiCl2 in solution = 159/659 × 100

% Mass of NiCl2 in solution = 0.2413 × 100

= 24.13%

B) % Mass of water in solution = mass of water/mass of solution × 100

% Mass of water in solution = 500/659 × 100

% Mass of water in solution = 0.7587 × 100

% Mass of water in solution = 75.87%

Answer:

C . Kinetic Molecular Theory

Answer:

pH = 3.02

Explanation:

Acetic Acid is a weak acid (HOAc) that ionizes only ~1.5% as follows:

HOAc ⇄ H⁺ + OAc⁻.

In pure water the hydronium ion concentration [H⁺] equals the acetate ion concentration [OAc⁻] and can be determined* using the formula [H⁺] = [OAc⁻] = SqrRt(Ka·[acid]) = SqrRt(1.8x10⁻⁵ x 0.0500)M = 9.5x10⁻⁴M.

By definition, pH = -log[H⁺] = -log(9.5x10⁻⁴) = 3.02

______________________________________________________

*This formula can be used to determine the [H⁺] & [Anion⁻] concentrations for any weak acid in pure water given its Ka-value and the molar concentration of acid in solution.

Answer:

11.76 g/cm^3

Explanation:

Mass of empty flask and stopper = 64.232g

Mass of flask filled with water = 153.617 g

Mass of water = 153.617 g - 64.232g = 89.385 g

Mass of flask, stopper and metal = 143.557 g

Mass of metal = 143.557 g - 64.232g = 79.325 g

Mass of water, flask, stopper and metal = 226.196 g

Mass of water = 226.196 g - 143.557 g = 82.639 g

Since mass of water =volume of water

Volume occupied by metal = 89.385 cm^3 - 82.639 cm^3 = 6.746 cm^3

Density of metal = mass/volume = 79.325 g/6.746 cm^3

= 11.76 g/cm^3

Explanation:

The required concentration of [tex]HNO_3[/tex] M1 =0.222 M.

The required volume of [tex]HNO_3[/tex] is V1 =225 mL.

The standard solution of [tex]HNO_3[/tex] is M2 =16 M.

The volume of standard solution required can be calculated as shown below:

Since the number of moles of solute does not change on dilution.

The number of moles [tex]n=molarity * volume[/tex]

[tex]M_1.V_1=M_2.V_2[/tex]

[tex]V2=\frac{M_1.V_1}{M_2} \\=0.222M x 225 mL / 16 M\\=3.12 mL[/tex]

Hence, 3.12 mL of 16 m nitric acid is required to prepare 0.222 M and 225 mL of nitric acid.

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[tex]SilentNature[/tex]

Answer:

B. 0.0000005461m

I used the method of moving the decimal.

Answer:

23592U+10n→14454Xe+9038Sr+210n

Explanation:

a nuclear reaction for the neutron-induced fission of U−235 to form Xe−144 and Sr−90.

Answer:

A. filtration

Hope it helps

In the original solution you have the mixture of a weak acid (Hypochlorous acid) and its conjugate base (Sodium hypochlorite). That is a buffer.

The barium hydroxide will react with hypochlorous acid. If this reaction cause the complete reaction of hypochlorous acid, the buffer break its capacity and the pH change in several units. In this case:

The addition of barium hydroxide will raise the pH slightly because the buffer still working.

The initial moles of those species are:

Hypochlorous acid:

[tex]1L * \frac{0.370mol}{1L} = 0.370 moles[/tex]

Sodium hypochlorite:

[tex]1L * \frac{0.493mol}{1L} = 0.493 moles[/tex]

Now, a strong acid as barium hydroxide (Ba(OH)₂) reacts with a weak acid as hypochlorous acid (HClO) as follows:

Ba(OH)₂ + 2HClO → Ba(ClO)₂ + 2H₂O

For a complete reaction of 0.092 moles of barium hydroxide are required:

[tex]0.092 moles Ba(OH)_2*\frac{2mol HClO}{1molBa(OH)_2} = 0.184 moles HClO[/tex]

As there are 0.370 moles, the moles of HClO after the reaction are:

0.370 moles - 0.184 moles = 0.186 moles of HClO will remain

As you still have hypochlorite and hypochlorous acid you still have a buffer.

Thus, the pH will raise slightly because the amount of acid is decreasing and slightly because the buffer can keep the pH.

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Complete Question

A sample of aluminum, which has a specific heat capacity of 0.897 JB loc ! is put into a calorimeter (see sketch at right) that contains 200.0 g of water. The aluminum sample starts off at 85.6 °C and the temperature of the water starts off at 16.0 °C. When the temperature of the water stops changing it's 20.1 °C. The pressure remains constant at 1 atm. Calculate the mass of the aluminum sample.

Answer:

[tex]M=58g[/tex]

Explanation:

From the question we are told that:

Heat Capacity [tex]H=0.897[/tex]

Mass of water [tex]M=200g[/tex]

Initial Temperature of Aluminium [tex]T_a=85.6[/tex]

Initial Temperature of Water [tex]T_{w1}=16.0[/tex]

Final Temperature of Water  [tex]T_{w2}=16.0[/tex]

Generally

Heat loss=Heat Gain

Therefore

[tex]M*0.897*(85.6-20.1) =200*4.184*(20.1-16)[/tex]

[tex]M=58g[/tex]

An indicator usually signals the endpoint of a neutralization reaction by undergoing a color change. They aid in discovering the point of equivalence of a titration.

The kind of indicator used depends on the nature of acid/base reacted.

In the case of  CH3NH2 with HBr which  strong acid and weak base titration, suitable indicators include; bromophenol blue, methyl  orange, methyl red, and chlorophenol blue.

In the case of HNO3 with NaOH, this is a strong acid, strong base titration hence phenolphthalein, methyl red, chlorophenol, and bromothymol blue cresol red blue are suitable indicators.

In the case of  HNO2 with KOH, this a weak acid, strong base titration and the suitable indicators are cresol red and  phenolphthalein.  

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Answer:

it is 11.55 and ik because I just had that question

18.0 grams of copper(II) phthalocyanine would be produced by the complete cyclotetramerization of 0.125 moles of phthalonitrile in the presence of excess copper(II) chloride.

Let's consider the following balanced equation.

4 C₈H₄N₂(l) + CuCl₂(s) → Cu(C₃₂H₁₆N₈)(s) + Cl₂(g)

The molar ratio of C₈H₄N₂ to Cu(C₃₂H₁₆N₈) is 4:1. The moles of Cu(C₃₂H₁₆N₈) produced from 0.125 moles of C₈H₄N₂ are:

[tex]0.125 mol C_8H_4N_2 \times \frac{1molCu(C_{32}H_{16}N_8)}{4mol C_8H_4N_2} = 0.0313 molCu(C_{32}H_{16}N_8)[/tex]

The molar mass of Cu(C₃₂H₁₆N₈) is 576.07 g/mol. The mass corresponding to 0.0313 moles of Cu(C₃₂H₁₆N₈) is:

[tex]0.0313 moles \times \frac{576.07g}{mol} = 18.0 g[/tex]

18.0 grams of copper(II) phthalocyanine would be produced by the complete cyclotetramerization of 0.125 moles of phthalonitrile in the presence of excess copper(II) chloride.

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Answer:

Part A

HCOOH(aq) + NaOH(aq) → HCOONa(aq) + H2O(l)

Part B

ClCH2CH2CO2H(aq) + NaOH(aq) ------> ClCH2CH2CO2Na(aq) + H2O(l)

Explanation:

The reaction between an alkanoic acid and a base is a neutralization reaction. The reaction occurs as follows;

RCOOH + NaOH ----> RCOONa + H2O

We have to note the fact that the net ionic reaction still remains;

H^+(aq) + OH^-(aq) ---> H2O(l)

In both cases, the reaction can occur and they actually do occur as written.

Answer:

Fe(NO3)3, Cr(NO3)3, Co(NO3)3

Explanation:

According to the question, no precipitate is observed when HCl was added. This means that we must rule out AgNO3.

Again, the sulphides of Cu^2+, Bi^3+ are soluble in acidic medium but according to the question, the sulphides do not precipitate at low pH hence Cu(NO3)2 and Bi(NO3)3 are both ruled out.

The sulphides of Fe^3+, Cr^3+ and Co^3+ all form precipitate in basic solution hence Fe(NO3)3, Cr(NO3)3, Co(NO3)3 may be present.

The presence of Ca(NO3)2 and KNO3 may be confirmed by flame tests.

I think the answer most be d

In chemistry like dissolves like hence hexane will dissolve in gasoline.

Dissolution stems from intermolecular interaction between solute and solvent molecules.

If this interaction can not occur, dissolution of one substance in another is impossible.

Hexane dissolves in gasoline because the both substances are non-polar and can interact with each other effectively.

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