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3 years ago

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Chemistry

2 answers:

antiseptic1488 [7]3 years ago

5 0

Answer:

what is your question friend

telo118 [61]3 years ago

5 0

DUDE WHAT'S MUSKAN?LIKE BRO?

You might be interested in

How many grams of NH3 can be produced from the reaction of 17.8 moles of H2 and a sufficient supply of N2? N2 + 3 H2 → 2 NH3

alexandr402 [8]

Answer:

201.73g

Explanation:

From the chemical reaction above, it can be seen that 3 moles of hydrogen yielded 2 moles of ammonia.

Now, 17.8 moles of hydrogen will yield a certain number of moles of ammonia. To get this certain number of moles: we have (17.8 * 2)/3 = 11.87 moles

Hence, 17.8 moles of hydrogen will produce 11.87 moles of ammonia. Now, we proceed to calculate the mass of ammonia produced.

The mass of ammonia produced is the molar mass of ammonia multiplied by the number of moles of ammonia.

The molar mass of ammonia is 14 +3(1) = 17g/mol

Now the mass produced is thus 17 * 11.87 = 201.73g

6 0

4 years ago

Select the keyword or phrase that will best complete each sentence.

pishuonlain [190]

Answer:

1) acetylide

2) enol

3) aldehydes

4) tautomers

5) alkynes

6) Hydroboration

7) Keto

8) methyl ketones

Explanation:

Acetylide anions (R-C≡C^-) is a strong nucleophile. Being a strong nucleophile, we can use it to open up an epoxide ring by SN2 mechanism. The attack of the acetylide ion occurs from the backside of the epoxide ring. It must attack at the less substituted side of the epoxide.

Oxomercuration of alkynes and hydroboration of alkynes are similar reactions in that they both yield carbonyl compounds that often exhibit keto-enol tautomerism.

The equilibrium position may lie towards the Keto form of the compound. Usually, if terminal alkynes are used, the product of the reaction is a methyl ketone.

3 0

3 years ago

The involuntary muscles of the arteries are controlled by the_________ system. respiratory,skeletal,nervous,circulaory

Soloha48 [4]

The involuntary muscles of the arteries are controlled by the nervous system. The correct option among all the options that are given in the question is the third option or the penultimate option. these muscles are not directly under the control of any person and so are called involuntary muscles. I hope the answer helps you.

3 0

3 years ago

Determine the oxidation numbers of all the elements in the unbalanced reactions. Then, balance each redox reaction in a basic so

givi [52]

Answer for 1:

- Oxidation number of Mn in MnO4^-1= +7

- Oxidation number of O in MnO4^-1= -2

- Oxidation number of C in C2O4^-2= +3

- Oxidation number of O in C2O4^-2= -2

- Oxidation number of Mn in MnO2= +4

- Oxidation number of O in MnO2= -2

- Oxidation number of C in CO2= +4

- Balanced redox equation in basic solution:

$3C_2O_4^{-2}+2MnO_4^{-1}+4H_2O\operatorname{\rightarrow}6CO_2\text{ +}2MnO_2\text{ + }8OH^-$

Explanation for 1:

1st) To determine the oxidation numbers, it is necessary that the total sum of the charges is equal to that of the molecule or ion in the equation.

• Oxidation numbers in MnO4-,:

We know that the oxidation number for oxygen is -2, then we multiply it by the subscript 4 to find the whole charge of oxygen in this molecule.

Oxygen oxidation number: -2 x 4 = -8

Since the total charge of the molecule is -1, by difference, we will know that the oxidation number of Mn will be +7:

$\begin{gathered} Mn^{+7}O_4^{-2} \\ +7+[(-2*4)]=-1 \end{gathered}$

To confirm that the oxidation number that we determined exists for that element, we can check the Periodic Table of Elements.

We proceed in the same way with all molecules.

• Oxidation numbers in C2O4-2,:

$\begin{gathered} C_2^{+3}O_4^{-2} \\ (+3*2)+[(-2)*4]= \\ +6-8=-2 \end{gathered}$

• Oxidation numbers in MnO2,:

$\begin{gathered} Mn^{+4}O_2^{-2} \\ +4+[(-2)*2]= \\ +4-4=0 \end{gathered}$

• Oxidation numbers in CO2,:

$\begin{gathered} C^{+4}O_2^{-2}_ \\ +4+[(-2)*2]= \\ +4-4=0 \end{gathered}$

2nd) Now that we know the oxidation number os each atom in the reaction, then we can find the element that is oxidized and the element that is reduced.

We can see that Mn goes from +7 to +4, the Mn atom is reduced. And, the carbon atom goes from +3 to +4 so it oxidizes.

3rd) It is necessary to write the oxidation reaction and the reduction reaction separately and balancing all elements except oxygen and hydrogen:

Oxidation:

$C_2O_4^{-2}\text{ }\rightarrow\text{ 2}CO_2\text{ + 2}e^-_$

Reduction:

$MnO_4^{-1}\text{ + 3}e^-\rightarrow MnO_2$

4th) Since the reaction occurs in a basic solution, we must add water (H2O) to balance the oxygen atoms and hydroxyl ion (OH-) to balance the hydrogen atoms. In this case, the reduction reaction is the only one that needs to be balanced with water and hydroxyl ion.

$MnO_4^{-1}\text{ + 3}e^-\text{ +2}H_2O\operatorname{\rightarrow}MnO_2\text{ + 4}OH^-$

5th) It is necessary to balance the electrons in each half-reaction. So, we multiply each half-reaction by the number of electrons in the other half-reaction:

Oxidation:

$\begin{gathered} (C_2O_4^{-2}\operatorname{\rightarrow}\text{2}CO_2\text{ + 2}e_^-)*3 \\ 3C_2O_4^{-2}\operatorname{\rightarrow}6CO_2\text{ + 6}e^- \end{gathered}$

Reduction:

$\begin{gathered} (MnO_4^{-1}\text{ + 3}e^-\text{ + 2}H_2O\operatorname{\rightarrow}\text{ MnO}_2\text{ }+4OH^-)*2 \\ 2MnO_4^{-1}\text{ + 6}e^-\text{ + 4}H_2O\operatorname{\rightarrow}\text{ 2MnO}_2\text{ }+8OH^- \end{gathered}$

6th) We need to cancel out everything that is repeated on opposite sides of the reactions, including the electrons.

Finally, we can write the balanced redox equation:

$3C_2O_4^{-2}+2MnO_4^{-1}+4H_2O\operatorname{\rightarrow}6CO_2\text{ +}2MnO_2\text{ + }8OH^-$

6 0

1 year ago

What will create a insoluble salt with a potassium ion

navik [9.2K]

Explanation:

1) The dissolution of the salt potassium sulfite:

K₂SO₃(aq) → 2K⁺(aq) + SO₃²⁻(aq).

Potassium has +1 charge because it lost one electron to accomplish stabile electron configuration of noble gas argon.

2) From dissolution reaction: n(K⁺) : n(SO₃²⁻) = 2 : 1.

n(K⁺) = 0.700 mol.

0.700 mol : n(SO₃²⁻) = 2 : 1.

n(SO₃²⁻) = 0.700 mol ÷ 2.

n(SO₃²⁻) = 0.350 mol; amount of sulfite anions.

4 0

3 years ago

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