The correct geometry around oxygen in CH3OCH3 is
(a). linear. (b). bent. C). tetrahedral/(a). trigonal planar​

Using boyles law

[tex]\boxed{\sf v\propto \dfrac{1}{p}}[/tex]

[tex]\\ \sf\longmapsto P_1V_1=P_2V_2[/tex]

[tex]\\ \sf\longmapsto P_2=\dfrac{P_1V_1}{V_2}[/tex]

[tex]\\ \sf\longmapsto P_2=\dfrac{2.8\times 1.5}{0.92}[/tex]

[tex]\\ \sf\longmapsto P_2=\dfrac{4.2}{0.92}[/tex]

[tex]\\ \sf\longmapsto P_2=4.56atm[/tex]

[tex]\\ \sf\longmapsto P_2\approx 4.6atm[/tex]

Answer:

[tex]\boxed {\boxed {\sf 4.6 \ atm}}[/tex]

Explanation:

We are asked to find the new pressure given a change in volume. We will use Boyle's Law, which states the volume of a gas is inversely proportional to the pressure. The formula for this law is:

[tex]P_1V_1= P_2V_2[/tex]

Initially, the gas occupies 1.5 liters at a pressure of 2.8 atmospheres.

[tex]1.5 \ L * 2.8 \ atm = P_2V_2[/tex]

The volume is changed to 0.92 liters, but the pressure is unknown.

[tex]1.5 \ L * 2.8 \ atm = P_2* 0.92 \ L[/tex]

We are solving for the final pressure, so we must isolate the variable P₂. It is being multiplied by 0.92 liters. The inverse operation of multiplication is division, so we divide both sides by 0.92 L.

[tex]\frac {1.5 \ L * 2.8 \ atm}{0.92 \ L} = \frac{P_2* 0.92 \ L}{0.92 \ L}[/tex]

[tex]\frac {1.5 \ L * 2.8 \ atm}{0.92 \ L}= P_2[/tex]

The units of liters cancel each other out.

[tex]\frac {1.5 * 2.8 \ atm}{0.92 }=P_2[/tex]

[tex]\frac {4.2}{0.92} \ atm= P_2[/tex]

[tex]4.565217391 \ atm = P_2[/tex]

The original measurements of pressure and volume have 2 significant figures, so our answer must have the same. For the number we calculated, that is the tenths place. The 6 in the hundredth place tells us to round the 5 up to a 6.

[tex]4.6 \ atm \approx P_2[/tex]

The pressure is approximately 4.6 atmospheres.

Answer:

3

Explanation:

The electron shells are labelled as K,L,M,N,O,P, and Q or 1,2,3,4,5,6, and 7.

As we go from innermost shell outwards, this number denotes the number of subshell in the shell. Electrons in outer shells have higher average energy and travel farther from the nucleus than those in inner shells.

Hence, M shell contains s,p and d subshells.

Answer:

(B). it's metallic bonding

Answer:

Explanation:

maleic acid ⇒ fumaric acid

ΔHreaction = ΔHproduct - ΔHreactant

ΔHproduct = -1336.0 kJ mol⁻¹

ΔHreactant = - 1359.2 kJ mol⁻¹.

ΔHreaction = -1336.0 kJ mol⁻¹ - ( - 1359.2 kJ mol⁻¹.)

=   1359.2 kJ mol⁻¹   -1336.0 kJ mol⁻¹

= 23.2 kJ mol⁻¹ .

Enthalpy of isomerization from maleic to fumaric acid is 23.2 kJ per mol.

The number of replaceable electrons in an atom is called its valency.

Examples

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Answer:

the combining capacity if an atom is know as valency.

the property of an element that determines the number of other atimd with an aton if the element can combine.

Answer:

c. C3H9N2O2

Explanation:

The empirical formula of a compound is defined as the simplest whole number ratio of atoms present in a molecule. To solve this question we need to convert the mass of each atom to moles. With the moles we can find the ratio as follows:

Moles N -Molar mass: 14.01g/mol-

0.420g N * (1mol/14.01g) = 0.0300 moles N

Moles O -Molar mass: 16g/mol-

0.480g O * (1mol/16g) = 0.0300 moles O

Moles C -Molar mass: 12.01g/mol-

0.540g C * (1mol/12.01g) = 0.0450 moles C

Moles H -Molar mass: 1.0g/mol-

0.135g H * (1mol/1g) = 0.135moles H

Dividing in the moles of N (Lower number of moles) the ratio of atoms is:

N = 0.0300 moles N / 0.0300 moles N = 1

O = 0.0300 moles O / 0.0300 moles N = 1

C = 0.0450 moles C / 0.0300 moles N = 1.5

H = 0.135 moles H / 0.0300 moles N = 4.5

As the empirical formula requires whole numbers, multiplying each ratio twice:

N = 2, O = 2, C = 3 and H = 9

And the empirical formula is:

Answer:

a) [tex]K_{2} S[/tex] and [tex]NH_{4} Cl[/tex] :

There are no insoluble precipitate forms.

b) [tex]Ca Cl_{2}[/tex] and [tex](NH_{4} )_{2} Co_{3}[/tex] :

There are the insoluble precipitates of [tex]CaCo_{3}[/tex]  forms.

c) [tex]Li_{2}S[/tex] and [tex]MnBr_{2}[/tex] :

There are the insoluble precipitates of [tex]MnS[/tex]  forms.

d) [tex]Ba(No_{3} )_{2}[/tex] and [tex]Ag_{2} So_{4}[/tex] :                        

As [tex]Ag_{2} So_{4}[/tex] is insoluble, therefore no precipitate forms.

e) [tex]Rb_{2}Co_{3}[/tex] and [tex]NaCl[/tex]:

There are no insoluble precipitates forms.

Explanation:

a)

Solubility rule suggests:- [tex]K_{2} S[/tex] ⇒ soluble, [tex]NH_{4} Cl[/tex] ⇒ soluble.

                                          KCl ⇒ soluble, [tex](NH_{4})_{2} S[/tex]  ⇒ soluble.

There are no insoluble precipitate forms.

b)

Solubility rule suggests:- [tex]Ca Cl_{2}[/tex] ⇒ soluble, [tex](NH_{4} )_{2} Co_{3}[/tex] ⇒ soluble.

                                        [tex]CaCo_{3}[/tex] ⇒ insoluble, [tex]NH_{4} Cl[/tex]  ⇒ soluble.

There are the insoluble precipitates of [tex]CaCo_{3}[/tex]  forms.

c)

Solubility rule suggests:- [tex]Li_{2}S[/tex] ⇒ soluble, [tex]MnBr_{2}[/tex] ⇒ soluble.

                                        [tex]LiBr[/tex] ⇒ soluble, [tex]MnS[/tex]  ⇒ insoluble.

There are the insoluble precipitates of [tex]MnS[/tex]  forms.

d)

Solubility rule suggests:- [tex]Ba(No_{3} )_{2}[/tex] ⇒ soluble, [tex]Ag_{2} So_{4}[/tex] ⇒insoluble.

As [tex]Ag_{2} So_{4}[/tex] is insoluble, therefore no precipitate forms.

e)

Solubility rule suggests:- [tex]Rb_{2}Co_{3}[/tex] ⇒ soluble, [tex]NaCl[/tex] ⇒ soluble.

                                        [tex]RbCl[/tex] ⇒ soluble, [tex]Na_{2} Co_{3}[/tex]  ⇒ soluble.

There are no insoluble precipitates forms.

Answer:

4.858 g

Explanation:

Start with the formula

density = [tex]\frac{mass}{volume}[/tex]

density = 1.98 g/mL

volume = 2.45 mL

mass = ??

rearrange the formula to solve for mass

(density) x (volume) = mass

Add in the substitutes and solve for mass

1.98 g/mL x 2.45 mL = 4.858 g

Answer:

There is no difference between the two.

Explanation:

They both show the same volume. But, adding decimal places shows the least count of the instrument used and is more acceptable when recording values in scientific experiments

The structures are shown in the image attached.

A structural formula is the representation of the molecule in which all atoms and bonds in the molecule are shown.

Since the question requires that all the lone pairs, formal charges and sigma and pi bonds should be shown, then the simple condensed structural formula becomes insufficient in this case.

I have attached images of the structural formula of sodium benzoate (image 1), benzoic acid (image 2)  and tetrahydrofuran (image 3).

All the formal charges, lone pairs as well as sigma and pi bonds are fully shown.

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Answer:

[tex]K=1.7x10^{-3}[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve this problem by firstly setting up the equilibrium expression for the given reaction, in agreement to the law of mass action:

[tex]K=\frac{[NO]^2}{[N_2][O_2]}[/tex]

Next, we plug in the given concentrations on the data table to obtain:

[tex]K=\frac{(0.034)^2}{(0.69)(0.98)}\\\\K=1.7x10^{-3}[/tex]

Regards!

Answer:

b

Explanation:

deposition is when water turns from gas to solid. b is the only one that fits

Deposition is frost forms when cold evening temperatures convert the humidity in the air to thin layers of ice on the ground.

Deposition is a process that involves collection of large mass or when mean distance between molecules are reduced. It can also be explained as gathering of substances together to form a larger mass.

Therefore, the phenomenon explained in the given example about deposition is frost forms when cold evening temperatures convert the humidity in the air to thin layers of ice on the ground.

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Answer:

0.696 m

Explanation:

We'll begin by calculating the number of mole in 16.7 g of C₆H₁₂O₆. This can be obtained as follow:

Mass of C₆H₁₂O₆ = 16.7 g

Molar mass of C₆H₁₂O₆ = (6×12) + (12×1) + (6×16)

= 72 + 12 + 96

= 180 g/mol

Mole of C₆H₁₂O₆ =?

Mole = mass / molar mass

Mole of C₆H₁₂O₆ = 16.7 / 180

Mole of C₆H₁₂O₆ = 0.093 mole

Next, we shall convert 133.6 g of water to Kg. This can be obtained as follow:

1000 g = 1 Kg

Therefore,

133.6 g = 133.6 g × 1 Kg / 1000 g

133.6 g = 0.1336 Kg

Thus, 133.6 g is equivalent to 0.1336 Kg.

Finally, we shall determine the molality of the solution. This can be obtained as illustrated below:

Mole of C₆H₁₂O₆ = 0.093 mole

Mass of water = 0.1336 Kg

Molality =?

Molality = mole / mass of water (in Kg)

Molality = 0.093 / 0.1336

Molality = 0.696 m

Therefore, the molality of the solution is 0.696 m

Answer:

a. Decrease

b. Increase

c. Increase

d. No effect

Explanation:

Glycolysis is present in muscle cells which converts glucose to pyruvate, water and NADH. It produces two molecules of ATP. Cellular respiration produces more molecules of ATP from pyruvate in mitochondria. Glycolysis increases in pyruvate kinase.

a. Loss of binding site for fructose 1,6-bisphosphate in pyruvate kinase: Decrease

b. Loss of allosteric binding site for ATP in pyruvate kinase: No effect

c. Loss of allosteric binding site for AMP in phosphofructokinase: Increase

d. Loss of regulatory binding site for ATP in phosphofructokinase: Increase

A. An important substrate in the glycolysis pathway is fructose 1,6-bisphosphate. It stimulates pyruvate kinase, an essential enzyme in glycolysis. The amount of pyruvate kinase that is activated will decrease if the fructose 1,6-bisphosphate binding site in pyruvate kinase is eliminated. As a result the rate of glycolysis in the muscle cells will probably decrease.

B. The allosteric ATP binding site of pyruvate kinase controls how active the enzyme is. However, pyruvate kinase is not significantly regulated by ATP in muscle cells. Therefore, it is unlikely that deletion of the ATP-binding allosteric site in pyruvate kinase would have no effect on the rate of glycolysis in muscle cells.

C. The rate-limiting enzyme in glycolysis, phosphofructokinase, is activated from all forms by AMP. It increases the rate of glycolysis by stimulating the activity of phosphofructokinase. If the allosteric binding site for AMP is eliminated, phosphofructokinase activation will be reduced. As a result, the rate of glycolysis in muscle cells will decrease.

D. Phosphofructokinase is inhibited allosterically by ATP. It regulates the rate of glycolysis by a feedback mechanism. High ATP concentrations cause phosphofructokinase to bind to its regulatory site, limiting its activity and delaying glycolysis. If the regulatory binding site for ATP is eliminated, the inhibitory action of ATP on phosphofructokinase would be lost. As a result, muscle cells will glycolysis at a faster rate.

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Explanation:

Add water, Na2CO3 dissolves, filter, PbCO3 stays in the paper and dissolved Na2CO3 goes through as the solution. Dry the PbCO3 and you have the dry solid.

OR

Add water to dissolve then filter to obtain PbCo3 as you're residue and Na2Co3 as the filtrate. Dry the insoluble PbCo3 between filter papers and you obtain solid PbCo3

Solution :

A substrate is defined as the chemical species that are being observed in the chemical reaction where the substrate reacts with a reagent and forms a product. It can also be referred to the surface where some other chemical reactions are performed.

Stereochemistry is defined as the study of relative spatial arrangement of the atoms which forms the structure of the molecules and their respective manipulations.

In the context, the products including the stereochemical designations for any chirality centers starting from the  (R)-3-methylhex-1-yne as the substrate are attached below.  

The mass of magnesium chloride produced from 2.30 moles of chlorine gas is 218.99 grams.

Stoichiometry refers to the study and calculation of quantitative (measurable) relationships of the reactants and products in chemical reactions.

According to this question, magnesium reacts with chlorine gas to form magnesium chloride as follows:

Mg + Cl₂ → MgCl₂

Based on the above chemical equation, 1 mole of chlorine gas forms 1 mole of magnesium chloride.

This means that 2.30 moles of chlorine gas will 2.30 moles of magnesium chloride.

Next, we convert moles of magnesium chloride to mass as follows:

molar mass of magnesium chloride = 95.211g/mol

mass of magnesium chloride = 95.211 × 2.30 = 218.99 grams.

Therefore, 218.99 grams of magnesium chloride will be formed.

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Answer:

State functions represent quantities or properties of a thermodynamic system, while non-state functions represent a process during which the state functions change. For example, the state function PV is proportional to the internal energy of an ideal gas, but the work W is the amount of energy transferred as the system performs work.

Explanation:

Explanation:

1. boiling and melting point

2. electrical conductivity

3. Bond strength

4. bond length

A covalent bond consists of negative electrons that are shared in between atoms. Because of this bond, they possess and manifest physical abilities, including electrical pressure/conductivity and lower melting points compared to ionic compounds.

Answer:

MoO2

Explanation:

The empirical formula is defined as the simplest whole number ratio of atoms present in a molecule.

To solve this question we need to find the moles of Mo2O3. Twice these moles = Moles of Mo. With the moles of Mo we can find its mass.

The difference in masses between mass of new oxide and mass of Mo = Mass of oxygen. With the mass of oxygen we can find its moles and the empirical formula as follows:

Moles Mo2O3 -Molar mass: 239.9g/mol-

12.37g * (1mol / 239.9g) = 0.05156 moles Mo2O3 * (2mol Mo / 1mol Mo2O3) = 0.1031 moles of Mo

Mass Mo -95.95g/mol-:

0.1031 moles of Mo * (95.95g/mol) = 9.895g of Mo

Mass oxygen in the oxide:

13.197 - 9.895g = 3.302g Oxygen

Moles oxygen -Molar mass: 16g/mol-:

3.302g Oxygen * (1mol / 16g) = 0.206 moles O

Now, the ratio of moles O / moles Mo is:

0.206 moles O / 0.1031 moles Mo = 2

That means there are 2 moles of O per mole of Mo and the empirical formula of the new oxide is:

Answer:

11.9g remains after 48.2 days

Explanation:

All isotope decay follows the equation:

ln [A] = -kt + ln [A]₀

Where [A] is actual amount of the isotope after time t, k is decay constant and [A]₀ the initial amount of the isotope

We can find k from half-life as follows:

k = ln 2 / Half-Life

k = ln2 / 27.7 days

k = 0.025 days⁻¹

t = 48.2 days

[A]  = ?

[A]₀ = 39.7mg

ln [A] = -0.025 days⁻¹*48.2 days + ln [39.7mg]

ln[A] = 2.476

[A] = 11.9g remains after 48.2 days

Answer:

1. Nitrate ions, NaNO3 - Sodium nitrate.

2. Sulphide ions, K2S - Potassium sulphide.

3. Sulphate ions, CaSO4 - Calcium sulphate.

4. Hydrogensulphite ions, NaHSO3 - Sodium hydrogensulphite.

5. Carbonate ions, CaCO3 - Calcium carbonate.

6. Hydrogencarbonate ions, KHCO3 - Potassium hydrogencarbonate.

7. Phosphite ions, PH3 - Hydrogen phosphite.

8. Nitride ions, NH3 - Hydrogen nitride ( ammonia ).

9. Ethanoate ions, CH3COONa - Sodium ethanoate.

10. Methanoate ions, HCOONa - Sodium methanoate.

11. Fluoride ions, HF - Hydrogen fluoride.

12. Chloride ions, KCl - Potassium chloride.

13. Bromide ions, HBr - Hydrogen bromide.

14. Iodide ions, NaI - Sodium iodide.

15. Phosphate ions, K3PO3 - potassium phosphate.

Answer:

0.0063 mol

Explanation:

Step 1: Write the balanced combustion equation

C₈H₁₈(l) + 12.5 O₂(g) ⇒ 8 CO₂(g) + 9 H₂O(g)

Step 2: Establish the appropriate molar ratio

According to the balanced equation, the molar ratio of C₈H₁₈ to CO₂ is 1:8.

Step 3: Calculate the number of moles of C₈H₁₈ needed to produce 0.050 moles of CO₂

0.050 mol CO₂ × 1 mol C₈H₁₈/8 mol CO₂ = 0.0063 mol C₈H₁₈

Explanation:

here's the answer to your question

Answer:

Stalactites and stalagmites form as ________ precipitates out of the water evaporating in underground caves.

Group of answer choices

hydrochloric acid

sodium bicarbonate

calcium carbonate

sodium chloride

sodium hydroxide

Answer:

calcium carbonate

Explanation:

A stalactite is an icicle-looking mould that is formed by the precipitation of natural minerals as a result of water dripping from the ceiling, hanging from a cave.

A stalagmites in the other hand, grows upwards and is also a mound that is formed by the deposits of minerals gotten by the water dripping on the floor of a cave.

Therefore, stalactites and stalagmites form as calcium carbonate precipitates out of the water evaporating in underground caves.

Answer :

volume of a gas = weight * 22.4 l / gram molecular weight

volume of o2 = ?

weight given = 20.5 g

gram molecular weight of oxygen = 32 (because of 2 oxygen atoms )

volume of oxygen = 20.5 * 22.4 / 32

volume of oxygen = 14.35 liters  

Explanation:

hope this helps you

if wrong just correct me

Answer:

See explanation

Explanation:

The relationship between the activation energy and rate of reaction is best captured by the Arrhenius equation;

k= Ae^-Ea/RT

Where;

k= rate constant

A= pre-exponential factor

Ea=activation energy

R= gas constant

T= temperature

We can see from the foregoing that, as the activation energy increases, the rate of reaction decreases and vice versa. reactions that have a very high activation energy are markedly slow.

Since the activation energy for the malonic acid reaction is found to be greater than the activation energy for the tartaric acid reaction, then the rate of the malonic acid reaction(k) will be slower than that of the tartaric acid reaction.

The study of chemistry and bonds is called chemistry. There are two types of elements metal and nonmetals.

The correct answer is mentioned below.

The equation is as follows:-

[tex]k= Ae^{-Ea/RT[/tex] Where;

We can see from the foregoing that, as the activation energy increases, the rate of reaction decreases and vice versa. reactions that have very high activation energy are markedly slow. Since the activation energy for the malonic acid reaction is found to be greater than the activation energy for the tartaric acid reaction, then the rate of the malonic acid reaction(k) will be slower than that of the tartaric acid reaction.

Hence, the correct answer is mentioned above.

For more information about the equation, refer to the link:-

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Explanation:

From the question given above, the following data were obtained:

Half-life (t½) = 9.7 days

Time (t) = 60 days

Next, we shall determine the number of half-lives that has elapsed. This can be obtained as follow:

Half-life (t½) = 9.7 days

Time (t) = 60 days

Number of half-lives (n) =?

n = t / t½

Finally, we shall determine the percentage remaining. This can be obtained as follow:

Let the original amount be N₀

Let the amount remaining be N

Number of half-lives (n) = 60 / 9.7

N = N₀ / 2ⁿ

Divide both side by N₀

N/N₀ = 1/2ⁿ

N/N₀ = 1 / 2⁽⁶⁰÷⁹•⁷⁾

N/N₀ = 0.0137

Multiply by 100 to express in percentage

N/N₀ = 0.0137 × 100

Therefore, the percentage remaining after 60 days is 1.37%

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