Draw a structural formula for the organic product formed by treating butanal with the following reagent: NaBH4 in CH3OH/H2O You do not have to consider stereochemistry. You do not have to explicitly draw H atoms. Do not include lone pairs in your answer. They will not be considered in the grading. Include counter-ions, e.g., Na , I-, in your submission, but draw them in their own separate sketcher. Do not draw organic or inorganic by-products.

Answer:

[tex]\boxed {\boxed {\sf A. \ 2.48 \ mol \ Na_2O}}[/tex]

Explanation:

We are asked to convert an amount in grams to moles. To do this, we use the molar mass. This is the number of grams in one mole of a substance. It is the same value numerically as the atomic mass on the Periodic Table, however the units are grams per mole, not atomic mass units.

Look up the molar masses for the individual elements.

Look back at the formula: Na₂O. Notice there is a subscript of 2 after sodium. This means there are 2 atoms of sodium in every molecule, so we have to multiply sodium's molar mass by 2 before adding oxygen's.

Set up a ratio using the molar mass.

[tex]\frac {61.9785386 \ g \ Na_2O}{1 \ mol \ Na_2O}[/tex]

Multiply by the given number of grams.

[tex]154 \ g \ Na_2O*\frac {61.9785386 \ g \ Na_2O}{1 \ mol \ Na_2O}[/tex]

Flip the ratio so the grams of sodium oxide can cancel each other out.

[tex]154 \ g \ Na_2O*\frac {1 \ mol \ Na_2O}{61.9785386 \ g \ Na_2O}[/tex]

[tex]154 *\frac {1 \ mol \ Na_2O}{61.9785386 }[/tex]

[tex]\frac {154}{61.9785386 } \ mol \ Na_2O[/tex]

[tex]2.48473106141 \ mol \ Na_2O[/tex]

The original measurement of grams given has 3 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place.

The 4 in the thousandth place tells us to leave the 8.

[tex]2.48 \ mol \ Na_2O[/tex]

There are 2.48 moles of sodium oxide in 154 grams, so choice A is correct.

Explanation:

(I only know this thing vaguely lol)

1.) It's the study of the chemistry that is happening here on earth that is not bounded by a laboratory.

2.) The early gas clouds compressed and formed a star we now call the sun and the massive gravitational pull of this star grabbed nearby debris and the debris formed and compressed into the planets and moons that we know today.

Answer:

i think its magma ajani i looked it up... and chelsey said u had corona so if its true hope u get better bro

Explanation:

Ans：2.3

Explanation:

pH = -log[H+(aq)]

= -log[0.00476]

=2.3

As HCl solution is monobasic. Each molecule ionizes in water to form ONE hydrogen ion. So the H+(aq) is equal to its molarity.

2.3 is he pH of a 0.00476 M HCI solution and this value tells that solution is acidic in nature.

pH of any solution can be define as the negative log of the concentration of the H⁺ ions.

In the question given that:

Concentration of HCl solution = 0.00476 M

Concentration of H⁺ ions is equal to concentration of HCl solution as it completely ionize into their ions. So, pH can be calculated as:

pH = -log[H⁺] = -log[0.00476] = 2.3

Hence, pH of the solution is 2.3.

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Answer:

The final pressure is 0.725 atm.

Explanation:

Gay Lussac's Law establishes the relationship between pressure and temperature of a gas when the volume is constant. This law says that when there is a constant volume, as the temperature increases, the pressure of the gas increases. And when the temperature is decreased, the pressure of the gas decreases. That is, pressure and temperature are directly proportional quantities.

Mathematically, Gay-Lussac's law states that, when a gas undergoes a constant volume transformation, the quotient of the pressure exerted by the gas temperature remains constant:

[tex]\frac{P}{T} =k[/tex]

When analyzing an initial state 1 and a final state 2, the following is satisfied:

[tex]\frac{P1}{T1} =\frac{P2}{T2}[/tex]

In this case:

Replacing:

[tex]\frac{0.81 atm}{306 K} =\frac{P2}{274 K}[/tex]

Solving:

[tex]P2=274 K*\frac{0.81 atm}{306 K}[/tex]

P2= 0.725 atm

The final pressure is 0.725 atm.

Answer:

[tex]=1.76gCa[/tex]

Explanation:

Hello there!

In this case, according to the given atoms of calcium, it is possible to calculate the mass of this element by considering that the definition of mole is in terms of atoms and also the atomic mass of calcium:

[tex]1molCa=6.022x10^{23}atoms Ca=40.08gCa[/tex]

In such a way, by considering the following setup, we can obtain:

[tex]=2.65x10^{22}atomsCa*\frac{1molCa}{6.022x10^{23}atoms} *\frac{40.08gCa}{1molCa}\\\\=1.76gCa[/tex]

Best regards!

Answer:

The type of reaction is a single-replacement reaction.

Explanation:

Mg switches places with H, leaving H by itself.

Answer:

46.41%

Explanation:

Find the total r.m.m of AL(CLO3)3 put it equals to 100%.....The take 106.5,the total r.m.m of Cl and find the missing which is the percentage by cross multiplication

Answer:

See explanation

Explanation:

According to Bronsted-Lowry, an acid is a proton donor while a base is a proton acceptor.

Hence, if we consider the reaction above, we will notice that for each base there is a conjugate acid and for each acid there is a conjugate base.

For the acid HNO3, its conjugate base is NO3^- while for the acid H3O^+, its conjugate base is H2O.

Answer:

Hello There!!

Explanation:

I think the answer is possibly C. It identified the charge on a single electron.

hope this helps,have a great day!!

~Pinky~

[tex]\huge{\textbf{\textsf{{\color{navy}{An}}{\purple{sw}}{\pink{er}} {\color{pink}{:}}}}}[/tex]

C. It identified the charge on a single electron.

Answer:

o diferença chave entre a radioatividade e a transmutação é que a radioatividade se refere à transmutação natural, enquanto a transmutação se refere à mudança de um elemento químico em outro por meios naturais ou artciais.Tanto a radioatividade quanto a transmutação são processos químicos que envolvem a mudança de núcleos atômicos para formar um novo elemento químico a partir de um elemento químico existente. A radioatividade é um tipo de processo de transmutação

Explanation:

mais informação em; ahttps://pt.strephonsays.com/radioactivity-and-transmutation-9500

Answer:

[tex]k =0.101 \ min^{-1}[/tex]

Explanation:

From the given information:

The initial concentration of a chemical [tex]C_{AO} = 15 \ mg/L[/tex]

The final concentration is [tex]C_A = 5 \ g/m^3[/tex] = 5 mg/L

Volume flow rate [tex]V_o = 0.42 \ m^3/sec[/tex]

Volume of the tank V = 500 000 L = 500 m³

The time t is determined by using the formula:

[tex]t= \dfrac{V}{V_o}[/tex]

[tex]t= \dfrac{500 \ m^3}{0.42 \ m^3/sec}[/tex]

t = 1190.47 sec

t ≅ 19.8 min

∴

The rate of the decay constant is:

[tex]kt= \dfrac{C_{AO}-C_{A}}{C_A} \\ \\ k = \dfrac{1}{19.8}( \dfrac{15-5}{5})[/tex]

[tex]k =0.101 \ min^{-1}[/tex]

Answer:

answer po Yan hope it help

Answer:

9.43 L Br₂ (g)

Explanation:

First, write a balanced reaction equation. Then use dimensional analysis to convert from grams of aluminum chloride to moles of aluminum chloride (with molar mass from the periodic table), from moles of aluminum chloride to moles of bromine gas (using the balanced equation), and from moles of bromine gas to liters of bromine gas (remember at STP, 1 mol = 22.4 L). See the attached image for the work out. Finally, account for sig figs to get 9.43 L of bromine gas

Explanation:

its hard to explain unless we know what the question fully asks..

Answer:

See explanation.

Explanation:

Hello there!

In this case, for the chemical reaction:

[tex]H_2O(g) CO(g)\rightleftharpoons CO_2(g) H_2(g)[/tex]

It is widely known that the the only factor altering the equilibrium constant is the temperature because of its thermodynamic definition:

[tex]K=exp(-\frac{\Delta G(T)}{RT} )[/tex]

Which is temperature dependent on the denominator and on the Gibbs free energy of reaction. In such a way, the answers can be just f and g, as the other factors modify Q (reaction quotient) rather than K. Thus, we can discuss about f and g:

f. This would increase K if the Gibbs free energy of reaction is positive (nonspontaneous).

g. This would increase K if the Gibbs free energy of reaction is negative (spontaneous).

It is important to discuss those two because the Gibbs free energy is not given.

Regards!

Answer:

0.31g KCl

Explanation:

Before we can start, we have to wonder, how many moles of KCl is formed from KClO3? To figure that out we have to make a balanced equation.

KClO3 -> KCl + O2

How do I know it makes O2? Looking back at the problem they're asking for KCl, which has less ions than KClO3 so it must have been broken down. So we can say it's a decomposition type of reaction. We can't just slap on O3, we know oxygen is one of our diatomic elements that usually exists as O2 so we'll put it down as O2 and balance our equation.

2KClO3 -> 2KCl + 3O2

Now that we know every 2 moles of KClO3 makes 2 moles of KCl, this is our mole to mole ratio. Knowing the ratio between the 2 compounds in our problem is the stepping stone from converting from grams KClO3 -> moles KClO3 -> moles KCl -> grams KCl.

1. Let's first convert grams KClO3 -> moles KClO3

[tex]0.500g KClO3 x \frac{1 mol}{123g KClO3} = 0.0041 mol KClO3[/tex]

2. Convert moles KClO3 -> moles of KCl using our mole to mole ratio

[tex]0.0041 mol KClO3 x \frac{2 mol KCl}{2 mol KClO3} = 0.0041 mols KCl[/tex]

3. Convert moles KCl -> grams KCl

[tex]0.0041 mol KCl x\frac{75g KCl}{1 mol} = 0.31g KCl[/tex]

Answer:

[tex]\Delta H_{comb}=2043.85kJ/mol[/tex]

Explanation:

Hello there!

In this case, according to the given chemical reaction, it possible for us to set up the expression for the calculation of the enthalpy change as shown below:

[tex]\Delta H_r=-\Delta H_{comb}=3\Delta _fH_{CO_2}+4\Delta _fH_{H_2O}-\Delta _fH_{C_3H_8}-3\Delta _fH_{O_2}[/tex]

Thus, given the values of the enthalpies of formation on the attached file, we obtain:[tex]-\Delta H_{comb}=3(-393.5kJ/mol)+4(-241.8kJ/mol)-(-103.85kJ/mol)-3(0kJ/mol)\\\\-\Delta H_{comb}=-2043.85kJ/mol\\\\\Delta H_{comb}=2043.85kJ/mol[/tex]

Best regards!

Explanation:

ur question is incomplete what elements make the compound and we talk of mass not weight and what are the units kjmol ,grams ,kgs

Answer:

74.08%

Explanation:

We'll begin by calculating the mass of

Oxygen in 4 g of N₂O₅. This can be obtained as follow:

Molar mass of N₂O₅ = (14×2) + (16×5)

= 28 + 80

= 108 g/mol

Mass of oxygen in N₂O₅ = 5 × 16 = 80 g

Mass of oxygen in 4 g of N₂O₅

= 80 / 108 × 4

= 2.963 g

Finally, we shall determine the percentage of oxygen in 4 g of N₂O₅. This can be obtained as follow:

Mass of oxygen = 2.963 g

Mass of N₂O₅ = 4 g

Percentage of oxygen =?

Percentage of oxygen = mass of oxygen / mass of N₂O₅ × 100

Percentage of oxygen = 2.963 / 4 × 100

Percentage of oxygen = 74.08%

Answer:

C₃H₈O

Explanation:

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

Mass of Carbon (C) = 0.599 g

Mass of Hydrogen (H) = 0.134 g

Mass of Oxygen (O) = 0.266 g

Empirical formula =?

The empirical formula of the compound can be obtained as follow:

C = 0.599 g

H = 0.134 g

O = 0.266 g

Divide by their molar mass

C = 0.599 / 12 = 0.050

H = 0.134 / 1 = 0.134

O = 0.266 / 16 = 0.017

Divide by the smallest

C = 0.050 / 0.017 = 3

H = 0.134 / 0.017 = 8

O = 0.017 / 0.017 = 1

Empirical formula => C₃H₈O

Answer:

a gray substance and colourless liquid

Explanation:

lead oxide is reduced to lead while hydrogen is oxidized to water

Answer:

Convection

Explanation:

Convection definition:

the movement caused within a fluid by the tendency of hotter and therefore less dense material to rise, and colder, denser material to sink under the influence of gravity, which consequently results in transfer of heat.

The method of heat transfer that occurs when particles move through a fluid is called convection.

Convection relies on the principle of fluid flow, where hotter particles become less dense and rise while cooler particles sink due to their increased density.

This movement creates a continuous circulation of the fluid, enabling the transfer of thermal energy. As the hot particles rise, they carry their thermal energy with them, and as the cooler particles sink, they absorb heat from their surroundings.

Convection is also observed in natural phenomena such as ocean currents, air currents, and even the movement of magma in the Earth's mantle.

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