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1 year ago

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Heating galactose, a monosaccharide sugar, in the presence of excess oxygen produces carbon dioxide gas and water vapor. Classif

y
this reaction.
C6H1206(s) + 602(9) - 6C02(0) + 6H20 (9)

1 answer:

artcher [175]1 year ago

6 0

Answer:

it's a combustion reaction

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According to the quantum-mechanical model for the hydrogen atom, which electron transition produces light with the longer wavele

LekaFEV [45]

Answer:

Explanation:

The main task here is that there are some missing gaps in the above question that needs to be filled with the appropriate answers. So, we are just going to do rewrite the answer below as we indicate the missing gaps by underlining them and making them in bold format.

SO; In the quantum-mechanical model of the hydrogen atom.

As the n level increases. the energy <u>increases</u> and thus levels are <u>closer to </u>each other. Therefore, the transition <u>3p→2s</u> would have a greater energy difference than the transition from <u>4p→3p.</u>

$Energy \ and \ wavelength \ are \ inversely \ proportional , \ so \ the \ \mathbf{ 4p\to 3p} \ transition \ would$ $produce \ a \ longer \ wavelength.$

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1 year ago

What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig

Mashutka [201]

Answer:

$P_H =2.86$

$c=1.4\times 10^{-4}$

Explanation:

first write the equilibrium equaion ,

$C_3H_6O_3$ ⇄ $C_3H_5O_3^{-} +H^{+}$

assuming degree of dissociation $\alpha$ =1/10;

and initial concentraion of $C_3H_6O_3$ =c;

At equlibrium ;

concentration of $C_3H_6O_3 = c-c\alpha$

$[C_3H_5O_3^{-} ]= c\alpha$

$[H^{+}] = c\alpha$

$K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}$

$\alpha$ is very small so $1-\alpha$ can be neglected

and equation is;

$K_a = {c\alpha \times \alpha}$

$[H^{+}] = c\alpha$ = $\frac{K_a}{\alpha}$

$P_H =- log[H^{+} ]$

$P_H =-logK_a + log\alpha$

$K_a =1.38\times10^{-4}$

$\alpha = \frac{1}{10}$

$P_H= 3.86-1$

$P_H =2.86$

composiion ;

$c=\frac{1}{\alpha} \times [H^{+}]$

$[H^{+}] =antilog(-P_H)$

$[H^{+} ] =0.0014$

$c=0.0014\times \frac{1}{10}$

$c=1.4\times 10^{-4}$

6 0

2 years ago

According to the law of conservation of mass, if an element A has an atomic mass of 2 mass units and element B has an atomic mas

LenaWriter [7]

The law of conservation of mass states that mass is neither created nor destroyed. Since we have 2 g/mol of A and 3 g/mol of B then AB should be equal to the sum of their molar mass that is

2 g/mol + 3 g/mol = 5 g/mol AB

for the case of A2B3

A2 = 2 * 2 = 4 g/mol
B3 = 3 * 3 = 9 g/mol

therefore A2B3 = 13 g/mol

5 0

2 years ago

if a sample of gas is intially at 1.8 atm,22.0 l, and 26.4 c, what will be the volume if the pressure is reduced by 0.8 atm and

Tems11 [23]

<span>pv=nrT Initial state (1.8atm)(22.0 l)=n(0.082057)(26.4+273.15); r=.082057, and converting C to K Solving for n = (1.8)(22)/(.082057*(26.4+273.15) moles n = 1.611 moles in initial state Now we solve for new volume pv=nrT (.8atm)v=(1.611)(.082057)(20.3+273.15) v=(1.611)(.082057)(20.3+273.15)/.8 v=48.49 l</span>

8 0

2 years ago

Read 2 more answers

Consider the reaction: 2clf3(g) + 2nh3(g) → n2(g) + 6hf(g) + cl2(g) when calculating the δh°rxn, why is the δhf° for n2 not impo

Leni [432]

<span>Actually, the heat of reaction hrxn s calculated by taking the sum of the heats of formation of the products minus the sum of the heats of formation of the reactants. However, at heat of formations of pure elements at atmospheric conditions is zero, therefore the hf of N2 is not important since it is zero anyway.</span>

8 0

2 years ago