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3 years ago

13

1. The fourth principle electronic level has:

1 answer:

attashe74 [19]3 years ago

5 0

Answer:

the fourth level can hold up to 32 electrons: 2 in the s orbital, 6 in the three p orbitals, 10 in the five d orbitals, and 14 in the seven f orbitals.

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For the problem: Mg (s) + O₂ (g) ➞ MgO (g) if 4.3 g of magnesium reacts with 83 g of oxygen (O₂), which is the limiting reagent?

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Explanation:

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3 years ago

Identify at least three control variables that could be used with this investigation: What temperature allows for bacterial grow

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Answer:

Time allowed for incubation, size of the dish, the amount of light, amount of agar, the type of agar…etc

Explanation:

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3 years ago

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If an object has a density of 0.55 g/mL, what is its density in cg/L?

otez555 [7]

if 1 g is equal to 100 cg

then 0.55 g are equal to X cg

X = (0.55 × 100 ) / 1 = 55 cg

The density of the object is 55 cg/L.

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4 years ago

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HELP WILL GIVE BRAINLIEST NO WEIRD ANSWERS PLSS

Mice21 [21]

Answer:

Solid

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The reason is because the liquid aspirin has a plastic cover with a liquid inside and it doesn't take long for out stomach acid to dissolve the cover unlike the Solid aspirin where is slowly dissolves.

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3 years ago

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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

3 years ago