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3 years ago

11 Which is a bright streak of light in Earth's atmosphere?

Chemistry

2 answers:

wolverine [178]3 years ago

7 0

A Comet a comet leaves a bright streak following its trail

LuckyWell [14K]3 years ago

4 0

Answer:

B. a meteor

Explanation:

Meteors, also known as shooting stars, are pieces of dust and debris from space that burn up in Earth's atmosphere, where they can create bright streaks across the night sky.

Hope this helped!!!

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The cost of painting the circular traffic sign shown below is $3.50 per square foot. How much, to the nearest dollar, will it co

KiRa [710]

The cost of painting the traffic sign of 7.065 square feet is approximately equal to 25 dollar.

<u>Explanation:</u>

The cost of painting the circular traffic sign is given as 3.50 dollar per square feet. The diameter of the traffic sign is 36 inch, then its radius will be $\frac{36}{2} = 18$ inches.

But as the cost is given in unit of feet, we have to convert the radius from inches to feet.

1 inches = 0.0833 feet

18 inches = 18 × 0.0833 feet

So, the radius of the traffic sign will be approximately equal to 1.5 feet.

The area of the traffic sign will be $\pi r^{2} = 3.14 \times 1.5 \times 1.5 = 7.065$ square feet.

So, if the cost of painting 1 square feet of traffic sign is 3.50 dollar, then

cost of painting 7.065 square feet of traffic sign = 3.50 × 7.065 = 24.7 dollar.

Thus, the cost of painting the traffic sign of 7.065 square feet is approximately equal to 25 dollar.

8 0

3 years ago

An atom X has atomic number 15,it accepts 3 electrons in other to become stable,determine the,*Number of protons of the atom. *D

Lisa [10]

Answer:

15 protons

8 electrons in valence shell after accepting 3 electrons

Neutral/no charge before accepting 3 electrons

-3 charge after accepting 3 electrons

4 0

3 years ago

How many grams of NaOH are needed to make 500 mL of a 2.5 M NaOH solution

Veronika [31]

The number of grams of NaOH that are needed to make 500 ml of 2.5 M NaOH solution

calculate the number of moles =molarity x volume/1000

= 2.5 x 500/1000 = 1.25 moles

mass = moles x molar mass of NaOH

= 1.25 x40= 50 grams of NaOH

4 0

4 years ago

50.0 mL solution of 0.160 M potassium alaninate ( H 2 NC 2 H 5 CO 2 K ) is titrated with 0.160 M HCl . The p K a values for the

scZoUnD [109]

Answer:

a) 6.12

b) 1.87

Explanation:

At the onset of the equivalence point (i.e the first equivalence point); alaninate is being converted to alanine.

$H_2NC_2H_5CO^-_2$ $+$ $H^+$ ------> $H_3}^+NC_2H_5CO^-_2$

1 mole of alaninate react with 1 mole of acid to give 1 mole of alanine;

therefore 50.0 mL of 0.160 M alaninate required 50.0 mL of 0.160M HCl to reach the first equivalence point.

The concentration of alanine can be gotten via the following process as shown below;

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{initial moles of alaninate}{total volume}$

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{(50.0mL)*(0.160M)}{(50.0mL+50.0mL)}$

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{8}{100mL}$

$[H_3}^+NC_2H_5CO^-_2]$ = 0.08 M

Alanine serves as an intermediary form, however the concentration of $H^+$ and the pH can be determined as follows;

$[H^+]$ = $\sqrt{\frac{K_{a1}K_{a2}{[H_3}^+NC_2H_5CO^-_2]+K_{a1}K_w}{ K_{a1}{[H_3}^+NC_2H_5CO^-_2] } }$

$[H^+]$ = $\sqrt{\frac{ (10^{-pK_{a1})}(10^{-pK_{a2})}(0.08)+(10^{-pK_{a1})}(1.0*10^{-14})} {(10^{-pK_{a1}})+(0.08)} }$

$[H^+]$ = $\sqrt{\frac{ (10^{-2.344})(10^{-9.868})(0.08)+(10^{-2.344})(1.0*10^{-14})} {(10^{-2.344})+(0.08)} }$

$[H^+]$ = $7.63*10^{-7}M$

pH = - log $[H^+]$

pH = $-log[7.63*10^{-7}]$

pH= 6.12

Therefore, the pH of the first equivalent point = 6.12

b) At the second equivalence point; all alaninate is converted into protonated alanine.

$H_2NC_2H_5CO^-_2$ $+$ $H^+$ -----> $H^+_3NC_2H_5CO^-_2$

$H^+_3NC_2H_5CO^-_2$ $+$ $H^+$ -----> $H^+_3NC_2H_5CO_2H$

Here; we have a situation where 1 mole of alaninate react with 2 moles of acid to give 1 mole of protonated alanine;

Moreover, 50.0 mL of 0.160 M alaninate is needed to produce 100.0mL of 0.160 M HCl in order to achieve the second equivalence point.

Thus, the concentration of protonated alanine can be determined as:

$[H^+_3NC_2H_5CO_2H]$ = $\frac{initial moles of alaninate}{total volume}$

$[H^+_3NC_2H_5CO_2H]$ = $\frac{(50.0mL)*(0.160M)}{(50.0mL+100.0mL)}$

$[H^+_3NC_2H_5CO_2H]$ = $\frac{8}{150}$

$[H^+_3NC_2H_5CO_2H]$ = 0.053 M

The pH at the second equivalence point can be calculated via the dissociation of protonated alanine at equilibrium which is represented as:

$H^+_3NC_2H_5CO_2H$ ⇄ $H^+_3NC_2H_5CO^-_2$ $+$ $H^+$

(0.053 - x) x x

$K_{a1}$ = $\frac{[H^+] [H^+_3NC_2H_5CO^-_2]}{[H^+_3NC_2H_5CO_2H]}$

$10^{-PK_{a1}}$ = $\frac{x*x}{(0.053-x)}$

$10^{-2.344} =\frac{x^2}{(0.053-x)}$

$0.00453 = \frac{x^2}{(0.053-x)}$

$0.00453(0.053-x) =x^2$

$x^2+0.00453x-(2.4009*10^{-4})$

Using quadratic equation formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

we have:

$\frac{-0.00453+\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)}$ OR $\frac{-0.00453-\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)}$

= 0.0134 OR -0.0179

So; we go by the positive integer which says

x = 0.0134

So $[H^+]=[H_3^+NC_2H_5CO^-_2]= 0.0134 M$

pH = $-log[H^+]$

pH = $-log[0.0134]$

pH = 1.87

Thus, the pH of the second equivalent point = 1.87

3 0

4 years ago

Which of the following is the best comparison of the density of basalt and granite? Basalt is more dense than granite Granite is

LUCKY_DIMON [66]

Basalt is more dense than granite because granite is made of lighter mass.

8 0

3 years ago

Read 2 more answers