All categories

3 years ago

What causes the moon to be illuminated?

Chemistry

1 answer:

I am Lyosha [343]3 years ago

5 0

Answer:

Reflection

Explanation:

The light from Earth's surface, reflected back towards the Moon, can illuminate it all the same. This is why, when the Moon is in a crescent phase, you can still see features on the dark portion of the Moon: we call this reflected, illuminating light the phenomenon of Earthshine.

You might be interested in

Which of these best explains why a paper clip can float on water?

Tems11 [23]

The water has surface tension

3 0

3 years ago

Read 2 more answers

How many grams of sodium phosphate monobasic would we add to a liter and how many grams of sodium phosphate dibasic would we add

Nostrana [21]

Answer :

The correct answer for Mass of Na₂HPO₄ = 4.457 g and mass of NaH₂PO₄ = 8.23 g

Given : pH = 6.86

Total concentration of Phosphate buffer = 0.1 M

Asked : Mass of Sodium phosphate monobasic (NaH₂PO₄) = ?

Mass of Sodium phosphate dibasic(Na₂HPO₄)= ?

Following steps can be done to find the masses of NaH₂PO₄ and Na₂HPO₄ :

(In phosphate buffer , Na+ ion from NaH₂PO₄ and Na₂HPO₄ acts as spectator ion , so only H₂PO₄⁻ and HPO₄²⁻ will be considered )

Step 1 : To find pka

H₂PO₄⁻ <=> HPO₄²⁻

The above reaction has pka = 7.2 ( from image shown )

Step 2 : Plug values in Hasselbalch- Henderson equation .

Hasselbalch -Henderson equation is to find pH for buffer solution which is as follows :

$pH = pka + log\frac{[A^-]}{[HA]}$

pH = 6.86 pKa = 7.2

$6.86 = 7.2 + log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}$

Subtracting both side by 7.2

$6.86-7.2 = 7.2 -7.2+ log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}$

$-0.34 = log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}$

Removing log

$10^-^0^.^3^4 = \frac{[HPO_4^2^-]}{[H_2PO_4^-]}$

$\frac{[HPO_4^2^-]}{[ H_2PO_4^-]} = 0.457$ ---------------- equation (1)

Step 3 : To find molarity of H₂PO₄⁻ and HPO₄²⁻

Total concentration of buffer = [H₂PO₄⁻] + [HPO₄²⁻] = 0.1 M

Hence, [H₂PO₄⁻ ] + [ HPO₄²⁻ ] = 0.1 M

Assume [H₂PO₄⁻ ] = x

So , [x ] + [ HPO₄²⁻ ] = 0.1 M

[ HPO₄²⁻ ] = 0.1 - x

Step 4 : Plugging value of [H₂PO₄⁻ ] and [ HPO₄²⁻ ]

[H₂PO₄⁻ ] = x

[ HPO₄²⁻ ] = 0.1 - x

Equation (1) = > $\frac{[HPO_4^2^-]}{[ H_2PO_4^-]} = 0.457$

Plug value of [H₂PO₄⁻ ] and [ HPO₄²⁻ ] ( from step 3 ) into equation (1) as :

$\frac{[0.1 - x ]}{[ x]} = 0.457$

Cross multiplying

$0.1 - x = 0.457 x$

Adding x on both side

$0.1 -x + x = 0.457 x + x$

$0.1 = 1.457 x$

Dividing both side by 1.457

$\frac{0.1}{1.457} = \frac{1.457 x }{1.457}$

x = 0.0686 M

Hence , [H₂PO₄⁻ ] = x = 0.0686 M

[ HPO₄²⁻ ] = 0.1 - x

[ HPO₄²⁻ ] = 0.1 - 0.0686

[ HPO₄²⁻ ] = 0.0314 M

Step 5 : To find moles of H₂PO₄⁻ ( NaH₂PO₄) and HPO₄²⁻ (Na₂HPO₄ ) .

Molarity is defined as mole of solute per 1 L volume of solution .

Molarity of NaH₂PO₄ = 0.0686 M or 0.0686 mole per 1 L

Molarity of Na₂HPO₄ = 0.0314 M or 0.0314 mole per 1 L

Since that volume of buffer solution is 1 L , so Molarity = mole

Hence Mole of NaH₂PO₄ = 0.0686 mol

Mole of Na₂HPO₄ = 0.0314 mol

Step 6 : To find mass of Na₂HPO₄ and NaH₂PO₄

Moles of Na₂HPO₄ and NaH₂PO₄ can be converted to their masses using molar mass as follows :

Molar mass of Na₂HPO₄ = $141.96 \frac{g}{mol}$

Molar mass of NaH₂PO₄ = $119.98 \frac{g}{mol}$

$Mass (g) = mole (mol)* molar mass(\frac{g}{mol})$

$Mass of Na_2HPO_4 = 0.0314 mol * 141.96 \frac{g}{mol}$

Mass of Na₂HPO₄ = 4.457 g

$Mass of NaH_2PO_4 = 0.0686 mol * 119.98 \frac{g}{mol}$

Mass of NaH₂PO₄ = 8.23 g

5 0

3 years ago

What is the mass in grams of 3.75 × 10 21 atoms of Li?

adelina 88 [10]

Answer:

The mass

Explanation:

the mass grams of 3.75 × 10 21 atoms of Li is

40. 21 atom of Li

6 0

3 years ago

When 0.873 grams of a protein were dissolved in 48.6 mL of solution at 15.6 degrees C, the osmotic pressure was found to be 0.06

Stels [109]

Answer:

697 g / mol

Explanation:

The osmotic pressure of a protein ( non electrolyte ) is given by:

π V= nRT where V is the volume, n is the number of moles, R is the gas constant ( 0.08206 L·atm/Kmol ), and T is the temperature (K).

n= mass/ MW protein ⇒ MW protein = mass / n

Thus,

π V = ( mass/ MW ) RT

MW = mass x R xT/ ( π V )

mass = 0.873 g

R = 0.08206 L·atm/K·mol

T = ( 15.6 + 273 ) K= 288.6 K

π = 0.061 atm

V = 48.6 mL = 48.6 mL x ( 1 L/ 1000 mL ) = 0.0486 L

We just need to plug our values into the aqbove equation for MW:

MW = 0.873 g x 0.08205 L· atm /K·mol x 288.6 K / ( 0.061 atm x0.0486 L )

= 697 g/mol

6 0

3 years ago

1.00L of a gas at STP is compressed to 473 ml. What is the new pressure of the gas?

I am Lyosha [343]

P₁V₁ = P₂V₂

Convert 473ml > L $P = \frac{(1ATM)(1.00L)}{(0.473L)}$

(1atm)(1.00L) = (0.473L)P₂

Divide by (0.473L)

$P = \frac{(1atm)(1.00L)}{(0.473L)}$

Solve for 2.11atm

4 0

3 years ago

Read 2 more answers