A) fluorite B) orthoclase C) apatite D) gypsum

If a laser operating at a wavelength of 488 nm and a power of 123.0 mW is turned on for 18.73 minutes, how many photons has it e

Verizon [17]

Answer:

3.39e+20

Explanation:

7 0

2 years ago

The equation that shows the relationship between speed, wavelength, and frequency of electromagnetic waves is

lorasvet [3.4K]

Answer:

c = λ f

c is the speed of light, λ the wavelength in meters, and f equals the frequency in cycles per second.

Explanation:

3 0

2 years ago

You are asked to prepare 500. mL 0.150 M acetate buffer at pH 5.10 using only pure acetic acid ( MW = 60.05 g/mol, p K a = 4.76

Amanda [17]

Answer:

4.504g of acetic acid

Explanation:

The acetic acid in reaction with NaOH produce acetate ion, thus:

CH₃COOH + NaOH → CH₃COO⁻ + H₂O + Na⁺

That means the moles of acetate buffer comes, in the first, from the acetic acid

As you need 500mL (0,500L) of a 0.150M acetate buffer, moles are:

0.500L × (0.150mol / 1L) = 0.075 moles of acetate. That is:

0.075mol = [CH₃COO⁻] + [CH₃COOH]

Thus, grams of acetic acid you need to prepare the buffer are:

0.075 moles acetic acid × (60.05g / 1mol) = 4.504g of acetic acid

6 0

3 years ago

Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl ( aq ) , as describe

Nataly [62]

Answer:

0.9483 grams of manganese dioxide should be added to excess HCl.

Explanation:

Pressure of the chlorine gas = P = 795 Torr = 1.046 atm (1 atm = 760 Torr)

Volume of the chlorine gas = V = 255 ml = 0.255 L

Temperature of the chlorine gas = T = 25°C= 298.15 K

Moles of chlorine gas = n

Using ideal gas equation:

PV = nRT

$n=\frac{PV}{RT}=\frac{1.046 atm\times 0.255 L}{0.0821 atm l/mol k\times 298.15 K}$

n = 0.01090 mol

$MnO_2 ( s ) + 4 HCl ( aq)\rioghtarrow MnCl_2 ( aq ) + 2 H_2O ( l ) + Cl_2 ( g )$

According to reaction , 1 mole of chlorine gas is obtained from 1 mole of manganese dioxide.

Then 0.01090 moles of chlorine gas will be obtained from:

$\frac{1}{1}\times 0.01090 mol=0.01090 mol$ manganese dioxide

Mass of 0.01090 moles of manganese dioxide:

0.01090 mol × 87 g/mol = 0.9483 g

8 0

3 years ago

Plz!!!!!!!!!!!!!!!!! name the given substances and calculate their molecular mass

azamat

Where is it then? i cant see it

4 0

3 years ago