What is the atomic number? what is the atomic mass (a.k.a. mass number?

Given the speed of light as 3.0 × 108 m/s, calculate the wavelength of the electromagnetic radiation whose frequency is 7.5 × 10

arlik [135]

Wavelength= velocity/frequency
wavelength= (3.0 x 10^8m/s) / 7.5 x 10^12 Hz)
you can do the math
I am assuming u that 108 is 10^8 and the 1012 is 10^12

3 0

3 years ago

Olympic cyclists fill their tires with helium to make them lighter. Assume that the volume of the tire is 860 mL , that it is fi

nata0808 [166]

Answer:

a) the mass of air = 8.24 grams

b) the mass of helium = 1.14 grams

c) the mass difference = 7.10 grams

Explanation:

Step 1: Data given

Volume of the tire = 860 mL

Total pressure = 120 psi

Temperature = 26°C

molar mass of air = 28.8 g/mol

Step 2: Convert psi to atm

(

120 psi) (1 atm / 14.7 psi) = 8.163

Step 4: Calculate moles

PV = nRT

⇒ with P = the pressure = 8.163 atm

⇒ with V = the volume = 860 mL = 0.860 L

⇒ with n = the number of moles = TO BE DETERMINED

⇒ with R = the universal gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 26 °C = 299 Kelvin

n = (8163*0.860)/(0.08206*299)

n = 0.2861 moles of gas

Step 5: Calculate the mass of air in an air-filled tire.

Mass = moles * molar mass

Mass = (0.2861 moles of gas) (28.8 g/mol)

Mass = 8.24 grams

Step 6: Calculate the mass of helium in a helium- filled tire.

mass of helium = 0.2861 moles of gas * 4 g/mol)

mass of helium = 1.14 grams

Step 7: What is the mass difference between the two?

Δmass= 8.24 grams - 1.14 grams

Δmass= 7.10 grams

8 0

3 years ago

A silver cube with an edge length of 2.42 cm and a gold cube with an edge length of 2.75 cm are both heated to 85.4 ∘C and place

kakasveta [241]

Answer:

Explanation:

Volume of silver cube = 2.42³ = 14.17 cm³

mass of silver cube = volume x density

= 14.17 x 10.49 = 148.64 gm

Volume of gold cube = 2.75³ = 20.8 cm³

mass of gold cube = 20.8 x 19.3 = 401.44 gm

specific heat of silver and gold are .24 and .129 J /g°C

mass of 112 mL water = 112 g

Heat absorbed = heat lost = mass x specific heat x temperature fall or rise

Heat lost by metals

= 148.64 x .24 x ( 85.4 -T) + 401.44 x .129 x ( 85.4 - T )

= (35.67 + 51.78 ) x ( 85.4 - T )

87.45 x ( 85.4 - T )

= 7468.23 - 87.45 T

Heat gained by water

= 112 x 1 x ( T - 20.5 )

= 112 T - 2296

Heat lost = heat gained

7468.23 - 87.45 T = 112 T - 2296

199.45 T = 9764.23

T = 48.95° C

7 0

3 years ago

For the titration of 25.00 mL of 0.150 M HCl with 0.250 M NaOH, calculate:

Leviafan [203]

1) Chemical reaction

HCl        +       NaOH      --->      NaCl + H2O

25.0 ml
0.150 M            0.250M

2) 50% completion => 0.025 l * 0.150 M * (1/2) = 0.001875 mol HCl consumed and 0.001875 mol HCl in solution

0.001875 mol HCl => 0.001875 mol H(+)

Volume = Volume of HCl solution + Volumen of NaOH solution added

Volume of HCl solution = 0.0250 l

Volume of NaOH = n / M = 0.001875 mol / 0.250M = 0.0075 l

Total volume = 0.0250 l + 0.0075 l = 0.0325 l

[H+] = 0.001875 mol / 0.0325 l = 0.05769 M

pH = - log [H+] = - log (0.05769) = 1.23

Answer: 1.23

3) Equivalence point

0.02500 l * 0.150 M = 0.250M * V

=> V = 0.02500 * 0.150 / 0.250 = 0.015 l

4) 1.00 ml NaOH added beyond the equivalence point

1.00 ml * 1 l / 1000 ml * 0.250 M = 0.00025 mol NaOH in excess

0.00025 mol NaOH = 0.00025 mol OH-

Volume of the solution = 0.02500 l + 0.015 l + 1.00/1000 l = 0.041 l

[OH-] = 0.00025 mol / 0.041 l = 0.00610 M

pOH = - log (0.00610) = 2.21

pH + pOH = 14 => pH = 14 - pOH = 14 - 2.21 = 11.76

Answer: 11.76

6 0

3 years ago

If this decay has a half life of 2.60 years, what mass of 72.5 g of sodium 22 will remain after 15.6 years

Vlad1618 [11]

Sodium-22 remain : 1.13 g

<h3>Further explanation </h3>

The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.

Usually, radioactive elements have an unstable atomic nucleus.

General formulas used in decay:

$\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{T/t\frac{1}{2} }}}$

T = duration of decay

t 1/2 = half-life

N₀ = the number of initial radioactive atoms

Nt = the number of radioactive atoms left after decaying during T time

half-life = t 1/2=2.6 years

T=15.6 years

No=72.5 g

$\tt Nt=72.5.\dfrac{1}{2}^{15.6/2.6}\\\\Nt=72.5.\dfrac{1}{2}^6\\\\Nt=1.13~g$

8 0

3 years ago