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4 years ago

What is the mass of 5.35×1021 platinum atoms?

1 answer:

6 0

Atomic weight of a Platinum:
195.078 u = 195.078 * 1.66 * 10^(-27) kg = 3.2382948 * 10^(-25) kg
m ( Pt ) = 3.2382948 * 10^(-25) * 5.35 * 10^(21) =
= 1.732488 * 10^(-3) kg = 1.732448 g

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which of the following will stay constant, no matter if the substance is in the solid, liquid, or gas state?

pogonyaev

Liquid I think it might be

6 0

3 years ago

H2SO4 is a strong acid because the first proton ionizes 100%. The Ka of the second proton is 1.1x10-2. What would be the pH of a

forsale [732]

Question options:

a) 2.05

b) 0.963

c) 0.955

d) 1.00

Answer:

b) 0.963

Explanation:

H2SO4→ HSO4- + H3O+

HSO4- + H2O ⇌ SO42- + H3O+

Construct ICE table:

HSO4- (aq) + H2O ⇌ SO42- (aq) + H3O+ (aq)

I 0.1 solid & 0 0.1

C -x liquid + x + x

E 0.1 - x are ignored x 0.1 + x

Calculate x

Ka = products/reactants

= $\frac{[SO42-] [H3O+]}{[HSO4-]}$

0.011 = $\frac{x (0.1 + x)}{0.1 - x}$

0.011 x (0.1 -x) = o.1x + x^2

0.0011 - 0.011 x - o.1x - x^2 = 0

0.0011 - 0.011 x - x^2 = 0

Use formula to solve for quadratic equation

x = ${ -b +,-\sqrt{b^2 - 4ac$ / 2a

a = -1, b = -0.111, c = 0.001

Solve for x

x = $\sqrt[-(-o.111)]{(-0.111)^2 - 4(-1) (0.0011) }$ / 2(-1)

x = 0.111 +,- $\sqrt{0.012321 + 0.0044}$ / -2

x = 0.111 +,- $\sqrt{0.016721}$ / -2

x = $\frac{0.111 +, - 0.1293}{-2}$

x = $\frac{0.111 + 0.1293}{-2}$ , x = $\frac{0.111 - 0.1293}{-2}$

x = $\frac{0.2403}{-2}$ , x = $\frac{0.0183}{-2}$

x = - 0.12015 , x = 0.00915

x cannot be negative, so

x = 0.00915 M

Calculate [H3O+]

[H3O+] = 0.1 M + x

[H3O+] = 0.1 M + 0.00915 M

[H3O+] = 0.10915 M

Clculate pH

pH = - log [ H3O+]

pH = - log [ 0.10915]

pH = 0.963

8 0

3 years ago

Be sure to answer all parts. Solving the Rydberg equation for energy change gives ΔE = R[infinity]hc [ 1 n12 − 1 n22 ] where the

icang [17]

Answer:

Explanation:

Utilizing Rydber's equation:

ΔE = Z²Rh ( 1/n₁² - 1/n₂²) and substituting the values given ( using the Rydbers constant value in Joules ), we have

n=1 to n= infinity

ΔE = 3² x (1/1² - 0) x 2.18 x 10⁻^18 J = 2.0 x 10⁻¹⁷ J (1/infinity is zero)

n= 3 to n= infinity

ΔE = 3² x (1/3² - 0) x 2.18 x 10⁻^18 J = 2.28 x 10^-18 J

b. The wavelength of the emitted can be obtained again by using Rydberg's equation but this time use the constant value 1.097 x 10⁷ m⁻¹ given in the problem .

1/λ = Z²Rh (1/n₁² - 1/n₂²) 10 ⁻¹ = 3² x 1.097 x 10⁷ m⁻¹ x (1/1² - 1/3²) m⁻¹

1/λ =8.8 x 10⁷ m⁻¹ ⇒ λ =1.1 x 10^-8 m

λ = 1.1 x 10^-8 m x 1 x 10⁹ nm/m = 11 nm

7 0

3 years ago

Read 2 more answers

I need help with this pls help !!

Doss [256]

Answer:

70g of HCl

Check the attachment below. Locate 25C and then follow the line where it meets the curve.

4 0

3 years ago

Calculate the volume of .5 molecsample of gas at 140 k and a pressure of 1.2 atm

nlexa [21]

582 ml is the volume of gas.

Answer:

Explanation:

Using the ideal law of gases, the kinetic theory of gas can be stated as

PV = nRT

Here P is the pressure and it is given as 1.2 atm, V is the volume, which we need to find. Then R is the gas constant and T is the temperature while n is the concentration present.

As per the give question, P = 1.2 atm, T = 140 K and n = 0.5 moles, R = 8.314

V = $\frac{nRT}{P}$

V = $\frac{(0.5*8.314*140}{1.2}= 582 l.$

So 582 ml is the volume of gas.

6 0

3 years ago