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Sunny_sXe [5.5K]
3 years ago
11

What is the wavelength of the matter wave associated with an electron (me= 9.1 x 10-31 kg) moving with a speed of 2.5 x 107 m/s?

Chemistry
1 answer:
riadik2000 [5.3K]3 years ago
3 0
And h value is constant 6.624 x 10^-34

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The technetium-99 nuclide radioactively decays by beta emission. write a balanced nuclear chemical equation that describes this
nekit [7.7K]
Tc-99m<span> is a </span>metastable  isomer<span> of </span>Tc-99. It finds widespread applications in <span>medical diagnostic procedures.
</span>
Tc-99 is also a radioactive element. It's half-life is 2,11,000 years. Upon radioactive decay, it emits beta particles and gets converted into stable compound Ruthenium-99

This process of radioactive decay is shown below.
99 43Tc    →       99 44Ru     +     0 -1e
                            (stable)           (β particle)
7 0
3 years ago
15. What volume of CCI, (d = 1.6 g/cc) contain
anastassius [24]

Answer:

\boxed{\text{(3) 9.6 L}}

Explanation:

1. Moles of CCl₄

n = 6.02 \times 10^{25} \text{ molecules} \times \dfrac{\text{1 mol}}{6.022 \times 10^{23}\text{ molecules}} = \text{100.0 mol}

2. Molar mass of CCl₄

MM = 1 × 12.01 + 4 × 35.5 = 12.01 + 142 = 154.0 g/mol

3. Mass of CCl₄

m =\text{100.0 mol} \times \dfrac{\text{154.0 g}}{\text{1 mol}} = \text{15 400 g}

4. Volume of CCl₄

V = \text{15 400 g} \times \dfrac{\text{1 cm}^{3}}{\text{1.6 g}} = \text{9600 cm}^{3}\\\\V = \text{9600 cm}^{3} \times \dfrac{\text{1 L}}{\text{1000 cm}^{3}} = \mathbf{{9.6 L}}\\\\\text{The volume of CCl$_{4}$ is } \boxed{\textbf{9.6 L}}

4 0
3 years ago
PLEASE HELP
Sergio [31]

Answer:

C. The Protons

5 0
3 years ago
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When an object falls toward the ground due to gravity, what type of energy becomes kinetic energy?
Goshia [24]

OPTION C

<h2>POTENTIAL ENERGY </h2>

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6 0
3 years ago
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47.9 ml hrdrogen is collected at 26° Celsius and 718 torr. Find the volume occupied at STP
Temka [501]

Answer:

41.45 mL

Explanation:

Applying the general gas equation,

PV/T = P'V'/T'............... Equation 1

Where P = Initial pressure of hydrogen, V = Initial volume of hydrogen, T= Initial Temperature of hydrogen, P' = Final pressure of hydrogen, V' = Final Volume of Hydrogen, T' = Final Temperature.

make V' the subject of the equation

V' = PVT'/TP'................ Equation 2

Given: P = 718 torr = (718×133.322) N/m² = 95725.196 N/m², V = 47.9 mL = 0.0479 dm³, T = 26 °C = (26+273) = 299 K, T' = 273 K, P' = 101000 N/m²

Substitute these values into equation 2

V' = ( 95725.196×0.0479×273)/(299×101000)

V' = 0.04145 dm³

V' = 41.45 mL

4 0
3 years ago
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