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2 years ago

The effective nuclear charge of an atom is primarily affected by ________.

1 answer:

5 0

The answer to this question would be: inner electrons

The effective nuclear charge is the net positive charge that applied to an electron of a polyelectronic atom. It was written with the formula of Zeff= Z-S where Zeff means the effective nuclear charge, Z is the nuclear charge and S is the average number of the electron between the electron and nucleus.
Since only inner electron is counted on S, the outer electron has no effect.

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Estimate your de Broglie wavelength when you are running. (For this problem use h = 10^−34 in SI units and 1 lb is equivalent to

AlekseyPX

Answer:

A. your running speed 1.5 m/s

B. your mass 70 kg

C. your de Broglie wavelength $6.32x10^{-36}$ m

Explanation:

Hello there!

In this case, since the equation for the calculation of the Broglie wavelength is:

$\lambda =\frac{h}{m*v}$

We can assume a running speed of about 1.5 m/s and a mass of 70 kg, so the resulting Broglie wavelength is:

$\lambda =\frac{6.626x10^{-34}kg\frac{m}{s} }{70kg*1.5\frac{m}{s} }\\\\\lambda =6.32x10^-36m$

Best regards!

7 0

3 years ago

The decomposition of carbon disulfide to carbon monosulfide and sulfur is first order with k=2.8 ×10^-7 at 1000°C .What is the h

RoseWind [281]

Answer:

2.5×10⁶ s

Explanation:

From the question given above, the following data were obtained:

Rate constant (K) = 2.8×10¯⁷ s¯¹

Half-life (t½) =?

The half-life of a first order reaction is given by:

Half-life (t½) = 0.693 / Rate constant (K)

t½ = 0.693 / K

With the above formula, we can obtain the half-life of the reaction as follow:

Rate constant (K) = 2.8×10¯⁷ s¯¹

Half-life (t½) =?

t½ = 0.693 / K

t½ = 0.693 / 2.8×10¯⁷

t½ = 2.5×10⁶ s

Therefore, the half-life of the reaction is 2.5×10⁶ s

6 0

2 years ago

A light or optical microscope most commonly uused is a

kirill115 [55]

A bright feild is most commonly used

4 0

3 years ago

Magnesium and oxygen balanced

11Alexandr11 [23.1K]

2Mg+ O₂= 2MgO
The charge of Mg is+2 and the charge of Oxygen is -2. 2 oxygen atoms will have a charge of -4 so the Mg atoms have to equal +4. A coefficient of 2 for the Mg atoms will balance it out

8 0

2 years ago

The normal freezing point of water is 0.00 ⁰C. What is the freezing point of a solution containing450.0 mg of ethylene glycol (M

anyanavicka [17]

Answer:

Freezing T° of solution = - 8.98°C

Explanation:

We apply Freezing point depression to solve this problem, the colligative property that has this formula:

Freezing T° of pure solvent - Freezing T° of solution = Kf . m

Kf = 1.86°C/m, this is a constant which is unique for each solvent. In this case, we are using water

m = molality (moles of solute / 1kg of solvent)

We convert the mass of solvent from g to kg

1.5 g . 1kg/1000g = 0.0015 kg

We convert the mass of solute, to moles. Firstly we make this conversion, from mg to g → 450mg . 1g/1000mg = 0.450 g

0.450 g. 1mol / 62.07g = 7.25×10⁻³ moles

Molality → 7.25×10⁻³ mol / 0.0015 kg = 4.83 m

- Freezing T° of solution = 1.86°C /m . 4.83 m - Freezing T° of pure solvent

-Freezing T° of solution = 1.86°C /m . 4.83 m - 0°C

Freezing T° of solution = - 8.98°C

8 0

2 years ago

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