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3 years ago

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A neutron has a mass A. that is less than the mass of an electron. B. that is about equal to the mass of an electron. C. that is

about equal to the mass of a proton. D. that is less than the mass of a proton.

1 answer:

soldi70 [24.7K]3 years ago

6 0

Answer:

Option C:- that is equal to mass of an proton.

Explanation:

Protons and neutrons have approximately the same mass, about 1.67 × 10-24 grams, which scientists define as one atomic mass unit (amu) or one Dalton. While electron has mass of 9.31 ×10⁻¹⁹.

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Practice entering numbers that include a power of 10 by entering the diameter of a hydrogen atom in its ground state, dH=1.06×10

kkurt [141]

Answer:

The diameter of the hydrogen $\mathbf{d =1.0605 \times 10^{-10}\ m}$

Explanation:

From the given information:

Using the concept of Bohr's Model, the equation for the angular momentum can be expressed as:

$L = \dfrac{nh}{2 \pi}$

Where the generic expression for angular momentum is:

L = mvr.

replacing the value of L into the previous equation, we have:

$mvr= \dfrac{nh}{2 \pi}$

$v= \dfrac{nh}{2 \pi mr}$ ----- (1)

The electron in the hydrogen atom posses an electrostatic force which gives a centripetal force.

$\dfrac{ke^2}{r^2} = \dfrac{mv^2}{r}$ ----- (2)

replacing the value of v in equation (1) into (2), and taking r as the subject of the formula, we have:

$\dfrac{ke^2}{r} = m (\dfrac{nh}{2 \pi mr})^2$

$ke^2=\dfrac{n^2h^2}{4 \pi^2 mr}$

$r =\dfrac{n^2h^2}{4 \pi^2 mke^2}$

For ground-state n = 1

$h = (6.625 \times 10^{-34} \ J.s)^2$

$m =( 9.1 \times 10^{-31} \ kg)(9 \times 10^9 \ N .m^2/C^2)$

$Ke = (1.6 \times 10^{-19} \ C)^2$

$r =\dfrac{(1)^2(6.625 \times 10^{-34})^2}{4 \pi^2 (9.1 \times 10^{-31} )(9 \times 10^9 ) (1.6 \times 10^{-19})^2}$

$r =\dfrac{4.3890625 \times 10^{-67}}{8.27720295 \times 10^{-57}}$

$\mathbf{r = 5.3025 \times 10^{-11} \ m}$

Therefore, the diameter of hydrogen d = 2r

$\mathbf{d = ( 2 \times 5.3025 \times 10^{-11} \ m})}$

$\mathbf{d =1.0605 \times 10^{-10}\ m}}$

4 0

3 years ago

10. A 38.0-g sample of NaOH is dissolved in water, and the solution is diluted to give a final

Sonbull [250]

Answer:

B.0.558M

Explanation:

M=n/L

n=m/Mm

Mm=NaOH

=23+16+1

=40g/mol

n=m/Mm

= 38/40

=0.95

M=n/L

=0.95/1.70

=0.558

8 0

3 years ago

Compute Reaction Rates for All Seven Trials

jeka94

Answer : The reaction rate at $3^oC,24^oC,40^oC\text{ and }65^oC$ are $36mg/L/sec,142mg/L/sec,190mg/L/sec\text{ and }352mg/L/sec$

Solution : Given,

Mass of tablet = 1000 mg

Volume of water = 0.200 L

The given formula will be,

$\text{Reaction rate}=\frac{\text{Mass of tablet}/\text{Volume of water}}{\text{Reaction time}}$

Now we have to calculate the reaction rate at different temperatures and reaction time.

$\text{Reaction rate at }3^oC=\frac{1000mg/0.200L}{138.5sec}=\frac{5000mg/L}{138.5sec}=36mg/L/sec$

$\text{Reaction rate at }24^oC=\frac{1000mg/0.200L}{34.2sec}=\frac{5000mg/L}{34.2sec}=142mg/L/sec$

$\text{Reaction rate at }40^oC=\frac{1000mg/0.200L}{26.3sec}=\frac{5000mg/L}{26.3sec}=190mg/L/sec$

$\text{Reaction rate at }65^oC=\frac{1000mg/0.200L}{14.2sec}=\frac{5000mg/L}{14.2sec}=352mg/L/sec$

Therefore, the reaction rate at $3^oC,24^oC,40^oC\text{ and }65^oC$ are $36mg/L/sec,142mg/L/sec,190mg/L/sec\text{ and }352mg/L/sec$

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