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3 years ago

What is NOT part of Rutherford's model of the atom?

Chemistry

1 answer:

zloy xaker [14]3 years ago

8 0

All the electrons orbit the nucleus in specific orbitals

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SCIENCEEEE!!!!! HELPPPP!!!!

IRINA_888 [86]

pretty sure it's both are physical changes.

5 0

3 years ago

Read 2 more answers

While heating up a 25 gram sample of concrete (specific heat = 0.210-cal/g°C), your initial tempărature is room temperature (25°

Lana71 [14]

Answer:

Final temperature = 83.1 °C

Explanation:

Given data:

Mass of concrete = 25 g

Specific heat capacity = 0.210 cal/g. °C

Initial temperature = 25°C

Calories gain = 305 cal

Final temperature = ?

Solution:

Q = m. c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

305 cal = 25 g ×0.210 cal/g.°C × T2 - 25°C

305 cal = 5.25cal/°C × T2 - 25°C

305 cal / 5.25cal/°C = T2 - 25°C

58.1 °C = T2 - 25°C

T2 = 58.1 °C + 25°C

T2 = 83.1 °C

7 0

3 years ago

What is the wavelength of a light of frequency 6.42 x 1014 Hz?

Dahasolnce [82]

Answer:

467 nm is the answer

Explanation:

IT'S the answer for

8 0

3 years ago

Which rock is an example of a clastic sedimentary rock

KATRIN_1 [288]

Answer:

common sedimentary rocks include sandstone, limestone, and shale.

8 0

3 years ago

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I need help on both a and b of question 1

marishachu [46]

Answer:

(a) -0.00017 M/s;

(b) 0.00034 M/s

Explanation:

(a) Rate of a reaction is defined as change in molarity in a unit time, that is:

$r = \frac{\Delta c}{\Delta t}$

Given the following reaction:

$2 N_2O_5 (g)\rightleftharpoons 4 NO_2 (g) + O_2 (g)$

We may write the rate expression in terms of reactants firstly. Since reactants are decreasing in molarity, we're adding a negative sign. Similarly, if we wish to look at the overall reaction rate, we need to divide by stoichiometric coefficients:

$r = -\frac{\Delta [N_2O_5]}{2 \Delta t}$

Reaction rate is also equal to the rate of formation of products divided by their coefficients:

$r = \frac{\Delta [NO_2]}{4 \Delta t} = \frac{\Delta [O_2]}{\Delta t}$

Let's find the rate of disappearance of the reactant firstly. This would be found dividing the change in molarity by the change in time:

$r_{N_2O_5} = \frac{0.066 M - 0.100 M}{200.00 s - 0.00 s} = -0.00017 M/s$

(b) Using the relationship derived previously, we know that:

$-\frac{\Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{4 \Delta t}$

Rate of appearance of nitrogen dioxide is given by:

$r_{NO_2} = \frac{\Delta [NO_2]}{\Delta t}$

Which is obtained from the equation:

$-\frac{\Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{4 \Delta t}$

If we multiply both sides by 4, that is:

$-\frac{4 \Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{\Delta t}$

This yields:

[tex]r_{NO_2} = \frac{\Delta [NO_2]}{\Delta t} = -2\frac{\Delta [N_2O_5]}{ \Delta t} = -2\cdot (-0.00017 M/s) = 0.00034 M/s[tex]

5 0

3 years ago