All categories

2 years ago

Moving from boron to carbon, the intensity of the bulb __

Chemistry

1 answer:

AleksandrR [38]2 years ago

7 0

Explanation:

The thickness of the frosting Moving from boron to carbon, the intensity of the bulb blank because Z increases from blank to blank. The thickness of the frosting blank because the core electron configuration is the same for both atoms. because the core electron configuration is the same for both atoms.

You might be interested in

A 220.0 gram piece of copper is dropped into 500.0 grams of water 24.00 °C. If the final temperature of water is 42.00 °C, what

FrozenT [24]

Answer:

C. 481 °C.

Explanation:

At equilibrium:

The amount of heat absorbed by water = the amount of heat released by copper.

To find the amount of heat, we can use the relation:

where, Q is the amount of energy.

m is the mass of substance.

c is the specific heat capacity.

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T).

<em>∵ Q of copper = Q of water</em>

∴ - (m.c.ΔT) of copper = (m.c.ΔT) of water

m of copper = 220.0 g, c of copper = 0.39 J/g °C, ΔT of copper = final T - initial T = 42.00 °C - initial T.

m of water = 500.0 g, c of water = 4.18 J/g °C, ΔT of water = final T - initial T = 42.00 °C - 24.00 °C = 18.00 °C.

∴ - (220.0 g)( 0.39 J/g °C)(42.00 °C - Ti) = (500.0 g)(4.18 J/g °C)(18.00 °C)

∴ - (85.8)(42.00 °C - Ti) = 37620.

∴ (42.00 °C - Ti) = 37620/(- 85.8) = - 438.5.

∴ Ti = 42.00 °C + 438.5 = 480.5°C ≅ 481°C.

<em>So, the right choice is: C. 481 °C.</em>

4 0

2 years ago

What is a common factor associated with most severe storms?

Volgvan

A common factor is low pressure system.

4 0

2 years ago

Read 2 more answers

mastering chem If the experiment below is run for 60 s, 0.16 mol A remain. Which of the following statements is or are true? At

Dvinal [7]

Answer: All of the statements are true.

Explanation:

(a) Considering the system mentioned in the equation:-

The sum of total moles in the flask will always be equal to 1 which leads to confirmation of this statement as for 60 secs= 0.16 mol A and 0.84 mol B

(b) 0<t< 20s, mole A got reduced from 1 mole to 0.54 moles while at 40s to 60s A got decreased from 0.30 moles to 0.16 moles.

0 to 20s is 0.46 (1 - 0.54 = 0.46)mol whereas,

40 to 60s is 0.14 (0.30-.16 = 0.14) mol

(0.46 > 0.14) mol leading this statement to be true as well.

(c) Average rate from t1 = 40 to t2 = 60 s is given by:

$\delta moles/\delta time = 0.30-0.16/60-40 = 0.007 Mol/s$ which is true as well

7 0

2 years ago

Consider the chemical reaction: N2 3H2 yields 2NH3. If the concentration of the reactant H2 was increased from 1.0 x 10-2 M to 2

Brilliant_brown [7]

The equilibrium constant of a reaction is defined as:

"The ratio between equilibrium concentrations of products powered to their reaction quotient and equilibrium concentration of reactants powered to thier reaction quotient".

The reaction quotient, Q, has the same algebraic expressions but use the actual concentrations of reactants.

To solve this question we need this additional information:

<em>For this reaction, K = 6.0x10⁻² and the initial concentrations of the reactants are:</em>

<em />

Thus, for the reaction:

N₂ + 3H₂ ⇄ 2NH₃

The equilibrium constant, K, of this reaction, is defined as:

$K = 6.0x10^{-2} = \frac{[NH_3]^2}{[N_2][H_2]^3}$

And Q, is:

$Q = \frac{[NH_3]^2}{[N_2][H_2]^3}$

Where actual concentrations are:

[NH₃] = 1.0x10⁻⁴M

[N₂] = 4.0M

[H₂] = 2.5x10⁻¹M

Replacing:

$Q = \frac{[1.0x10^{-4}]^2}{[4.0][2.5x10^{-1}]^3}$

As Q < K,

<h3>The chemical system will shift to the right in order to produce more NH₃</h3>

Learn more about chemical equililbrium in:

brainly.com/question/24301138

8 0

2 years ago

Part a how many grams of xef6 are required to react with 0.579 l of hydrogen gas at 6.46 atm and 45°c in the reaction shown belo

Nina [5.8K]

First, let us find the corresponding amount of moles H₂ assuming ideal gas behavior.

PV = nRT
Solving for n,
n = PV/RT
n = (6.46 atm)(0.579 L)/(0.0821 L-atm/mol-K)(45 + 273 K)
n = 0.143 mol H₂

The stoichiometric calculations is as follows (MW for XeF₆ = 245.28 g/mol)
Mass XeF₆ = (0.143 mol H₂)(1 mol XeF₆/3 mol H₂)(245.28 g/mol) = <em>11.69 g</em>

6 0

2 years ago