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3 years ago

8

1. What did Henri Becquerel discover about radiation emitted from uranium salts?

1 answer:

Dmitriy789 [7]3 years ago

6 0

Explanation:

When Henri Becquerel investigated the newly discovered X-rays in 1896, it led to studies of how uranium salts are affected by light. By accident, he discovered that uranium salts spontaneously emit a penetrating radiation that can be registered on a photographic plate.

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How are catapults an example of a third class lever? Please explain!

Katen [24]

I hope that you can understand this!!! Lol

The thing that you have to pull back to release with, that would be considered a third class lever.

I hope this helps. :)

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3 years ago

How long will it take for a 750 mg sample of radium with a half life of 15 days to decay to exactly 68mg?

weqwewe [10]

Answer:

52 da

Step-by-step explanation:

Whenever a question asks you, "How long to reach a certain concentration?" or something similar, you must use the appropriate integrated rate law expression.

The i<em>ntegrated rate law for a first-order reaction </em>is

ln([A₀]/[A] ) = kt

Data:

[A]₀ = 750 mg

[A] = 68 mg

t_ ½ = 15 da

Step 1. Calculate the value of the rate constant.

t_½ = ln2/k Multiply each side by k

kt_½ = ln2 Divide each side by t_½

k = ln2/t_½

= ln2/15

= 0.0462 da⁻¹

Step 2. Calculate the time

ln(750/68) = 0.0462t

ln11.0 = 0.0462t

2.40 = 0.0462t Divide each side by 0.0462

t = 52 da

8 0

3 years ago

The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures

QveST [7]

<u>Answer:</u>

<u>For A:</u> The $K_p$ for the given reaction is $4.0\times 10^1$

<u>For B:</u> The $K_c$ for the given reaction is 1642.

<u>Explanation:</u>

The given chemical reaction follows:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

<u>For A:</u>

The expression of $K_p$ for the above reaction follows:

$K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

$p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm$

Putting values in above equation, we get:

$K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1$

Hence, the $K_p$ for the given reaction is $4.0\times 10^1$

<u>For B:</u>

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = $4.0\times 10^1$

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642$

Hence, the $K_c$ for the given reaction is 1642.

7 0

3 years ago

In chapter 51, Pip, Wemmick, and Jaggers discuss Estella's parentage. What does Jaggers advise Pip to do with this information?

Aleks04 [339]

What’s this story called

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3 years ago

How is the periodic table helpful in determining the types of bonds an element will form?

BigorU [14]

1.)b
2.)true
3.)false
are the answer don't take me on my word

6 0

3 years ago