Ask question

Ask question

All categories

2 years ago

11

What are valence electrons? The number of neutrons in a nucleus. The number of protons in the nucleus. The atomic mass is the sa

me as the number of valence electrons. The number of electron in the outer most shell of an atom.

1 answer:

Tanzania [10]2 years ago

3 0

Valence Electrons are The number of electrons in the outer most shell of an atom. These shells have a maximum amount before being filled by electrons, in which remaining electrons will fill the next remaining valence shell.

You might be interested in

SOMEONE PLEASE HELP ME! MEDAL, FAN, AND TESTIMONIAL

yKpoI14uk [10]

As we can see the chemical equation is balanced.K3PO4 + Al(NO3)3 → 3KNO3 + AlPO4

So, by principle of conservation of mass when 1 mole of K3PO4 reacts with 1 mol of Al(NO3)3 , it peoduces 3 mol of KNO3 and 1 mol of AlPO4

So, when 2.5 moles of potassium phosphate react and Al(NO3)3 is present in excess , 2.5*3= 7.5 mol of KNO3 is formed

8 0

3 years ago

____________ is a point source of water pollution; _____________ is a nonpoint source of water pollution.

Deffense [45]

Answer:

good afternoon

Explanation:

thanks I will send

6 0

2 years ago

A chemical ____ is a process in which some substances change chemically into different substances

garik1379 [7]

Is reaction a choice?

6 0

1 year ago

Read 2 more answers

The compound Xe(CF3)2 decomposes in afirst-order reaction to elemental Xe with a half-life of 30. min.If you place 7.50 mg of Xe

Irina-Kira [14]

Answer:

$t=147.24\ min$

Explanation:

Given that:

Half life = 30 min

$t_{1/2}=\frac{\ln2}{k}$

Where, k is rate constant

So,

$k=\frac{\ln2}{t_{1/2}}$

$k=\frac{\ln2}{30}\ min^{-1}$

The rate constant, k = 0.0231 min⁻¹

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given that:

The rate constant, k = 0.0231 min⁻¹

Initial concentration $[A_0]$ = 7.50 mg

Final concentration $[A_t]$ = 0.25 mg

Time = ?

Applying in the above equation, we get that:-

$0.25=7.50e^{-0.0231\times t}$

$750e^{-0.0231t}=25$

$750e^{-0.0231t}=25$

$x=\frac{\ln \left(30\right)}{0.0231}$

$t=147.24\ min$

6 0

3 years ago

A 15.0 g sample of nickel metal is heated to 100.0 degrees C and dropped into 55.0 g of water, initially at 23.0 degrees C. Assu

OLEGan [10]

Answer: The final temperature of nickel and water is $25.2^{o}C$ .

Explanation:

The given data is as follows.

Mass of water, m = 55.0 g,

Initial temp, $(t_{i}) = 23^{o}C$ ,

Final temp, $(t_{f})$ = ?,

Specific heat of water = 4.184 $J/g^{o}C$ ,

Now, we will calculate the heat energy as follows.

q = $mS \Delta t$

= $55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C)$

Also,

mass of Ni, m = 15.0 g,

Initial temperature, $t_{i} = 100^{o}C$ ,

Final temperature, $t_{f}$ = ?

Specific heat of nickel = 0.444 $J/g^{o}C$

Hence, we will calculate the heat energy as follows.

q = $mS \Delta t$

= $15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C)$

Therefore, heat energy lost by the alloy is equal to the heat energy gained by the water.

$q_{water}(gain) = -q_{alloy}(lost)$

$55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C)$ = -( $15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C)$ )

$t_{f} = \frac{25.9^{o}C}{1.029}$

= $25.2^{o}C$

Thus, we can conclude that the final temperature of nickel and water is $25.2^{o}C$ .

6 0

2 years ago