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2 years ago

Why is the process of nuclear fusion important to life on Earth?

1 answer:

6 0

Nearly all life on Earth gets its energy from the sun, and the sun gets its energy through the process of nuclear fusion, which is why these type of energy is important to life on Earth.

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Consider the following reaction:

Sidana [21]

Your limiting is CuCI2 and the excess is KI (from what i’ve heard from my tc to find it just use the moles or look at the grams)Do you want me to do the qn and give u the ans or?

Explanation:You have more grams of KI than CuCI2
irl example : I need 200g of flour to bake 1 muffin and 100g of butter.But I have 300g of butter and only 200g of flour.This means I can only bake up to 1 muffin since I got excess grams of butter.But to use up all my 300g of butter I need 400g more of flour.Making my butter the excess while my flour the limiting since I have less of it and it also determines how much muffin would I get at the end of the bake.

im sorry if that example sounds clowny T-T

6 0

2 years ago

The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151

vitfil [10]

Answer:- The natural abundance of $^1^5^1_E_u$ is 0.478 or 47.8% and $^1^5^3_E_u$ is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for $^1^5^1_E_u$ and $^1^5^3_E_u$ as 150.919860 and 152.921243 amu, respectively. Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of $^1^5^1_E_u$ as n then the abundance of $^1^5^3_E_u$ would be 1-n .

Let's plug in the values in the formula:

$151.964=150.919860(n)+152.921243(1-n)$

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

$151.964-152.921243=150.919860n-152.921243n$

$-0.957243=-2.001383n$

negative sign is on both sides so it is canceled:

$0.957243=2.001383n$

$n=\frac{0.957243}{2.001383}$

$n=0.478$

The abundance of $^1^5^1_E_u$ is 0.478 which is 47.8%.

The abundance of $^1^5^3_E_u$ is = $1-0.478$

= 0.522 which is 52.2%

Hence, the natural abundance of $^1^5^1_E_u$ is 0.478 or 47.8% and $^1^5^3_E_u$ is 0.522 or 52.2% .

3 0

2 years ago

Consider the balanced equation.

vaieri [72.5K]

The balanced chemical reaction is:

N2 + 3H2 = 2NH3

We are given the amount of H2 being reacted. This will be our starting point.

26.3 g H2 (1 mol H2 / 2.02 g H2) 2 mol O2/3 mol H2) ( 17.04 g NH3 / 1mol NH3) = 147.90 g O2

Percent yield = actual yield / theoretical yield x 100

Percent yield = 79.0 g / 147.90 g x 100

Percent yield = 53.4%

8 0

2 years ago

Read 2 more answers

What is the charge for MnO4

lakkis [162]

Answer:

Mn = 2+, O = -2

Explanation:

3 0

2 years ago

the amount of heat involved in the synthesis of 1 mole of a compound from its elements with all substances in their standard sta

otez555 [7]

The standard enthalpy of formation is defined as the change in enthalpy when one mole of a substance in the standard state (1 atm of pressure and temperature of 298.15 K) is formed from its pure elements under the same conditions.

4 0

2 years ago

Read 2 more answers