Answer:your answer is 0.95
Explanation:
Because you put the number down if it is in the hundreds then put into a decimal that it is in the hundredths:)
Answer:
Ba²⁺(aq) + 2 NO₃⁻(aq) + 2 Rb⁺(aq) + 2 OH⁻(aq) = Ba(OH)₂(s) + 2 Rb⁺(aq) + 2NO₃⁻(aq)
Explanation:
Let's consider the molecular equation between barium nitrate and rubidium hydroxide to produce barium hydroxide and rubidium nitrate.
Ba(NO₃)₂(aq) + 2 RbOH(aq) = Ba(OH)₂(s) + 2 RbNO₃(aq)
The complete ionic equation includes all the ions and the molecular species.
Ba²⁺(aq) + 2 NO₃⁻(aq) + 2 Rb⁺(aq) + 2 OH⁻(aq) = Ba(OH)₂(s) + 2 Rb⁺(aq) + 2NO₃⁻(aq)
Two protons and two neutrons are emitted and trapped as materials like uranium and thorium deep underground decay into radium and thorium, respectively. These alpha-particles transform into stable helium atoms as they take on electrons from their surroundings.
<h3>
What elements go through alpha decay?</h3>
Alpha decay usually occurs in heavy nuclei such as uranium or plutonium, and therefore is a major part of the radioactive fallout from a nuclear explosion.
<h3>
Where does alpha decay occur?</h3>
Alpha decay occurs most often in massive nuclei that have too large a proton to neutron ratio. An alpha particle, with its two protons and two neutrons, is a very stable configuration of particles.
Learn more about alpha decay here:
brainly.com/question/1898040
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Answer:39.8375
Explanation:
The mole for the equation is 1:1
Then the molar mass of KCl is 74.5g
Molar mass of k is 39
74.5g of KCl gives 39g of k
76.10g of KCl gives xg of k
X= 76.10×39/74.5
X= 2967.9/7
X= 39.8375
<h3>
Answer:</h3>
1 x 10^13 stadiums
<h3>
Explanation:</h3>
From the question;
1 x 10^5 people can fill 1 stadium
We are given, 1 x 10^18 atoms of iron
We are required to determine the number of stadiums that 1 x 10^18 atoms of iron would occupy.
We are going to assume that a stadium would occupy a number of atoms equivalent to the number of people.
Therefore;
One stadium = 1 x 10^5 atoms
Then, to find the number of stadiums that will be occupied by 1 x 10^18 atoms;
No. of stadiums = Total number of atoms ÷ Atoms in a single stadium
= 1 x 10^18 atoms ÷ 1 x 10^5 atoms
= 1 x 10^13 stadiums
Therefore, 1 x 10^18 atoms of iron would occupy 1 x 10^13 stadiums
