Both figures are mixtures,
Figure II is a heterogenous mixture
Figure I is a homogenous mixture
Answer:
The condor has a wing span of 10 feet
Explanation:
This can be solved by a simple rule of three
In a rule of three problem, the first step is identifying the measures and how they are related, if their relationship is direct of inverse.
When the relationship between the measures is direct, as the value of one measure increases, the value of the other measure is going to increase too. In this case, the rule of three is a cross multiplication.
When the relationship between the measures is inverse, as the value of one measure increases, the value of the other measure will decrease. In this case, the rule of three is a line multiplication.
In this problem, our measures are the wing span of the condon in meters and the wing span of the condor is feet. As the value of one of these measures increases, the other is going to increase too.
We know that 1m has 3.281 feet,
So we have the following rule of three:
1m - 3.281 feet
3.05m - x feet
x = 3.821*3.05
x = 10 feet
The condor has a wing span of 10 feet
The mass number is the sum of the protons and neutrons in an atom. 235-92 = 143 neutrons.
Element 92 is Uranium.
Uranium is also the highest atomic number element which exists in nature. All of the rest above element 92 only exist nuclear high-energy reactions (the transuranium elements), and most experience nuclear decay very quickly and form other elements with fewer protons.
Answer:
1.209g of MgO participates
Explanation:
In this problem, we have 0.030 moles of MgO that participates in a particular reaction.
And we are asked to solve for the mass of MgO that participates, that means, we need to convert moles to grams.
To convert moles to grams we need to use molar mass of the compound:
<em>1 atom of Mg has a molar mass of 24.3g/mol</em>
<em>1 atom of O has a molar mass of 16g/mol</em>
<em />
That means molar mass of MgO is 24.3g/mol + 16g/mol = 40.3g/mol
And mass of 0.030 moles of MgO is:
0.030 moles MgO * (40.3g/mol) =
<h3>1.209g of MgO participates</h3>
