Its a covalent bond for this q
The formula that only shows the ratio of one atom to another is called the empirical formula.
Firstly, the density of any substance is represented by the mass (amount of matter) as divided by the volume(amount of space). According to external websites, the mass of a penny is 2.5 grams.However, the volume of a penny is .35cm to the power of 3 (due to the thickness of the penny being extremely minimal.Thus the amount of density is extremely little). Therefore, the density of a penny is 0.875 g/cm cubed (dimensional analysis).As for an invention that could be used, that is possible with the usage of a series of measurements that can both calculate mass and volume and directly allocate that to attain density
Answer:
The atomic weight of Rubidium (Rb) is 85.91 amu, using 4 significant figures.
Explanation:
The average atomic mass of Rubidium (Rb) is composed by the mass of the two given isotopes 85^Rb and 87^Rb but each isotope has a different abundance in nature expressed by percentage (natural occurrence).
Therefore, we need to calculate how much of this mass with its abundance percentage contribute to the average atomic mass for each isotope.
First step: Calculate the contribution in mass using the natural occurrence percentage.
85^Rb = 84.91 × = 61.73 amu
87^Rb = 86.91 × = 24.18 amu
The answers for each isotope are given using a typical percentage formula.
Second step: Add the two mass values founded.
24.18 amu + 61.73 amu = 85.91 amu
Third Step: Check the significant figures
The answer is 85.91 amu and it has 4 significant figures.
So,
Given the reaction:
The initial mass of Mg was 0.326, and, in the real experiment, we obtained 0.528 grams of MgO. Right?
Now, let's find the amount of MgO supposed to obtain according to the chemical equation:
Now, the amount that we were supposed to obtain, was 0.54g of MgO. (The result of mutiplying all the previous operations). This is the theoretical yield, what we obtained using the theory.
Now, as you discovered, the mass of MgO in the laboratory was 0.528g so that's the actual yield. This is, what you've found using the experimental process.
Finally, the percent yield can be found using the following equation:
And, we know both values, so let's just replace: