<span>Select the block of cells to be included in the scatter plot by clicking and dragging, then from the Insert ribbon under Chart drop down the Scatter or Bubble menu and select Scatter. A chart will appear on the spreadsheet.
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To set up a scatter plot in Excel, enter the pairs of data in two columns with each value of a pair on the same row. By default, Excel considers the column on the left to contain the horizontal (X) values and the column on the right to contain the vertical (Y) values.
</span><span>If you click on the + sign at the upper right of the chart, a list of checkboxes will appear. Check Axes, Axis Titles, and Trendline. Uncheck everything else. You should edit the Axis Titles to include the name of the factor and any units associated with it. Double-click on the Axis numbers to bring up the Format Axis dialog, then click on the bar-graph icon to access Axis Options. Set the bounds and units appropriately and set the tick marks to something sensible.</span><span>
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Answer : The mass of oxygen combined with 1.00 g of carbon in carbon dioxide will be, 2.66 grams.
Explanation :
Law of multiple proportion : It states that when two elements can combine to form two or more different compounds then the mass of one element compared to fixed mass of the other will always be in a ratio of small whole numbers.
As we are given that the mass of ratio of carbon and oxygen in CO is 1 gram and 1.33 gram.
Ratio of C and O in CO = 1 : 1
Ratio of C and O in CO₂ = 1 : 2
So, the mass of ratio of carbon and oxygen in CO₂ will be 1 gram and (2×1.33) 2.66 gram.
Thus, the mass of oxygen combined with 1.00 g of carbon in carbon dioxide will be, 2.66 grams.
Answer:
82.24% percent composition of N and 17.76% percent composition of H .
Explanation:
Answer:
Explanation:
Explanation:
As you know, the empirical formula tells you what the smallest whole number ratio that exists between the atoms that make up a compound is.
In your case, you know that the empirical formula is
NH Cl
2
, which means that the regardles of how many atoms of each element you get in the actual compound, the ratio that exists between them will always be
1:2:1.
What you actually need to determine is how many empirical formulas are needed to get to the molecular formula.
Notice that the problem provides you with the molar mass of the compound. This means that you can use the molar mass of the empirical formula to determine exactly how many atoms you need to form the compound's molecule.
molar mass empirical formula×n=molar mass compound
To get the molar mass of the empirical formula, use the molar masses of its constituent atoms
14.0067 g/mol+2×1.00794 g/mol+35.453 g/mol=51.48 g/mol≈
51.5 g/mol
This means that you have
51.5g/mol×n=51.5g/mol
As you can see, you have
n=1.
This means that the empirical formula and the molecular formula are equivalent,
NH Cl.
2
