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Korolek [52]
3 years ago
10

After viewing the tutorial, open a spreadsheet in the Excel application and type the following data gathered from mass and volum

e data of different samples of an unknown metal.
Data Table of Mass and Volume for Unknown Metal

Enter "Volume (mL)" in column A and "Mass (g)" in column B

Volume (mL) Mass (g)
0.259 5.00
0.776 15.0
1.24 24.0
2.69 52.0
3.31 64.0
4.19 81.0
4.92 95.0
5.28 102.0
7.35 142.0
8.59 166.0

After the data is entered - follow the steps in the tutorial and you will have a graph that is professional in appearance.

Be certain that the following is included on your scatter plot (line graph). If not, go back to the graph and enter the missing information. The legend is not needed on this graph.
I NEED HELP ASAP
Chemistry
1 answer:
Anna [14]3 years ago
6 0

Answer:

wow this is an interesting question

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Explain why c 12 6 and c 14 6 are istopes of carbon
sasho [114]
Bc the number of protons & electrons are the same but the number or neutrons are changed
5 0
3 years ago
Read 2 more answers
How do i find a ionic formula? <br><br>for example Ca2+ &amp; S2-​
jeka94

Example #1 write the chemical formula for Calcium Oxide

Step 1

-Find the Atomic symbol of the metal and non-metal on the periodic table

CaO

Step 2

Find the charges for Calcium and Oxygen

which are Ca 2+ and O 2-

Step 3

Balance out the charges

They are already balanced out

Here´s the chemical formula for Calcium Oxide

CaO

Example 2 write the chemical formula for Aluminum Oxide

Step 1

-Find the Atomic symbol of the metal and non-metal on the periodic table

A

l

O

Step 2

Find the charges for Aluminum and Oxygen

Which are Al 3+ and O 2-

Step 3

Balance out the charges

You need 2 Aluminum and 3 Oxygen to balance the charges

Al 3+ O 2-

Al 3+ O 2-

= 6 + O 2-

= 6-

Step 4

If you need more then one element to balance out the charges you identify that by using subscripts

Heres your chemical formula for Aluminum Oxide

Al_{2} O_{3}

8 0
3 years ago
A sample of nitrogen gas is stored in a 0.500 L flask at 101.3 atm. The gas is transferred to a 0.750
Paladinen [302]

Answer:

67.5 atm

Explanation:

To answer this problem we can use <em>Boyle's law</em>, which states that at constant temperature the pressure and volume of a gas can be described as:

P₁V₁=P₂V₂

In this case:

P₁ = 101.3 atm

V₁ = 0.500 L

P₂ = ?

V₂ = 0.750 L

We input the data:

101.3 atm * 0.500 L = P₂ * 0.750 L

And solve for P₂:

P₂ = 67.5 atm

7 0
3 years ago
A chemical engineer has determined by measurements that there are 69.0 moles of hydrogen in a sample of methyl tert-butyl ether.
Juli2301 [7.4K]
<span>5.75 moles The formula for methyl tert-butyl ether is (CH3)3COCH3, so a single molecule has 5 carbon, 12 hydrogen, and 1 oxygen atoms. So for every 12 moles of hydrogen, there's 1 mole of oxygen. So simply divide the number of moles of hydrogen by 12 to get the number of moles of oxygen. 69.0 / 12 = 5.75 Therefore there's 5.75 moles of oxygen in the sample.</span>
3 0
3 years ago
Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction: P4(g) + 10 Cl2(g) → 4PCl5(s) ΔH
Pie

Answer: -1835 kJ

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to Hess’s law, the chemical equation can be treated as algebraic expressions and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

PCl_5(s)\rightarrow PCl_3(g)+Cl_2(g)  \Delta H_1=+157kJ (1)

P_4(g)+6Cl_2\rightarrow 4PCl_3(g)  \Delta H_2=-1207J (2)

Net chemical equation:

P_4(g)+10Cl_2(g)\rightarrow 4PCl_5(s)  \Delta H=? (3)

Multiplying equation (1) by 4,  and reversing we get

4PCl_3(g)+4Cl_2(g)\rightarrow 4PCl_5(s)  \Delta H_4=4\times -157kJ=-628kJ (4)

Adding (2) and (4)

P_4(g)+10Cl_2(g)\rightarrow 4PCl_5(s)  \Delta H_3=\Delta H_2+\Delta H_4=-1207kJ-628kJ=-1835kJ  

Thus enthaply change for the reaction is -1835 kJ.

3 0
4 years ago
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