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Irina18 [472]
3 years ago
7

Question 911 point)

Chemistry
1 answer:
andrew-mc [135]3 years ago
8 0
Copper 2

Just like you how you would say Cu(I) as copper one, you would say Cu(II) as copper two. Nothing fancy.
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Can someone please show me how to do problem number 4? Please show work so I can try to understand it. Thanks!
larisa86 [58]
With the given formula, we can calculate the amount of CO₂ using the balance equation but we first need the moles of CH₄

1) to find the moles of CH₄, we need to use the ideal gas formula (PV= nRT). if we solve for n, we solve for the moles of CH₄, and then we can convert to CO₂. Remember that the units put in this formula depending on the R value units. I remember 0.0821 which means pressure (P) has to be in atm, volume (V) in liters, the amount (n) in moles, and temperature (T) in kelvin.

PV= nRT

P= 1.00 atm
V= 32.0 Liters
n= ?
R= 0.0821 atm L/mol K
T= 25 C= 298 K

let plug the values into the formula.

(1.00 x 32.0 L)= n x 0.0821 x 298K

n= (1.00 x 32.0 L )/ (0.0821 x 298)= 1.31 moles CH₄

2) now let's convert the mole of CH₄ to moles to CO₂ using the balance equation

1.31 mol CH₄ (1 mol CO₂/ 1 mol CH₄)= 1.31 mol CO₂

3) Now let's convert from moles to grams using the molar mass of CO₂ (find the mass of each atom in the periodic table and add them)

molar mass CO₂= 12.00 + (2 x 16.0)= 44.0 g/mol

1.31 mol CO₂ ( 44.0 g/ 1 mol)= 57.6 g CO₂

Note: let me know if you any question.



5 0
3 years ago
Visible light is __________________________________ the electromagnetic spectrum. the same as, only one part of ,the major part
viva [34]
Only one part of as there is a lot more that isn’t visible to us
4 0
3 years ago
A compound with molecular formula C6H9No that has an amide functional group and does not have an alkene group
Flauer [41]

Answer:

Following are the responses to the given question:

Explanation:

\to C_6H_9NO

calculating the HOI:

= \frac{2C+2+N-H-X}{2} \\\\ =\frac{2\times 6+2+1-9-0}{2}\\\\ =\frac{12-6}{2}\\\\=\frac{6}{2}\\\\=3

So, in the above compund there are three bounds or we can say that one is double and two bound is single.

Please find the attachment file of the Function group:

In the above given conditon compound doesn't include the alkene functional group that is absense of-C=C- , and compound include -C\equiv C-  So, please find the attached file of the structure of the compound:

3 0
2 years ago
If 5.58 g of iron reacts with sulfur to produce 8.79 g of iron sulfide, what is the mass of reacting sulfur? A) 3215 B) 14:37 C)
gogolik [260]

Answer: A) 3.21 g

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side.

Fe+S\rightarrow FeS

We are given:

Mass of iron = 5.58 g

Mass of iron sulphide = 8.79 g

Mass of sulphur = x g

Total mass on reactant side = 5.58 + x

Total mass on product side = 8.79 g

Applying law of conservation of mass, we get:

5.58+x=8.79\\\\x=3.21g

Hence, the mass of reacting sulfur is 3.21 g.

7 0
3 years ago
Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 with aqueous NaI . Include phases. chemical equation: Wha
Citrus2011 [14]

<u>Answer:</u> The mass of lead iodide produced is 9.22 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of NaI = 0.200 M

Volume of solution = 0.200 L

Putting values in above equation, we get:

0.200M=\frac{\text{Moles of NaI}}{0.200}\\\\\text{Moles of NaI}=(0.200mol/L\times 0.200L)=0.04moles

The chemical equation for the reaction of NaI and lead chlorate follows:

Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)

By Stoichiometry of the reaction:

2 moles of NaI reacts produces 1 mole of lead iodide

So, 0.04 moles of NaI will react with = \frac{1}{2}\times 0.04=0.02mol of lead iodide

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of lead iodide = 461 g/mol

Moles of lead iodide= 0.02 moles

Putting values in above equation, we get:

0.02mol=\frac{\text{Mass of lead iodide}}{461g/mol}\\\\\text{Mass of lead iodide}=(0.02mol\times 461g/mol)=9.22g

Hence, the mass of lead iodide produced is 9.22 grams

6 0
3 years ago
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