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bulgar [2K]
3 years ago
9

Directions: Show all of your work for each of the following problems.

Chemistry
1 answer:
AysviL [449]3 years ago
8 0

Answer:

<h3>3.Ba=2</h3><h3>4.Ni=2...PO4=3</h3><h3>5.Al=3...CrO4=2</h3><h3>6.NH4=1...SO4=2</h3>
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An electron ______ is an area around the nucleus of an atom where an electron is likely to be found.
jek_recluse [69]

Answer:

orbital

hope this helps! <3

4 0
2 years ago
What is handling of materials and chemicals
Serjik [45]
Exsperements? labs? chemesty? it could be a few things...
5 0
3 years ago
**PLATO QUESTION, PLEASE ANSWER CORRECTLY**
Ivenika [448]
The corect answers will be:
1) A
2) B
3) D

Hope this helped :)
7 0
2 years ago
When 125 mL of 0.150 M Pb(NO3)2 is mixed with 145 mL of 0.200 M KBr, 4.92 g of PbBr2 is collected. Calculate the percent yield.
Semenov [28]

Answer:

Y = 92.5 %

Explanation:

Hello there!

In this case, since the reaction between lead (II) nitrate and potassium bromide is:

Pb(NO_3)_2+2KBr\rightarrow PbBr_2+2KNO_3

Exhibits a 1:2 mole ratio of the former to the later, we can calculate the moles of lead (II) bromide product to figure out the limiting reactant:

0.125L*0.150\frac{molPb(NO_3)_2}{L} *\frac{1molPbBr_2}{1molPb(NO_3)_2} =0.01875molPbBr_2\\\\0.145L*0.200\frac{molKBr}{L} *\frac{1molPbBr_2}{2molKBr} =0.0145molPbBr_2

Thus, the limiting reactant is the KBr as it yields the fewest moles of PbBr2 product. Afterwards, we calculate the mass of product by using its molar mass:

0.0145molPbBr_2*\frac{367.01gPbBr_2}{1molPbBr_2} =5.32gPbBr_2

And the resulting percent yield:

Y=\frac{4.92g}{5.32g} *100\%\\\\Y=92.5\%

Regards!

4 0
3 years ago
What is the wavelength of infrared radiation with a frequency of 3.85 x 10^12 Hz?
ELEN [110]

Answer:

Wavelenght is 7,79x10⁻⁵ m

Explanation:

The equation that connects wavelentgh (λ) and frequency (ν) is:

λ=c/ν

Where c is speed of light (3x10⁸ m/sec) and λ is expressed in lenght´s units and ν is expressed in "time⁻¹ " units (for example, sec⁻¹)

According to the details, if we just replace the given value of frequency, we just obtaing wavelenght data:

λ= (3x10⁸ m/sec)/(3,85x10¹² sec⁻¹) = 7,79x10⁻⁵ m

3 0
3 years ago
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