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2 years ago

The number of _____ determines which element it is.

Chemistry

1 answer:

Alenkasestr [34]2 years ago

5 0

<span>electrons hope this helps</span>

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What is the oxidation number for Zn(OH)4 2-?

AleksAgata [21]

<span>Zn(OH)4= ; -2 = -4 + ?, Zn = +2 , the -4 is the charge on the 4 OH- ions.

NO3- + 6 H2O + 8 e- -------> NH3 + 9 OH-
( Zn + 4 OH- ----> Zh(OH)4= + 2e-) x 4
7 OH- + NO3- + 4 Zn + 6 H2O -------> NH3 + 4 Zn(OH)4</span>

8 0

2 years ago

Gaseous C2H4 reacts with O2 according to the following equation:

gavmur [86]

Each mole of C₂H₄ requires 3 moles of O₂ to react.

5.75 moles of C₂H₄ will require:
3 x 5.75 = 17.25 moles

Each mole of gas occupies 22.4 L at STP.
Liters of gas = 17.25 x 22.4
= 3.86 x 10² L

The answer is C.

6 0

2 years ago

Read 2 more answers

Name the following bases.

zheka24 [161]

Answer:

Explanation:

Ferric Hydroxide

3. 1

Fe. (OH)

Fe(OH)3

6 0

2 years ago

Tengo un volumen desconocido de gas mantenido a una temperatura de 115 K en un recipiente con una presión de 60.0 atm. Si al aum

Delvig [45]

Answer:

be nice bother high school bo

6 0

2 years ago

If 20.0 g of NaOH is added to 0.750 L of 1.00 M Cd(NO₃)₂, how many grams of Cd(OH)₂ will be formed in the following precipitatio

bulgar [2K]

Answer:

$m_{Cd(OH)_2}=36.6 gCd(OH)_2$

Explanation:

Hello.

In this case, for the given chemical reaction, in order to compute the grams of cadmium hydroxide that would be yielded, we must first identify the limiting reactant by computing the yielded moles of that same product, by 20.0 grams of NaOH (molar mass = 40 g/mol) and by 0.750 L of the 1.00-M solution of cadmium nitrate as shown below considering the 1:2:1 mole ratios respectively:

$n_{Cd(OH)_2}^{by\ NaOH}=20.0gNaOH*\frac{1molNaOH}{40gNaOH} *\frac{1molCd(OH)_2}{2molNaOH} =0.25molCd(OH)_2\\\\n_{Cd(OH)_2}^{by\ Cd(NO_3)_2}=0.750L*1.00\frac{molCd(NO_3)_2}{L}*\frac{1molCd(OH)_2}{1molCd(NO_3)_2} =0.75molCd(OH)_2$

Thus, since 20.0 grams of NaOH yielded less of moles of cadmium hydroxide, NaOH is the limiting reactant, therefore the mass of cadmium hydroxide (molar mass = 146.4 g/mol) is:

$m_{Cd(OH)_2}=0.25molCd(OH)_2*\frac{146.4gCd(OH)_2}{1molCd(OH)_2} \\\\m_{Cd(OH)_2}=36.6 gCd(OH)_2$

Best regards.

4 0

2 years ago