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madam [21]
3 years ago
6

Calculate the number of oxygen atoms in a 80.0 g sample of scheelite (CaWo).

Chemistry
1 answer:
castortr0y [4]3 years ago
3 0

Answer:

Explanation:

scheelite is CaWO₄

Mol weight = 288

80 g of scheelite = 80 / 288 = 27.77 x 10⁻² moles

27.77 x 10⁻² moles of scheelite = 27.77 x 10⁻² x 6.02 x 10²³  molecules of scheelite

= 167.17 x 10²¹ molecules of scheelite

1 molecule of scheelite contains 4 atoms of oxygen

167.77 x 10²¹ molecules of scheelite contains 4 x 167.77 x 10²¹ atoms of oxygen .

= 671.08 x 10²¹ atoms of oxygen .

= 671 x 10²¹ atoms .

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Cu + HNO3 --&gt; Cu(NO3)2 + NO + H2O<br> how to balance... WILL GIVE 15 POINTS
aivan3 [116]

Answer:

3 Cu + 8 HNO3 --> 3 Cu(NO3)2 + 2 NO + 4 H2O

Explanation:

Make sure both sides are equal

3 <u>Cu</u> + 8 H<u>N</u>O3 --> 3 <u>Cu</u>(NO3)2 + 2 <u>N</u>O + 4 H2O

// start by those elements that change their oxidation degree

<u>Cu</u> and <u>N</u>

// also you can write reduction-oxidation reactions

Cu^{0} + 2 e^{-} --> Cu^{-2} | 2

N^{+5} - 3 e^{-} -->  N^{+2} | 3

// write the numbers of electrons that are lost/gained as the coefficients of the opposite elements

// then check if H and O are the same on both sides

// adjust if they aren't.

7 0
3 years ago
Read 2 more answers
The question is in the pic
kumpel [21]
Ngl, there is no picture.
7 0
3 years ago
In what area of the titration curve for phosphoric acid (h3po4) is the molecule fully deprotonated?
Arada [10]

The titration curve should have flat regions centered around each of the

three halfway points (buffer zones) and sharp increases in pH around the

equivalence points.

Initial pH

This is determined by the most acidic of the Ka values and the initial

concentration of the acid. (Same as a monoprotic acid)

Half-way points

At each halfway point, the pH = pKa of the group you are titrating. At this point in

the titration curve, we are in a buffering region, and the curve will be relatively

flat.

Equivalence points

At each equivalence point, the pH is the average of the pKa values above and

below. At the last equivalence point (the endpoint), the pH is determined by the

Kb of the conjugate base of the weakest acid.

Plotting the titration of 100 mL of 0.10 M phosphoric acid with 1.0 M NaOH.

H3PO4 + H2O  H2PO4

-

+ H3O+

Ka1 = 7.5 x 10-3

H2PO4

-

+ H2O  HPO4

2- + H3O+

Ka2 = 6.2 x 10-8

HPO4

2- + H2O  PO4

3- + H3O+

Ka3 = 4.2 x 10-13

Plot these points and connect them to determine the titration curve of phosphoric acid. The curve should be relatively flat around each of the halfway points when we are in a buffering region.

The titration curve should have flat regions centered around each of the

three halfway points (buffer zones) and sharp increases in pH around the

equivalence points.

For more information on the titration curve click on the link below:

brainly.com/question/3130161

#SPJ4

4 0
2 years ago
What is the total mass of KNO3 that must be dissolved in 50. grams of H2O at 60.°C to make a saturated solution?
IRINA_888 [86]

at 60ºC:

106 g KNO₃ -------------- 100 g (H₂O)
? g KNO₃ ------------------ 50.0 g (H₂O)

mass KNO₃ = 50.0 * 106 / 100

mass KNO₃ = 5300 / 100

<span>= </span>53 g of KNO₃

answer (2)

<span>hope this helps!</span>

4 0
4 years ago
Read 2 more answers
H₂SO₄ is considered to be
uranmaximum [27]

Answer:

sulfuric acid

Explanation:

a strong acid that soluble in water.

6 0
3 years ago
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