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Rina8888 [55]
3 years ago
10

What metirial is this

Chemistry
2 answers:
BaLLatris [955]3 years ago
7 0
The pic is very unclear sorry
aleksandrvk [35]3 years ago
3 0
I'm not sure based on the picture, but if this is chewing gum I think it would be the material of chicle. (Sorry if this didn't help)
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From the combinations of substances listed below, which would most likely be miscible in
polet [3.4K]

Answer:

C

Explanation:

polar has unequal sharing of electrons that has the lone pairs which has the electronegativity difference. can be mixed with water.

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3 years ago
What statement would beast describe a digital signal
Assoli18 [71]
Best* and are there answer choic
6 0
3 years ago
Cr2o2−7(aq)+i−(aq)→cr3+(aq)+io−3(aq) (acidic solution) express your answer as a chemical equation. identify all of the phases in
Rashid [163]
<span>Answer: Nothing is balanced in your final equation: not H, not O, not Cr, not I and your charges aren't either. Start with your 2 half reactions: I- --> IO3- Cr2O72- --> 2 Cr3+ Balance O by adding H2O: I- + 3 H2O --> IO3- Cr2O72- --> 2 Cr3+ + 7H2O Balance H by adding H+: I- + 3 H2O --> IO3- + 6 H+ Cr2O72- + 14 H+ --> 2 Cr3+ + 7H2O Balance charge by adding e-: I- + 3 H2O --> IO3- + 6 H+ + 6 e- Cr2O72- + 14 H+ + 6 e- --> 2 Cr3+ + 7H2O Since the numbers of electrons in your two half reactions are the same, just add them and simplify to give: Cr2O72- + I- + 8 H+ --> IO3- + 2 Cr3+ + 4 H2O</span>
4 0
3 years ago
What is the maximum concentration of Ag⁺ that can be added to a 0.00750 M solution of Na₂CO₃ before a precipitate will form? (Ks
Firlakuza [10]

Answer:

\large \boxed{1.64\times 10^{-5}\text{ mol/L }}

Explanation:

Ag₂CO₃(s) ⇌2Ag⁺(aq) + CO₃²⁻(aq); Ksp = 8.10 × 10⁻¹²

                           2x      0.007 50 + x

K_{sp} =\text{[Ag$^{+}$]$^{2}$[CO$_{3}^{2-}$]} = (2x)^{2}\times 0.00750 = 8.10 \times 10^{-12}\\0.0300x^{2} = 8.10 \times 10^{-12}\\x^{2} = 2.70 \times 10^{-10}\\x = \sqrt{2.70 \times 10^{-10}} = \mathbf{1.64\times 10^{5}} \textbf{ mol/L}\\\text{The maximum concentration of Ag$^{+}$ is $\large \boxed{\mathbf{1.64\times 10^{-5}}\textbf{ mol/L }}$}

 

3 0
3 years ago
How many moles are in a 12.0 g sample of NiC12
Nady [450]

Answer:

0.17 moles

Explanation:

In the elements of the periodic table, the atomic mass = molar mass. <u>Ex:</u> Atomic mass of Carbon is 12.01 amu which means molar mass of Carbon is also 12.01g/mol.

In order to find the # of moles in a 12 g sample of NiC-12, we will need to multiply the number of each atom by its molar mass and then add the masses of both Nickel and C-12 found in the periodic table:

  • Molar Mass of Ni (Nickel): 58.69 g/mol
  • Molar Mass of C (Carbon): 12.01 g/mol

Since there's just one atom of both Carbon and Nickel, we just add up the masses to find the molar mass of the whole compound of NiC-12.

  • 58.69 g/mol of Nickel + 12.01 g/mol of Carbon = 70.7 g/mol of NiC-12

There's 12g of NiC-12, which is less than the molar mass of NiC-12, so the number of moles should be less than 1. In order to find the # of moles in NiC-12, we need to do some dimensional analysis:

  • 12g NiC-12 (1 mol of NiC-12/70.7g NiC-12) = 0.17 mol of NiC-12
  • The grams cancel, leaving us with moles of NiC-12, so the answer is 0.17 moles of NiC-12 in a 12 g sample.

<em>P.S. C-12 or C12 just means that the Carbon atom has an atomic mass of 12amu and a molar mass of 12g/mol, or just regular carbon.</em>

5 0
3 years ago
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