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Alinara [238K]
3 years ago
7

How many moles of mgci2 are there in 332 g of the compound

Chemistry
1 answer:
Agata [3.3K]3 years ago
3 0
Find molar mass of MgCl2 so M=<span>95.211 g/mol
N=m/M 
N=332/95.2
N=3.487
dont forget to round your sig figs </span>
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Identify the limiting reactant when 7.81 g of HCl reacts with 5.24 g of NaOH to produce NaCl and H2O
elena55 [62]

Answer:

NaOH is limiting

Explanation:

                                 Hcl   +    NaOH   ------------      Nacl   + H2O  

Given                     7.81g        5.24g

mol weights           1+35.5     23+16+1

                                 36.5            40

No of Moles        7.81/36.5       5.24/40  

                             0.2139              0.131

No of moles = weight/ molecular weight

No of moles of NaOH is  lower so it is limiting as according to equation one mole of Hcl requires mole of NaOH so equal number of moles are required. NaOH is 0.131 which is far lower than 0.2139

-------- represents arrow in forward direction in equation

7 0
3 years ago
Pls help and match them
slega [8]

Answer:

Simile: C

Personification: D

Symbol: A

Hyperbole: B

Metaphor: E

Explanation:

3 0
3 years ago
What can student do to increase the force of attraction of magnet on the iron(metal) block?
Fantom [35]
Rub the magnet on the iron an it will cause a stronger force of attraction. Hope this helped!
3 0
4 years ago
Read 2 more answers
The atomic number is
vovikov84 [41]
It’s D I’m pretty sure
6 0
3 years ago
When 0.040 mol of propionic acid, c2h5co2h, is dissolved in 750 ml of water, the equilibrium concentration of h3o+ ions is measu
torisob [31]

<span>Answer is: Ka for propinoic acid is 6,57·10</span>⁻⁵.<span>
Chemical reaction: C</span>₂H₅COOH(aq) + H₂O(l) ⇄ C₂H₅COO⁻(aq) + H₃O⁺(aq).<span>
n(C</span>₂H₅COOH) = 0,04 mol.<span>
V(C</span>₂H₅COOH) = 750 mL = 0,75 L.<span>
c(C</span>₂H₅COOH) = 0,04 mol ÷ 0,75 L.<span>
c(C</span>₂H₅COOH) = 0,053 mol/L = 0,053 M.<span>
[C</span>₂H₅COO⁻] = [H₃O⁺] = 1,84·10⁻³ M = 0,00184 M.<span>
[HCN] = 0,053 M - 0,00184 M = 0,0515 M.
Ka = [C</span>₂H₅COO⁻] · [H₃O⁺] / [C₂H₅COOH].<span>
Ka = (0,00184 M)² / 0,0515 M.
Ka = 6,57·10</span>⁻⁵.

3 0
3 years ago
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