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zmey [24]
3 years ago
12

I need to know the weighted average please!

Chemistry
1 answer:
maria [59]3 years ago
8 0

Answer:

63.5456 amu

Explanation:

0.6917*62.9296+0.3083*64.9278= 63.5456\ amu

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ZnSO4 +<br> +<br> LINO, → ---- Zn(NO3)2<br> +<br> Li,SO4<br> Balanced equation
Tanzania [10]

Answer:

ZnSO4 + 2LiNO3 → Zn(NO3)2 + Li2SO4

Explanation:

There's many resources on web that can assist you with this concept:

https://en.intl.chemicalaid.com/tools/equationbalancer.php

https://www.webqc.org/balance.php

4 0
3 years ago
C12H22O11 + 11 H2SO4 12 C + 11 H2SO4 + 11 H2O
Vladimir [108]

Answer:

B: The sulfuric acid is not consumed or react with the reactant.

4 0
2 years ago
How many multiple bonds in ch4
chubhunter [2.5K]
It has 8 bonding electrons
7 0
3 years ago
If 15.6 grams of copper (ii) chloride react with 20.2 grams of sodium nitrate how many grams of sodium chloride can be formed? W
olasank [31]

Answer:

- 13.56 g of sodium chloride are theoretically yielded.

- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.

- 0.50 g of sodium nitrate remain when the reaction stops.

- 92.9 % is the percent yield.

Explanation:

Hello!

In this case, according to the question, it is possible to set up the following chemical reaction:

CuCl_2+2NaNO_3\rightarrow 2NaCl+Cu(NO_3)_2

Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:

m_{NaCl}^{by\ CuCl_2}=15.6gCuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2} *\frac{2molNaCl}{1molCuCl_2} *\frac{58.44gNaCl}{1molNaCl} =13.56gNaCl\\\\m_{NaCl}^{by\ NaNO_3}=20.2gNaNO_3*\frac{1molNaNO_3}{84.99gNaNO_3} *\frac{2molNaCl}{2molNaNO_3} *\frac{58.44gNaCl}{1molNaCl} =13.89gNaCl

Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:

m_{NaNO_3}^{by\ NaCl}=13.56gNaCl*\frac{1molNaCl}{58.44gNaCl}*\frac{2molNaNO_3}{2molNaCl} *\frac{84.99gNaNO_3}{1molNaNO_3}=19.72gNaNO_3

Therefore, the leftover of sodium nitrate is:

m_{NaNO_3}^{leftover}=20.2g-19.7g=0.5gNaNO_3

Finally, the percent yield is computed via:

Y=\frac{12.6g}{13.56g} *100\%\\\\Y=92.9\%

Best regards!

6 0
2 years ago
PLEASE HELP this is chemistry 12 points
Inessa05 [86]

Answer:

6.791

Explanation:

For proper significant figures with addition, you would use the significant figures of the number with lowest decimal place.  6.298 goes to the 10⁻³ place.  0.492712 goes to the 10⁻⁶ place.  You will go out to the 10⁻³ place.

6.298 + 0.492712 = 6.790712 ≈ 6.791

8 0
3 years ago
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