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3 years ago

How to do balance chemical equiations

Chemistry

1 answer:

alex41 [277]3 years ago

4 0

Hey, lovely! It's a pretty lengthy process but here is a pretty clear video on how to do it. Hope this helps ya!

https://www.khanacademy.org/science/chemistry/chemical-reactions-stoichiome/balancing-chemical-equat...

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Metallic copper is formed when aluminum reacts with copper(ii) sulfate. how many grams of metallic copper can be obtained when 5

exis [7]

Answer is: 127 grams <span>rams of metallic copper can be obtained.
</span>Balanced chemical reaction: 2Al + 3CuSO₄ → Al₂(SO₄)₃ + 3Cu.
m(Al) = 54.0 g.
n(Al) = m(Al) ÷ M(Al).
n(Al) = 54 g ÷ 27 g/mol.
n(Al) = 2 mol.
m(CuSO₄) = 319 g.
n(CuSO₄) = 319 g ÷ 159.6 g/mol.
n(CuSO₄) = 2 mol; limiting reactant.
From chemical reaction: n(CuSO₄) : n(Cu) = 3 : 3 (1 : 1).
n(Cu) = 2 mol.
n(Cu) = 2 mol · 63.55 g/mol.
n(Cu) = 127.1 g.

8 0

3 years ago

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Varvara68 [4.7K]

Answer:

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6 0

2 years ago

What is the molarity of a solution that contains 20 grams of CaBr2<br> in 0.50 liter of solution?

ELEN [110]

Answer:

2 M

Explanation:

mole weight of CaBr2 = 40 + 2 * 79.9 = 199.8 gm

20 gm is then 20/199.8 =.1 mole

.1 mole / .50 liter = 2 M

8 0

2 years ago

A solution of permanganate is standardized by titration with oxalic acid, . To react completely with mol of oxalic acid required

Zielflug [23.3K]

Answer:

$M=0.120M$

Explanation:

Hello,

In this case, the undergone chemical reaction is:

$MnO_4^-(aq)+H_2C_2O_4(aq)\rightarrow Mn^{+2}+CO_2$

In such a way, the acidic redox balance turns out:

$(Mn^{+7}O_4)^-+5e^-+8H^+\rightarrow Mn^{+2}+4H_2O\\H_2C_2O_4\rightarrow2CO_2+2H^++2e^-$

Which leads to the total balanced equation as follows:

$2(MnO_4)^-(aq)+6H^+(aq)+5H_2C_2O_4(aq)\rightarrow2Mn^{+2}(aq)+8H_2O(l)+10CO_2(g)$

Thus, as the mass of oxalic acid is not given, one could suppose a value of 1 g (which you can modify based on the actual statement) in order to compute the oxalic acid moles as shwon below:

$1gH_2C_2O_4*\frac{1molH_2C_2O_4}{90.04gH_2C_2O_4} *\frac{2mol(MnO_4)^-}{5molH_2C_2O_4} =0.00444mol(MnO_4)^-$

Whereby the molality results:

$M=\frac{0.00444mol(MnO_4)^-}{0.03702L} =0.120M$

Remember you can modify the oxalic acid mass as you desire.

Best regards.

5 0

3 years ago

In protein synthesis, how many nitrogenous bases code for a single amino acid?

steposvetlana [31]

3 nitrogenous bases code a single amino acid , and it is called codon.

8 0

3 years ago