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2 years ago

What is Alkali metals

2 answers:

6 0

Any of the elements lithium, sodium, potassium, rubidium , cesium, and francium, occupying group 1A of the periodic table.

seropon [69]2 years ago

3 0

Answer:

The alkali metals consist of the chemical elements lithium, sodium, potassium, rubidium, caesium, and francium. Together with hydrogen, they constitute group 1, which lies in the s-block of the periodic table

The alkali metals are all shiny, soft, highly reactive metals at standard temperature and pressure and readily lose their outermost electron to form cations with charge +1.

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How many moles of mgs2o3 are in 193 g of the compound?

Mashutka [201]

Answer:

1.414 Moles

Solution:

Data Given:

Mass of MgS₂O₃ = 193 g

M.Mass of MgS₂O₃ = 136.43 g.mol⁻¹

Moles = ?

Formula Used:

Moles = Mass ÷ M.Mass

Putting values,

Moles = 193 g ÷ 136.43 g.mol⁻¹

Moles = 1.414 mol

8 0

3 years ago

How do i solve it and what is the answer?

Masja [62]

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6 0

2 years ago

How many grams of silver nitrate are needed to react with 156.2g of sodium sulfide to produce 595.8g of silver sulfide and 340.0

qaws [65]

Before proceeding, we should write the reaction equation to better understand what is happening:
2AgNO₃ + Na₂S → Ag₂S + 2NaNO₃

Now, we may apply the law of conservation of mass, due to which the total mass before a chemical reaction is equivalent to the total mass after a chemical reaction. Therefore:
Mass of silver nitrate + mass of sodium sulfide = mass of silver sulfide + mass of sodium nitrate

Mass of silver nitrate + 156.2 = 595.8 + 340
Mass of silver nitrate = 779.6 grams

3 0

3 years ago

A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.

irga5000 [103]

Answer:

1. pH = 1.23.

2. $H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

$H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+$

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

$pH=pKa+log(\frac{[base]}{[acid]} )$

Whereas the pKa is:

$pKa=-log(Ka)=-log(5.90x10^{-2})=1.23$

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

$pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23$

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

$n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol$

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

$H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Which is also shown in net ionic notation.

Best regards!

4 0

3 years ago

What is the formula of ideal gas law?

Debora [2.8K]

PV = nRT. Where P = pressure, V = volume, n = number of moles, R = universal gas constant and T = temperature. Hope this helps!

3 0

3 years ago

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