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3 years ago

A compound has 15.39 g of gold for every 2.77 g of chlorine.

Chemistry

1 answer:

Ugo [173]3 years ago

3 0

Answer:

There is 5.56 g of gold for every 1 g of chlorine

Explanation:

The question is to determine the simplest mass ratio of gold to Chlorine in the compound.

Since the mass of gold in the compound compared to chlorine is 15.39 g for every 2.77 g, the mass of gold per gram of chlorine is given as:

15.39 / 2.77 = 5.56 g of gold to two decimal places

<em>Therefore, for every 5.56 g of gold, there is 1 g of chlorine.</em>

<em>Note : The ratio in which different elements combine by mass to form a compound is given by the law of constant composition which states that, "all pure samples of a particular chemical compound contain the same elements combined in the same proportion by mass.</em>

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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!

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3 years ago

The reaction nacl(s) → nacl(aq) is performed in a coffee cup calorimeter, using 100 ml of h2o(l) and 5.00g of nacl. if the tempe

MrRissso [65]

Since the density of water is 1 g /mL, hence there is 100 g of H2O. So total mass is:

m = 100 g + 5 g = 105 g

=> The heat of reaction can be calculated using the formula:

δhrxn = m C ΔT

where m is mass, C is heap capacity and ΔT is change in temperature = negative since there is a decrease

δhrxn = 105 g * 4.18 J/g°C * (-2.30°C)

δhrxn = -1,009.47 J

=> However this is still in units of J, so calculate the number of moles of NaCl.

moles NaCl = 5 g / (58.44 g / mol)

moles NaCl = 0.0856 mol

=> So the heat of reaction per mole is:

δhrxn = -1,009.47 J / 0.0856 mol

δhrxn = -11,798.69 J/mol = -11.8 kJ/mol

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