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eimsori [14]
3 years ago
6

Find the volume of 0.100M hydrochloric acid necessary to react completely with 1.51g Al(OH)3.

Chemistry
1 answer:
shtirl [24]3 years ago
3 0
Reaction equation:
Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O
Moles of Al(OH)₃:
moles = mass/Mr
= 1.51 / (27 + 17 x 3)
= 0.019
Molar ratio Al(OH)₃ : HCl = 1 : 3
Moles of HCl required = 0.019 x 3
=0.057
concentration = moles/volume
volume = 0.057 / 0.1
= 0.57 dm³
= 570 ml
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The enzyme, carbonic anhydrase, is a large zinc-containing protein with a molar mass of 3.00 x10^4 g/mol. Zn is 0.218% by mass o
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The mathematical expression for mass percent is given by:

Mass percent = \frac{mass of the compound}{molar mass of the compound}\times 100

Put the values,

0.218 percent of zinc = \frac{mass of zinc}{3.00 \times 10^{4} g/mol}\times 100

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Now, number of moles of zinc  =\frac{given mass in g}{molar mass of zinc}

= \frac{65.4 g}{65.38 g/mol}

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Now, according to mole concept

6.022\times 10^{23} molecules of enzyme consists of 6.022\times 10^{23} atoms of zinc

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5 0
3 years ago
If a sample of neon gas at 25.00∘C and 0.500 atm is heated at a constant volume until the pressure is 0.850 atm, what must be th
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Answer:

T₂ = 506.6 K

Explanation:

Given data:

Initial pressure of gas = 25°C (25+273 =298 K)

Initial temperature = 0.500 atm

Final pressure = 0.850 atm

Final temperature = ?

Solution:

According to Gay-Lussac Law,

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

Now we will put the values in formula:

0.500 atm / 298 K = 0.850 atm /T₂

T₂ = 0.850 atm × 298 K / 0.500 atm

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T₂ = 506.6 K

8 0
3 years ago
You are given 10ml (M) 20 Naoh solution in a conical flask and asked to titrate with (M) 20 Hcl and (M) 20 H2so4 separately. cal
Solnce55 [7]

Answer:

n_{HCl}=0.2molHCl\\n_{H_2SO_4}=0.1molH_2SO_4

Explanation:

Hello!

In this case, since the reactions between NaOH and the acids are:

NaOH+HCl\rightarrow NaCl+H_2O\\\\2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O

Whereas we can see the 1:1 and 2:1 mole ratios between NaOH and HCl and H2SO4 respectively. In such a way, at the equivalence point we realize that:

n_{HCl}=n_{NaOH}=V_{NaOH}M_{NaOH}=0.01L*20mol/L=0.2molHCl\\\\2n_{H_2SO_4}=n_{NaOH}\\\\n_{H_2SO_4}=\frac{1}{2} V_{NaOH}M_{NaOH}=\frac{0.01L*20mol/L}{2} =0.1molH_2SO_4

Best regards!

8 0
3 years ago
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