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lana [24]
2 years ago
6

Trimix is a general name for a type of gas blend used by technical divers and contains nitrogen, oxygen and helium. In one Trimi

x blend, the partial pressures of each gas are 55.0 atm oxygen, 90.0 atm nitrogen, and 50.0 atm helium. What is the percent oxygen (by volume) in this trimex blend
Chemistry
1 answer:
gladu [14]2 years ago
8 0

Answer:

The correct answer is 28.2 %.

Explanation:

Based on the given question, the partial pressures of the gases present in the trimix blend is 55 atm oxygen, 50 atm helium, and 90 atm nitrogen. Therefore, the sum of the partial pressure of gases present in the blend is,  

Ptotal = PO2 + PN2 + PHe

= 55 + 90 + 50

= 195 atm

The percent volume of each gas in the trimix blend can be determined by using the Amagat's law of additive volume, that is, %Vx = (Px/Ptot) * 100

Here Px is the partial pressure of the gas, Ptot is the total pressure and % is the volume of the gas. Now,  

%VO2 = (55/195) * 100 = 28.2%

%VN2 = (90/195) * 100 = 46%

%VHe = (50/195) * 100 = 25.64%

Hence, the percent oxygen by volume present in the blend is 28.2 %.  

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Write an equation to show how HC2O4− can act as a base with HS− acting as an acid.
Artyom0805 [142]

An acid donates H^{+} ion in aqueous solution. A base accepts H^{+} ion in aqueous solution.

The equation representing the acid base reaction of HC_{2}O_{4}^{-} and HS^{-}:

HC_{2}O_{4}^{-}(aq) + HS^{-}(aq) ----> H_{2}C_{2}O_{4}(aq)+S^{2-}(aq)

In the above reaction, as HC_{2}O_{4}^{-} acts as a base it is accepting the hydrogen ion from HS^{-}. Similarly, HS^{-} donates its hydrogen ion to HC_{2}O_{4}^{-} acting as an acid.

7 0
2 years ago
In a chemistry lab, you have two vinegars. one is 5% acetic acid, and one is 6.5% acetic acid. you want to make 200 ml of a vine
nexus9112 [7]
Assume that the amount needed from the 5% acid is x and that the amount needed from the 6.5% acid is y.

We are given that:
The volume of the final solution is 200 ml
This means that:
x + y = 200
This can be rewritten as:
x = 200 - y .......> equation I

We are also given that:
The concentration of the final solution is 6%
This means that:
5%x + 6.5%y = 6% (x+y)
This can be rewritten as:
0.05 x + 0.065 y = 0.06 (x+y) ............> equation II

Substitute with equation I in equation II and solve for y as follows:
0.05 x + 0.065 y = 0.06 (x+y)
0.05 (200-y) + 0.065 y = 0.06 (200-y+y)
10 - 0.05 y + 0.065 y = 12
0.015y = 12-10 = 2
y = 2/0.015
y = 133.3334 ml

Substitute with the y in equation I to get the x as follows:
x = 200 - y
x = 200 - 133.3334
x = 66.6667 ml

Based on the above calculations:
The amount required from the 5% acid = x = 66.6667 ml
The amount required from the 6.5% acid = y = 133.3334 ml

Hope this helps :)

6 0
2 years ago
Read 2 more answers
BH+ClO4- is a salt formed from the base B (Kb = 1.00e-4) and perchloric acid. It dissociates into BH+, a weak acid, and ClO4-, w
Len [333]

Answer:

The pH of 0.1 M BH⁺ClO₄⁻ solution is <u>5.44</u>

Explanation:

Given: The base dissociation constant: K_{b} = 1 × 10⁻⁴, Concentration of salt: BH⁺ClO₄⁻ = 0.1 M

Also, water dissociation constant: K_{w} = 1 × 10⁻¹⁴

<em><u>The acid dissociation constant </u></em>(K_{a})<em><u> for the weak acid (BH⁺) can be calculated by the equation:</u></em>

K_{a}. K_{b} = K_{w}    

\Rightarrow K_{a} = \frac{K_{w}}{K_{b}}

\Rightarrow K_{a} = \frac{1\times 10^{-14}}{1\times 10^{-4}} = 1\times 10^{-10}

<em><u>Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:</u></em>

Reaction involved: BH⁺  +  H₂O  ⇌  B  +  H₃O+

Initial:                     0.1 M                    x         x            

Change:                   -x                      +x       +x

Equilibrium:           0.1 - x                    x         x

<u>The acid dissociation constant: </u>K_{a} = \frac{\left [B \right ] \left [H_{3}O^{+}\right ]}{\left [BH^{+} \right ]} = \frac{(x)(x)}{(0.1 - x)} = \frac{x^{2}}{0.1 - x}

\Rightarrow K_{a} = \frac{x^{2}}{0.1 - x}

\Rightarrow 1\times 10^{-10} = \frac{x^{2}}{0.1 - x}

As, x

\Rightarrow 0.1 - x = 0.1

\therefore 1\times 10^{-10} = \frac{x^{2}}{0.1 }

\Rightarrow x^{2} = (1\times 10^{-10})\times 0.1 = 1\times 10^{-11}

\Rightarrow x = \sqrt{1\times 10^{-11}} = 3.16 \times 10^{-6}

<u>Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M</u>

Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44

<u>Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44</u>

5 0
1 year ago
A well-insulated, closed device claims to be able to compress 100 mol of propylene, acting as a SoaveRedlich-Kwong gas and with
Setler79 [48]

Explanation:

The given data is as follows.

    Moles of propylene = 100 moles,    T_{i} = 300 K

    T_{f} = 800 K,    V_{i} = 2 m^{3}

    V_{f} = 0.02 m^{3},   C_{p} of propylene = 100 J/mol

Now, we assume the following assumptions:

Since, it is a compression process therefore, work will be done on the system. And, work done will be equal to the heat energy liberating without any friction.

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                 = 5 \times 10^{6} J

                 = 5 MJ

Thus, we can conclude that a minimum of 5 MJ work is required without any friction.

3 0
2 years ago
What is the oxidation number for iodine in Mg(IO3)2 ?
navik [9.2K]
The oxidation number of iodine is 5 in Mg(IO3)2 which can be calculated as 
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As we know that
Mg has +2
O has -2
So,
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    10 = 2I
     I =5

7 0
1 year ago
Read 2 more answers
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