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umka21 [38]
2 years ago
13

What is the pH of a 0.00001 M solution of HCI?

Chemistry
1 answer:
ale4655 [162]2 years ago
5 0

Answer:

0.00001= 1 x 10^-5. Since HCl is an acid, 1 x10^-5 is the H+ concentration. Write only the number of the exponent. Therefore, pH = 5.

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Which of these substances has the highest pOH? 0.10 M HCl, pH = 1 0.001 M HNO3, pH = 3 0.01 M NaOH, pH = 12 The answer is 0.10 M
Naya [18.7K]

Answer:On these combined scales of pH and pH it can be shown that because for water when pH = pH = 7 that pH + pH = 14. This relationship is useful in the inter conversion of values. For example, the pH at a 0.01 M solution of sodium hydroxide is 2, the pH of the same solution must be 14-2 = 12.

Explanation:

4 0
3 years ago
Write the rates for the following reactions in terms of the disappearance of reactants and appearance of products: (a) 302 .....
Oduvanchick [21]

Answer:

Explanation:

\mathbf{From \  the  \ information \  given:} \\ \\ \mathbf{The \  rates \  of  \ the \ f ollowing \  reactions \  can \  be \  expressed  \ as \  follows:}

(a)

\mathbf{3O_2 \to 2O_3} \\ \\ \\ \mathbf{-\dfrac{1}{3}\dfrac{d[O_2]}{dt}=\dfrac{2}{3} \dfrac{d[O_3]}{dt}}

(b)

\mathbf{C_2H_6 \to C_2H_4 + H_2}  \\ \\ \\ \mathbf{  -\dfrac{d[C_2H_6]}{dt}= \dfrac{d[C_2H_4]}{dt}=\dfrac{d[H_2]}{dt}}

(c)

\mathbf{ClO^-+Br^- \to BrO^-+Cl^-} \\ \\ \\ \mathbf{ -\dfrac{d[ClO^-]}{dt}= -\dfrac{d[Br^-]}{dt} =  \dfrac{d[BrO^-]}{dt} = \dfrac{d[Cl^-]}{dt}   }

(d)

\mathbf{(CH_3)_3 CCl+H_2O \to (CH_3)_3COH + H^+ + Cl^-} \\ \\ \\  \mathbf{- \dfrac{d[(CH_3)_3CCl}{dt}= - \dfrac{d[H_2O]}{dt}= \dfrac{d[CH_3)_3COH}{dt}= \dfrac{d[H^+]}{dt}= \dfrac{d[Cl^-]}{dt}}

(e)

\mathbf{2AsH_3 \to 2As + 3H_2} \\ \\ \\  \mathbf{-\dfrac{1}{2}\dfrac{d[AsH_3]}{dt}=\dfrac{1}{2}\dfrac{d[As]}{dt}=\dfrac{1}{3}\dfrac{d[H_2]}{dt}}

5 0
3 years ago
the pressure and temperature of 10 liters of gas are doubled. if the original condition are 2 atmosphere of pressure and 400k wh
Harman [31]

Answer:

\boxed{\text{10 L}}

Explanation:

We have two pressures, two temperatures, and one volume.

This looks like a question in which we can use the Combined Gas Law to calculate the volume.

\dfrac{p_{1}V_{1}}{T_{1}} = \dfrac{p_{2}V_{2} }{T_{2}}

Data:

\begin{array}{rclrclrcl}p_{1}& =& \text{2 atm}\qquad & V_{1} &= & \text{10 L}\qquad & T_{2}& =& \text{400 K}\\p_{2}& =& \text{4 atm}\qquad & V_{2} &= & \text{?}\qquad & T_{2}& =& \text{800 K}\\\end{array}

Calculation:

\begin{array}{rcl}\dfrac{2 \times 10}{400}& =& \dfrac{4V_{2} }{800}\\\\0.050& = &0.0050V_{2}\\V_{2}& = &\mathbf{10 L}\end{array}\\\text{The final volume is }\boxed{\textbf{10 L}}

4 0
3 years ago
naturally occurring bromine molecules, br2 have masses of 158, 160, and 162. they occur in the relative abundances 25.69%, 49.99
4vir4ik [10]
The average atomic mass of an element can be determined by multiplying the individual masses of the isotopes with their respective relative abundances, and adding them. 

Average atomic mass of Br = 158 amu(0.2569) + 160 amu(0.4999) + 162 amu(0.2431)
Average atomic mass = 159.96 amu

As described in the problem, the relative abundance for Br-79 is 25.69%. This is because 2 atoms of Br is equal to 79*2 = 158 amu. Similarly, the relative abundance of Br-81 is 81*2 = 162, which is 24.31%.
4 0
3 years ago
Some people must eat a low-sodium diet with no more than 2,000 mg of sodium per day. By eating 1 cracker, 1 pretzel, and 1 cooki
Leni [432]

Answer:

The amount of sodium is 32 mg per cracker, 49 mg per pretzel and 68 mg per cookie.

Explanation:

Let's assume amount of sodium is x mg per cracker, y mg per pretzel and z mg per cookie.

So, the following three equations can be written as per given information:

x+y+z = 149 ........(1)

8y+8z = 936 ........(2)

6x+7y = 535 .........(3)

From equation- (2), y+z = \frac{936}{8} = 117

By substituting the value of (y+z) in equation- (1) we get,

                          x = 149-(y+z) = 149-117 = 32

By substituting the value of x into equation- (3) we get,

                           y = \frac{535-(6\times 32)}{7} = 49

By substituting the value of y  into equation- (2) we get,

                           z = (117-49) = 68

So, the amount of sodium is 32 mg per cracker, 49 mg per pretzel and 68 mg per cookie.

6 0
3 years ago
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