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mart [117]
1 year ago
11

Mercury-197 has a half-life of 3 days. Starting

Chemistry
1 answer:
Ahat [919]1 year ago
4 0

Answer:

2.34 gms left

Explanation:

3 weeks = 21 days   =    7 half lives

300 * (1/2)^7 = 2.34 gms

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How many moles of H2SO4 are needed to completely react with 15.0 mole of Al?
mezya [45]

Answer:

The answer to your question is 22.5 moles of H₂SO₄

Explanation:

Data

moles of H₂SO₄ = ?

moles of Al = 15

Balanced chemical reaction

                2Al + 3H₂SO₄  ⇒  Al₂(SO₄)₃  +  3H₂

To solve this problem use proportions and cross multiplication. Use the coefficients of the balanced chemical reaction.

                2 moles of Al ------------------- 3 moles of H₂SO₄

               15 moles of Al ------------------  x

                                    x = (15 x 3) / 2

                                    x = 45/2

                                   x = 22.5 moles of H₂SO₄

-Conclusion

22.5 moles of H₂SO₄ react completely with 15 moles of Al.

8 0
3 years ago
Why do the more complex sugar disaccharides store more energy than<br> monosaccharides?
sertanlavr [38]

Answer:

c) disaccharides have more chemical bonds

Explanation:

disaccharides are made from the condensation of two monosaccharides.

This means there are more bonds in disaccharides. More bonds mean greater energy and vice versa .

7 0
3 years ago
We mix 0.08 moles of chloroacetic acid (ClCH2COOH) and 0.04 moles of
Arte-miy333 [17]

Answer:

A. pH using molar concentrations = 2.56

B. pH using activities                      = 2.46

C. pH of mixture                              = 2.56

Explanation:

A. pH using molar concentrations

ClCH₂COOH + H₂O ⇌ ClCH₂COO⁻ + H₃O⁺

        HA        + H₂O ⇌          A⁻         + H₃O⁺

We have a solution of 0.08 mol HA and 0.04 mol A⁻

We can use the Henderson-Hasselbalch equation to calculate the pH.

\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\& = & 2.865 +\log \left(\dfrac{0.04}{0.08}\right )\\\\& = & 2.865 + \log0.50 \\& = &2.865 - 0.30 \\& = & \mathbf{2.56}\\\end{array}

B. pH using activities

(i) Calculate [H⁺]

pH = -log[H⁺]

\text{[H$^{+}$]}  = 10^{-\text{pH}} \text{ mol/L} = 10^{-2.56}\text{ mol/L} = 2.73  \times 10^{-3}\text{ mol/L}

(ii) Calculate the ionic strength of the solution

We have a solution of 0.08 mol·L⁻¹ HA, 0.04 mol·L⁻¹ Na⁺, 0.04 mol·L⁻¹ A⁻, and 0.00273 mol·L⁻¹ H⁺.

The formula for ionic strength is  

I = \dfrac{1}{2} \sum_{i} {c_{i}z_{i}^{2}}\\\\I = \dfrac{1}{2}\left [0.04\times (+1)^{2} + 0.04\times(-1)^{2} +  0.00273\times(+1)^{2}\right]\\\\=  \dfrac{1}{2} (0.04 + 0.04 + 0.00273) = \dfrac{1}{2} \times 0.08273 = 0.041

(iii) Calculate the activity coefficients

\ln \gamma = -0.510z^{2}\sqrt{I} = -0.510(-1)^{2}\sqrt{0.041} = -0.510\times 0.20 = -0.10\\\gamma = 10^{-0.10} = 0.79

(iv) Calculate the initial activity of A⁻

a = γc = 0.79 × 0.04= 0.032

(v) Calculate the pH  

\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{a_{\text{A}^{-}}}{a_{\text{[HA]}}}\right )\\\\& = & 2.865 +\log \left(\dfrac{0.032}{0.08}\right )\\\\& = & 2.865 + \log0.40 \\& = & 2.865 -0.40\\& = & \mathbf{2.46}\\\end{array}\\

C. Calculate the pH of the mixture

The mixture initially contains 0.08 mol HA, 0.04 mol Na⁺, 0.04 mol A⁻, 0.05 mol HNO₃, and 0.06 mol NaOH.

The HNO₃ will react with the NaOH to form 0.05 mol Na⁺ and 0.05 mol NO₃⁻.

The excess NaOH will react with 0.01 mol HA to form 0.01 mol Na⁺ and 0.01 mol A⁻.

The final solution will contain 0.07 mol HA, 0.10 mol Na⁺, 0.05 mol A⁻, and 0.05 mol NO₃⁻.

(i) Calculate the ionic strength

I = \dfrac{1}{2}\left [0.10\times (+1)^{2} + 0.05 \times(-1)^{2} +  0.05\times(-1)^{2}\right]\\\\=  \dfrac{1}{2} (0.10 + 0.05 + 0.05) = \dfrac{1}{2} \times 0.20 = 0.10

(ii) Calculate the activity coefficients

\ln \gamma = -0.510z^{2}\sqrt{I} = -0.510(-1)^{2}\sqrt{0.10} = -0.510\times 0.32 = -0.16\\\gamma = 10^{-0.16} = 0.69

(iii) Calculate the initial activity of A⁻:

a = γc = 0.69 × 0.05= 0.034

(iv) Calculate the pH

\text{pH} = 2.865 + \log \left(\dfrac{0.034}{0.07}\right ) = 2.865 + \log 0.49 = 2.865 - 0.31 = \mathbf{2.56}

3 0
2 years ago
What is the pH of a 0.025 M [OH] solution?
Debora [2.8K]
The answer is 12.4.I think its correct answer.

8 0
3 years ago
A voltaic cell is constructed using the following half-reactions: Ag+(aq) + e- ---&gt; Ag(s) EoAg+ = +0.80 V Cu2+(aq) + 2e- ---&
Anton [14]
- Standard reduction potential of Ag/Ag⁺ is 0.80 v and that of Cu⁺²(aq)/Cu⁰ is +0.34 V.
- The couple with a greater value of standard reduction potential will oxidize the reduced form of the other couple.
Ag⁺ will be reduced to Ag(s) and Cu⁰ will be oxidized to Cu²⁺
Anode reaction: Cu⁰(s) →  Cu²⁺ + 2 e⁻     E⁰ = +0.34 V
Cathode reaction: Ag⁺(aq) + e → Ag(s)     E⁰ = +0.80 V 
Cell reaction: Cu⁰(s) + 2 Ag⁺(aq) → Cu⁺²(aq) + 2 Ag⁰(s)
E⁰ cell = E⁰ cathode + E⁰ anode
            = 0.80 + (-0.34) = + 0.46 V
   
5 0
2 years ago
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