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BabaBlast [244]
3 years ago
10

Can any one help on this one

Chemistry
1 answer:
enyata [817]3 years ago
8 0
0.004382166 Make sure to round to the right amount of Sig Figs
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Elements can be described by various properties, and identified by their boiling and melting points. For example, gold melts at
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Answer:

Explanation:

Melting and boiling point variations are not clear (do not have uniform pattern) in periodic table. But we can see, some elements have higher melting and boiling points and some have less. Here we study melting and boiling points of s, p, d blocks elements. IVAth group elements (C,Si) show high melting and boiling points because they have covalent gigantic lattice structures.

5 0
3 years ago
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What is the [h​3​o​+​] in a solution that consists of 1.0 m nh​3​ and 2.5 m nh​4​cl? [k​b​ (nh​3​) = 1.8 ×10​-5​]a) 1.1 × 10​-5​
Archy [21]

Answer:

[H₃O⁺] = 1.4 × 10⁻⁹ M.

Explanation:

NH₄Cl is a salt that dissolves well in water. The 2.5 M NH₄Cl will give an initial NH₄⁺ concentration of 2.5 M.

NH₃ is a weak base. It combines with water to produce NH₄⁺ and OH⁻. The opposite process can also take place. NH₄⁺ combines with OH⁻ to produce NH₃ and H₂O. The final H₃O⁺ concentration can be found from the OH⁻ concentration. What will be the final OH⁻ concentration?

Let the increase in OH⁻ concentration be x. The initial OH⁻ concentration at room temperature is 10⁻⁷ M.

Construct a RICE table for the equilibrium between NH₃ and NH₄⁺:

\begin{array}{c|ccccccc}\text{R}&\text{NH}_3 &+&\text{H}_2\text{O}&\rightleftharpoons &{\text{NH}_4}^{+}&+&\text{OH}^{-}\\\text{I}&1.0&&&&2.5&&1.0\times 10^{-7}\\\text{C}& -x &&&& +x &&+x\\\text{E} &1.0 - x &&&&2.5+x&&1.0\times 10^{-7}+x\end{array}.

The \text{K}_b value for ammonia is small. The value of x will be so small that at equilibrium, 1.0 - x \approx 1.0 and 2.5- x \approx 2.5.

\displaystyle \text{K}_b = \frac{[{\text{NH}_4}^{+}]\cdot [{\text{OH}}^{-}]}{[\text{NH}_3]} \approx \frac{2.5\;(x + 1.0\times 10^{-7})}{1.0}.

\displaystyle \frac{2.5\;(x + 1.0\times 10^{-7})}{1.0} = 1.8\times 10^{-5}.

\displaystyle [\text{OH}^{-}] = x+1.0\times 10^{-7} = 1.8\times 10^{-5} /\left(\frac{2.5}{1.0}\right) = 7.2\times 10^{-6}\;\text{mol}\cdot\text{L}^{-}.

Again, \text{K}_w = 1.0\times 10^{-14} at room temperature.

\displaystyle [\text{H}_3\text{O}^{+}] = \frac{\text{K}_w}{[\text{OH}^{-}]}=\frac{1.0\times 10^{-14}}{7.2\times 10^{-6}} = 1.4\times 10^{-9} \;\text{mol}\cdot\text{L}^{-1}

7 0
3 years ago
K3PO4(aq)+MgCl2(aq)⟶
Wittaler [7]

Answer:

6KCl(aq) +  Mg3(PO4)2(s)  ...  balanced equation

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7 0
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Sodium carbonate (na2co3) is used to neutralize the sulfuric acid spill. how many kilograms of sodium carbonate must be added to
irga5000 [103]

Sodium carbonate is used to neutralised sulfuric acid, H₂SO₄. Sodium carbonate is the salt of stron base (NaOH) and weak acid (H₂CO₃). The balanced chemical reaction for neutralization is as follows:

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From balanced chemical equation, it is clear that one mole of Na₂CO₃ is required to neutralize one mole of  H₂SO₄. Molar mass of Na₂CO₃= 106 g/mol=0.106 kg/mol and molar mass of H₂SO₄= 98 g/mol=0.098 kg/mol.

To neutralize 0.098 kg of H₂SO₄ amount of Na₂CO₃ required is 0.106 kg, so, to neutralize 5.04×10³ kg of H₂SO₄, Na₂CO₃ required is=\frac{0.106 X 5.04 X 10^{3} }{0.098} kg= 5.451 X 10³ kg.

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2Li+2H2O—>2LiOH+H2
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