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kirza4 [7]
3 years ago
8

In an experiment Teresa's measures 15.5 mL of water she must have used a

Chemistry
1 answer:
kotykmax [81]3 years ago
3 0

Answer:

transfer pipet that had markings every 0.1 mL.

Explanation:

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How are elements grouped according to the number of valence electrons in their outermost levels?
melisa1 [442]

Answer:

I think so this is the answer

8 0
3 years ago
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
Plsz do solv this question hpl​
emmasim [6.3K]
We are given the resistance and voltage of this lamp and we are asked to find the current; the equation that relates these together is Ohm’s Law, V = IR. Simply plug and solve:

V = IR
(220 V) = I(484 Ohms)
I = 0.4545 Amps

The lamp has a current of 0.4545 Amps passing through it under these conditions.

Hope this helps!
7 0
2 years ago
1. List the seasons from longest to shortest, including the number of days in each season.
GenaCL600 [577]
Search it up bro it’s on the internet lol
4 0
3 years ago
What is the solubility in moles/liter for silver chloride at 25 oC given a Ksp value of 1.6 x 10-10. Write using scientific nota
trapecia [35]

Answer:

Explanation:

AgCl      ⇄       Ag⁺       +       Cl⁻

m                       m                   m

If x mole of AgCl be dissolved in one litre .

[ Ag⁺ ] [ Cl⁻ ] = 1.6 x 10⁻¹⁰

m² =  1.6 x 10⁻¹⁰

m = 1.26 x 10⁻⁵ moles

So solubility of AgCl is 1.26 x 10⁻⁵ moles / L

5 0
3 years ago
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