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2 years ago

What is the correct term for water Atom element product reactant

Chemistry

2 answers:

alexgriva [62]2 years ago

7 0

Its either Atom or Product....

Water is a compound so idk the real answer there

LUCKY_DIMON [66]2 years ago

7 0

atom is most likely ...

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⦁ Find the concentration of H+, OH-, PH and POH of 0.03 M of magnesium hydroxide which ionizes to the extent of only 1 /3 in aqu

lions [1.4K]

Answer:

$pH=12.3\\\\pOH=1.7\\$

$[H^+]=5x10^{-13}M$

$[OH^-]=0.02M$

Explanation:

Hello there!

In this case, according to the given ionization of magnesium hydroxide, it is possible for us to set up the following reaction:

$Mg(OH)_2(s)\rightleftharpoons Mg^{2+}(aq)+2OH^-(aq)$

Thus, since the ionization occurs at an extent of 1/3, we can set up the following relationship:

$\frac{1}{3} =\frac{x}{[Mg(OH)_2]}$

Thus, x for this problem is:

$x=\frac{[Mg(OH)_2]}{3}=\frac{0.03M}{3}\\\\x= 0.01M$

Now, according to an ICE table, we have that:

$[OH^-]=2x=2*0.01M=0.02M$

Therefore, we can calculate the H^+, pH and pOH now:

$[H^+]=\frac{1x10^{-14}}{0.02}=5x10^{-13}M$

$pH=-log(5x10^{-13})=12.3\\\\pOH=14-pH=14-12.3=1.7$

Best regards!

4 0

2 years ago

What mass of NaC6H5COO should be added to 1.5 L of 0.40 M C6H5COOH solution at 25 °C to produce a solution with a pH of 3.87 giv

statuscvo [17]

Answer:

41 g

Explanation:

We have a buffer formed by a weak acid (C₆H₅COOH) and its conjugate base (C₆H₅COO⁻ coming from NaC₆H₅COO). We can find the concentration of C₆H₅COO⁻ (and therefore of NaC₆H₅COO) using the Henderson-Hasselbach equation.

pH = pKa + log [C₆H₅COO⁻]/[C₆H₅COOH]

pH - pKa = log [C₆H₅COO⁻] - log [C₆H₅COOH]

log [C₆H₅COO⁻] = pH - pKa + log [C₆H₅COOH]

log [C₆H₅COO⁻] = 3.87 - (-log 6.5 × 10⁻⁵) + log 0.40

[C₆H₅COO⁻] = [NaC₆H₅COO] = 0.19 M

We can find the mass of NaC₆H₅COO using the following expression.

M = mass NaC₆H₅COO / molar mass NaC₆H₅COO × liters of solution

mass NaC₆H₅COO = M × molar mass NaC₆H₅COO × liters of solution

mass NaC₆H₅COO = 0.19 mol/L × 144.1032 g/mol × 1.5 L

mass NaC₆H₅COO = 41 g

7 0

2 years ago

You add 100.0 g of water at 52.0 °C to 100.0 g of ice at 0.00 °C. Some of the ice melts and cools the water to 0.00 °C. When the

lara31 [8.8K]

Answer:

$m_{ice} = 65.336\,g$

Explanation:

Accoding to the First Law of Thermodynamics, the heat released by the water melts a portion of ice. That is to say:

$Q_{water} = Q_{ice}$

$(100\,g)\cdot \left(4.184\,\frac{J}{kg\cdot ^{\textdegree}C}\right)\cdot (52\,^{\textdegree}C - 0\,^{\textdegree}C) = m_{ice}\cdot \left(333\,\frac{J}{g} \right)$

The amount of ice that is melt is:

$m_{ice} = 65.336\,g$

5 0

3 years ago

Calculate the frequency of electromagnetic wave with the wave length 344nm ( hint: use dimensional analysis)

Svetlanka [38]

Frequency = speed of light
---------------------------
wavelength

= 3 x 10^8
------------------------
344 x 10^-9

= 8.72 x 10^14 Hz.

Hope this helps!

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2 years ago

the elements helium (he), neon (ne), and xenon (xe) are all part of the same on the periodic table plz answer correctly gets a b

bagirrra123 [75]

Noble gas. Noble gas, any of the seven chemical elements that make up Group 18 (VIIIa) of the periodic table. The elements are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), radon (Rn), and oganesson (Og)

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2 years ago

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