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const2013 [10]
3 years ago
13

Calculate the concentration (in M) of hydroxide ions in a solution with a pOH of 8.873.

Chemistry
1 answer:
Pachacha [2.7K]3 years ago
8 0

Answer:

it should be

Explanation:

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3 0
3 years ago
Pls help 25 pts
mote1985 [20]

Answer:

In the quantum-mechanical model of an atom, electrons in the same atom that have the same principal quantum number (n) or principal energy level are said to occupy an electron shell of the atom. Orbitals define regions in space where you are likely to find electrons.

6 0
2 years ago
Solid magnesium has mass of 1300g and a volume of 743cm . what's the density of magnesium?
Lilit [14]
D = m / V

d = 1300 g / 743 cm³

d = 1.749 g/cm³
3 0
3 years ago
How many joules of heat are needed to raise the temperature of 47.5 g of aluminum from 21°C to 94°C , if the specific heat of al
Montano1993 [528]

Answer:

3120.75J

Explanation:

So, we have the formula q = mc\Delta t. For this example, q is the heat energy in Joules, m is the mass in grams, c is the specific heat capacity in \frac{J}{{g\°C}}, and \Deltat is the change in temperature. In this case, m = 47.5g, c = 0.9  \frac{J}{{g\°C}}, and \Deltat  = 94-21 = 73°C. Plugging in the values, we get the joules of heat required to raise 47.5g of Al from 21°C to 94°C which is stated above. You can double check my answer but that should be it. An important thing to be aware of are the units. Sometimes, the heat capacity may not be \frac{J}{{g\°C}}. I may be in Kelvin or something. Anyways, hope this helps.

5 0
4 years ago
A solution is prepared by mixing 0.10 of 0.12 M sodium chloride with 0.23 L of a 0.18 M magnesium chloride solution. What is the
Ad libitum [116K]

Answer:

pCl⁻ = 0.54

Explanation:

First we <u>calculate how many Cl⁻ moles are coming from each substance</u>, using the <em>given volumes and concentrations</em>:

  • 0.12 M NaCl * 0.10 L = 0.012 mol NaCl = 0.012 mol Cl⁻
  • 0.18 M MgCl₂ * 0.23 L = 0.0414 mol MgCl₂ = (0.0414 * 2) 0.0828 mol Cl⁻

The final volume of the mixture is = 0.10 L + 0.23 L = 0.33 L

Now we <u>calculate [Cl⁻]</u>, using the<em> total number of Cl⁻ moles and the final volume:</em>

  • [Cl⁻] = (0.012 mol + 0.0828 mol) / 0.33 L = 0.29 M

Finally we <u>calculate the pCl⁻ of the resulting solution</u>:

  • pCl⁻ = -log[Cl⁻]
  • pCl⁻ = 0.54
7 0
3 years ago
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