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dexar [7]
3 years ago
15

What is the molar mass of CaCl2?

Chemistry
1 answer:
kherson [118]3 years ago
3 0

Answer: The correct option is 1.11\times 10^{2} g/mol

Explanation:

Atomic mass of calcium ,Ca = 40.07 g/mol

Atomic mass of chlorine ,Cl = 35.5  g/mol

Molar mass of CaCl_2:

= (Atomic mass of Ca+2 × (Atomic mass of Cl))

= (40.07 g/mol + (2 × 35.5))=111.07 g/mol

111.07=1.1107\times 10^{2} g/mol\approx 1.11\times 10^{2} g/mol

Hence, the correct option is 1.11\times 10^{2} g/mol



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I remember learning this last y’all i jus don’t remember it might be A or C
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2 years ago
tate whether the following changes are physical or chemical for rancidipication fixation of water 2 tearing of paper 3 rusting o
damaskus [11]

Answer: Physical change : tearing of paper, fixing of wtaer

Chemical change:  rusting of iron ,  electrolysis of water​, Rancidification

Explanation:

Physical change is a change in which there is no rearrangement of atoms and thus no new substance is formed. There is only change in physical state of the substance.

Example:  tearing of paper, fixing of wtaer

Chemical change is a change in which there is rearrangement of atoms and thus new substance is formed. There may or may not be a change in physical state.

Example: rusting of iron ,  electrolysis of water​, Rancidification

3 0
2 years ago
A buffer solution contains 0.345 M acetic acid and 0.377 M sodium acetate . If 0.0613 moles of potassium hydroxide are added to
melamori03 [73]

Answer:

pH = 5.54

Explanation:

The pH of a buffer solution is given by the <em>Henderson-Hasselbach (H-H) equation</em>:

  • pH = pKa + log\frac{[CH_3COO^-]}{[CH_3COOH]}

For acetic acid, pKa = 4.75.

We <u>calculate the original number of moles for acetic acid and acetate</u>, using the <em>given concentrations and volume</em>:

  • CH₃COO⁻ ⇒ 0.377 M * 0.250 L = 0.0942 mol CH₃COO⁻
  • CH₃COOH ⇒ 0.345 M * 0.250 L = 0.0862 mol CH₃COOH

The number of CH₃COO⁻ moles will increase with the added moles of KOH while the number of CH₃COOH moles will decrease by the same amount.

Now we use the H-H equation to <u>calculate the new pH</u>, by using the <em>new concentrations</em>:

  • pH = 4.75 + log\frac{(0.0942+0.0613)mol/0.250L}{(0.0862-0.0613)mol/0.250L} = 5.54
6 0
3 years ago
Submit What is the solubility of Cd3(POA) 2 in water? (Ksp of Cd3(PO4)2 is 2.5 x 10-33) | 1 2 3 +/- . 0 x100
vesna_86 [32]

<u>Answer:</u> The solubility of Cd_3(PO_4)_2 in water is 1.18\times 10^{-7}mol/L

<u>Explanation:</u>

The balanced equilibrium reaction for the ionization of cadmium phosphate follows:

Cd_3(PO_4)_2\rightleftharpoons 3Cd^{2+}+2PO_4^{3-}

                      3s       2s

The expression for solubility constant for this reaction will be:

K_{sp}=[Cd^{2+}]^3[PO_4^{3-}]^2

We are given:

K_{sp}=2.5\times 10^{-33}

Putting values in above equation, we get:

2.5\times 10^{-33}=(3s)^3\times (2s)^2\\\\2.5\times 10^{-33}=108s^5\\\\s=1.18\times 10^{-7}mol/L

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6 0
3 years ago
I need help with number 8 please help ASAP.
Rzqust [24]

Answer:

33.33% = 33%

Explanation:

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1 mole of MCO3 will produce → 1 mole of CO2

We need to get the number of mole of CO2:

and when we have 0.22 g of CO2, so number of mole = mass / molar mass

Moles = 0.22 g / 44 g/mol = 0.005 mole

Moles of Mg = moles of CO2 = 0.005 mole

Mass of Mg = moles * molar mass

= 0.005 * 84 /mol = 0.42 g

Percent of MgCO3 by mass of Mg = 0.42 g / 1.26 * 100

=33.33 %

8 0
3 years ago
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