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garik1379 [7]
3 years ago
12

What is the type 1 binary compound for MgCl2

Chemistry
1 answer:
Yuri [45]3 years ago
8 0

Answer:

Explanation:

magnesium chloride

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Does a reaction occur when aqueous solutions of chromium(III) sulfate and calcium nitrate are combined
kotykmax [81]

Answer: Cr_2(SO_4)_3(aq)+3Ca(NO_3)_2(aq)\rightarrow 2Cr(NO_3)_3(aq)+3CaSO_4(aq)

Explanation:

A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.  

The balanced chemical equation when when aqueous solutions of chromium(III) sulfate and calcium nitrate are combined is:

Cr_2(SO_4)_3(aq)+3Ca(NO_3)_2(aq)\rightarrow 2Cr(NO_3)_3(aq)+3CaSO_4(aq)

8 0
3 years ago
Consider a mixture of NaCl and NaNO3 that is 31.8% Na by mass. Calculate the percentage by mass of NaCl in the mixture.
lutik1710 [3]

<u>Answer:</u> The percentage by mass of NaCl in the mixture is 38.5 %

<u>Explanation:</u>

To calculate the mass percentage of element in compound, we use the equation:

\text{Mass percent of element}=\frac{\text{Mass of element}}{\text{Mass of compound}}\times 100      .......(1)

  • <u>For NaCl:</u>

Mass of sodium element = 23 g

Mass of NaCl = 58.5 g

Putting values in equation 1, we get:

\text{Mass percent of sodium element}=\frac{23g}{58.5g}\times 100=39.32\%

Mass fraction of sodium metal in NaCl = 0.3932

  • <u>For NaNO_3 :</u>

Mass of sodium element = 23 g

Mass of NaNO_3 = 85 g

Putting values in equation 1, we get:

\text{Mass percent of sodium element}=\frac{23g}{85g}\times 100=27.1\%

Mass fraction of sodium metal in sodium nitrate = 0.271

Let us assume the mass fraction of NaCl in the mixture is 'x'

So, the mass fraction of NaNO_3 in the mixture will be '(1-x)'

We are given:

Percent by mass of Na in the mixture = 31.8 %

Mass fraction of Na in the mixture = 0.318

Evaluating the mass fraction of NaCl in the mixture:

[(x\times 0.3932)+((1-x)\times 0.271)]=0.318

x = 0.385

Percent by mass of NaCl in the mixture will be = (0.385\times 100)=38.5\%

Hence, the percentage by mass of NaCl in the mixture is 38.5 %

4 0
3 years ago
List the ideas of dalton
Romashka-Z-Leto [24]
Atoms has never been seen,nor experimentally detected. Their existence remained hypothetical.
5 0
3 years ago
Please Hurry
Jet001 [13]

Answer:

B,C,F

Explanation:

6 0
3 years ago
Determine the ph of a 0. 35 m aqueous solution of CH3NH2 (methylamine). The kb of methylamine is?
guajiro [1.7K]

The pH of 0.35 M of CH₃NH₂ is 12.09 and the k_b for methylamine is 4.4 x 10⁻⁴.

<h3>What is pH?</h3>

pHis the quantitative measure of acidity and basicity of an aqueous or other liquid solution. The scale range goes from 0 to 14. Water has a pH of 7 and is neutral in nature.

<h3>What is dissociation constant?</h3>

The dissociation constant is an equilibrium constant that describes the dissociation or ionization of a base or an acid

k_a describes the dissociation of an acid

k_b describes the dissociation of a base

For methylamine,

CH_3NH_2 + H_2O\Leftrightarrow CH_3NH_3^+ + OH^-

Initial concentration of methylamine = 0.35 M

Initial concentration of products = 0

Let, at equilibrium concentration of CH₃NH₂ = 0.35 - x

Then, concentration of CH₃NH₃⁺ and OH⁻ is x and x respectively

k_b = \frac{[CH_3NH_3^+] [OH^-]}{[CH_3NH_2]}

The dissociation constant for methylamine, k_b = 4.4 x 10⁻⁴

4.4 \times 10^-^4 = \frac{x.x}{0.35} \\x^{2} = 4.4 \times 10^-^4 \times 0.35\\x^{2}  = 1.54  \times 10^-^4 \\

x =0.0124

pOH = -log[OH] = -log(0.0124) = 1.91

pH + pOH = 14

pH = 14 - 1.91 = 12.09

Thus, the pH of methylamine is 12.09 and k_b is 4.4 x 10⁻⁴

Learn more about pH:

brainly.com/question/172153

#SPJ4

7 0
2 years ago
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