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Anastaziya [24]
3 years ago
8

What are the products of a reaction between ammonium iodide and magnesium sulfate?

Chemistry
1 answer:
tatyana61 [14]3 years ago
7 0
<span>2NH4I(aq) + MgSO4(aq) =(NH4)2SO4(aq) + MgI2(aq) just balance</span>
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Which of the following lists the correct symbol for the ion that tin forms when it loses 2 electrons, the correct classification
ser-zykov [4K]

Answer:

\huge\boxed{\sf Option \ A}

Explanation:

The symbol for TIN is Sn.

When Sn loses 2 electrons, it gets a double positive charge ( +2 ) and becomes \sf Sn ^{+2}.

It becomes a cation.

The name of Ion is Tin ( II ) Ion.

\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3>
6 0
3 years ago
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Who's your celebrity crush and why?
dezoksy [38]

Answer:

Ian Somerhalder because he's JUST SO fine

Explanation:

6 0
2 years ago
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The density (d) of a substance is an intensive property that is defined as the ratio of its mass (m) to its volume (v).
vovangra [49]

Answer:

Density is an intensive property

Explanation:

The intensive properties don´t depend on mass or size, if you have a liter of water, this litter have a density equal 1000 kg/m^3. If zo have 250 liters of water, the density is 1000 kg/m^3 too.

The density does not change with the addition or substraction of matter.

I hope I have been clear.

8 0
3 years ago
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Suppose a student started with 142.0 mg of trans-cinnamic acid, 412 mg of pyridinium tribromide, and 2.30 mL of glacial acetic a
nirvana33 [79]

Answer: Theoretical Yield = 0.2952 g

               Percentage Yield = 75.3%

Explanation:

Calculation of limiting reactant:

n-trans-cinnamic acid moles = (142mg/1000) / 148.16 = 9.584*10⁻⁴ mol

pyridium tribromide moles = (412mg/1000) / 319.82= 1.288*10⁻³ mol

  • n-trans-cinnamic acid is the limiting reactant

The molar ratio according to the equation mentioned is equals to 1:1

The brominated product moles is also = 9.584*10⁻⁴ mol

Theoretical yield = (9.584*10⁻⁴ mol) * (Mr of brominated product)

                             =  (9.584*10⁻⁴ mol) * (307.97) = 0.2952 g

Percentage Yield is : Actual Yield / Theoretical Yield = 0.2223/0.2952

                                                                                           = 75.3%

4 0
3 years ago
How many kilograms of solvent would contain 0.43 mol of CaO in a 2.5 molality solution
dalvyx [7]
<h3>Answer:</h3>

0.024 kg CaO

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Aqueous Solutions</u>

  • Molarity = moles of solute / liters of solution

<u>Atomic Structure</u>

  • Reading a Periodic Tables
  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

0.41 mol CaO

2.5 M Solution

<u>Step 2: Identify Conversions</u>

1000 g = 1 kg

Molar Mass of Ca - 40.08 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CaO - 40.08 + 16.00 = 56.08 g/mol

<u>Step 3: Convert</u>

  1. Set up:                                \displaystyle 0.43 \ mol \ CaO(\frac{56.08 \ g \ Cao}{1 \ mol \ CaO})(\frac{1 \ kg \ CaO}{1000 \ g \ CaO})
  2. Multiply:                              \displaystyle 0.024114 \ kg \ CaO

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs as our lowest.</em>

0.024114 kg CaO ≈ 0.024 kg CaO

4 0
2 years ago
Read 2 more answers
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