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allochka39001 [22]
2 years ago
5

Which of the following most likely happens when the volume of a gas increases?

Chemistry
1 answer:
mafiozo [28]2 years ago
6 0

the pressure of the gas increases

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An antacid tablet contains 640.0 mg of magnesium oxide per tablet.
Salsk061 [2.6K]

Answer:

317.6 mL

Explanation:

Step 1: Write the balanced neutralization equation

MgO + 2 HCl ⇒ MgCl₂ + H₂O

Step 2: Calculate the mass corresponding to 640.0 mg of MgO

The molar mass of MgO is 40.30 g/mol. The moles corresponding to 640.0 mg (0.6400 g) of MgO are:

0.6400 g × (1 mol/40.30 g) = 0.01588 mol

Step 3: Calculate the moles of HCl that react with 0.01588 moles of MgO

The molar ratio of MgO to HCl is 1:2. The moles of HCl are 2/1 × 0.01588 mol = 0.03176 mol

Step 4: Calculate the volume of 0.1000 M HCl that contains 0.03176 moles

0.03176 mol × (1 L/0.1000 mol) = 0.3176 L = 317.6 mL

3 0
2 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
One beaker contains 100 mL of pure water and second beaker contains 100 mL of seawater. The two beakers are left side by side on
gayaneshka [121]

Explanation:

When we add a non-volatile solute in a solvent then due to the impurity added to the solution there will occur an increase in the boiling point of the solution.  

This increase in boiling point will be known as elevation in boiling point.

As one beaker contains seawater (water having NaCl) will have some impurity in it. So, more temperature is required by seawater to escape into the atmosphere.

Whereas another beaker has only pure water so it is able to easily escape into the atmosphere since, it contains no impurity.

Thus, we can conclude that level of pure water will decrease more due to non-volatile solute present in it as compared to seawater.

7 0
2 years ago
Limestone is made of predominantly calcium carbonate. Geologists can test for limestone using hydrochloric acid. The products ar
Blizzard [7]
I'm obsessed with me as much as you
Say you'd die for me I'd die for me too
7 0
3 years ago
Iodine-131 is a beta emitter used as a tracer in radio immunoassays in biological systems. It follows first order kinetics. The
allsm [11]

<u>Answer:</u> The amount of Iodine-131 remain after 39 days is 0.278 grams

<u>Explanation:</u>

The equation used to calculate rate constant from given half life for first order kinetics:

t_{1/2}=\frac{0.693}{k}

where,

t_{1/2} = half life of the reaction = 8.04 days

Putting values in above equation, we get:

k=\frac{0.693}{8.04days}=0.0862days^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,

k = rate constant = 0.0862days^{-1}

t = time taken for decay process = 39 days

[A_o] = initial amount of the sample = 8.0 grams

[A] = amount left after decay process = ?

Putting values in above equation, we get:

0.0862=\frac{2.303}{39}\log\frac{8.0}{[A]}

[A]=0.278g

Hence, the amount of Iodine-131 remain after 39 days is 0.278 grams

3 0
2 years ago
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