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kati45 [8]
2 years ago
5

PLEASE HELP I BEG

Chemistry
1 answer:
Gwar [14]2 years ago
5 0

Answer:

1.Velocity and pitch

2.The unit of vibration is hertz. The number of oscillations of an oscillating body per second is known as frequency. Complete step by step answer: The time taken by a vibrating body to complete one vibration is time period.

3.Frequencies above 20,000 Hz are known as ultrasound.

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A solution is formed by mixing 15.2 g KOH into
nikitadnepr [17]

Answer:

There are approximately 0.271\; \rm mol of formula units in that \rm 15.2\; g of \rm KOH (the solute of this solution.)

Explanation:

A solution includes two substances: the solute and the solvent. Note the solution here contains significantly more water than \rm KOH. Hence, assume that water is the solvent (as it is in many other solutions.)

The (molar) formula mass of \rm KOH is necessary for finding the number of moles of

  • One \rm K atom,
  • One \rm O atom, and
  • One \rm  H atom.

The formula mass of \rm KOH will thus be the sum of:

  • The mass of one mole of \rm K atoms,
  • The mass of one mole of \rm O atoms, and
  • The mass of one mole of \rm H atoms.

On the other hand, the mass (in grams) of one mole of atoms of an element is (numerically) the same as its relative atomic mass. The relative atomic mass data can be found on most modern periodic tables.

Relative atomic mass data from a modern periodic table:

  • \rm K: 39.098.
  • \rm O: 15.999.
  • \rm H: 1.008.

For example, the relative atomic mass of \rm K (potassium, atomic number 19) is 39.098 (3 sig. fig.) Hence, the mass of one mole of

The formula mass of \rm KOH is the sum of these three masses:

\begin{aligned}& M(\mathrm{KOH}) \\ &\approx 39.098 + 15.999 + 1.008 \\ &= 56.105\; \rm g \cdot mol^{-1}\end{aligned}.

The number of moles of \rm KOH formula units in this 15.2\; \rm g sample would be:

\begin{aligned}n &= \frac{m(\mathrm{KOH})}{M(\mathrm{KOH})} \\ &\approx \frac{15.2\; \rm g}{56.105\; \rm g \cdot mol^{-1}} \approx 0.271\; \rm mol \end{aligned}.

6 0
2 years ago
The ease with which a raw material can be molded , flattened , or bent is known as its ?
Leya [2.2K]
A. Malleability

Ductility is being able to be drawn into wires
Elasticity is being able to resist stress
Resilience is being able to spring back into shape (kind of like elasticity)
6 0
3 years ago
Read 2 more answers
Given these reactions, where X represents a generic metal or metalloid 1 ) H 2 ( g ) + 1 2 O 2 ( g ) ⟶ H 2 O ( g ) Δ H 1 = − 241
Mazyrski [523]

Answer:

-767,2kJ

Explanation:

It is possible to sum enthalpies of half-reactions to obtain the enthalpy of a global reaction using Hess's law. For the reactions:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁= −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂= +361.7 kJ

3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl (g) ΔH₃= −92.3 kJ

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄= − 607.9 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅= − 44.0 kJ

The sum of (4) - (2) produce:

6) XCl₄(s) + O₂(g) ⟶ XO₂(s) + 2Cl₂(g) ΔH₆ = ΔH₄ - ΔH₂ = -969,6 kJ

(6) + 4×(3):

7) XCl₄(s) + 2H₂(g) + O₂(g) ⟶ XO₂(s) + 4HCl(g) ΔH₇ = ΔH₆ + 4ΔH₃= -1338,8 kJ

(7) - 2×(1):

8) XCl₄(s) + 2H₂O(g) ⟶ XO₂(s) + 4HCl(g) ΔH₈ = ΔH₇ - 2ΔH₁= -855,2kJ

(8) - 2×(5):

9) XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g) ΔH₉ = ΔH₈ - 2ΔH₅= <em>-767,2kJ</em>

I hope it helps!

6 0
3 years ago
If the percent (mass/mass) for a solute is 8% and the mass of the solution is 200 g, what is the mass of solute in solution?
tatyana61 [14]

Answer:

The answer to your question is  16 g

Explanation:

Data

Percent by mass = 8%

Mass of the solution = 200 g

Mass of solute = ?

Formula

Percent by mass = mass of solute / mass of solution x 100

- Solve for mass of solute

  Mass of solute = Percent by mass x mass of solution / 100

- Substitution

   Mass of solute = 8 x 200 / 100

- Simplification

    Mass of solute = 1600 / 100

- Result

    Mass of solute = 16 g

3 0
3 years ago
Read 2 more answers
It is a large body of air that has the same properties as the Earth's surface over which it develops.what is it?
Debora [2.8K]
The answer is AIR MASS.

4 0
3 years ago
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