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SCORPION-xisa [38]
2 years ago
14

In this project, you will practice calculating work and speed using the proper formula. You will also use a data table to record

and analyze your data. For each question, you will complete a row in your data table to record the details, and your results. Don't forget to use proper units of measurements in all of the parts of your table!
pls help asap mark brainyist can't spell sorry
Chemistry
2 answers:
Gala2k [10]2 years ago
6 0

Answer:

What is the project about ?

Explanation:

kirill115 [55]2 years ago
4 0

Answer:

I don't understand what your project is about do you have any ss/ Screenshots?

Explanation:

You might be interested in
a 20ml sample of hcl was titrated with the 0.0220 M Naoh. to reach the endpoint required 23.72 mL of the NaOh. Calculate the mol
Y_Kistochka [10]

Answer:

Molarity of HCl=0.026092M

Explanation:

The equation for the reaction is;

HCl + NaOH ⇒ NaCl + H2O

Using the formular, \frac{C_{A}V_{A}}{C_{B}V_{B} }=\frac{nA}{nB}    ..........equ1

whereC_{A} is the concentration of Acid,

          V_{A} is volume of acid

          C_{B} is concentration of the base

          V_{B} is volume of the base

          nA is the number of moles of Acid

          nB is number of moles of base

nA = 1,    nB=1 , V_{A}=20ml, C_{B}=0.022M, V_{B}=23.72mL

we will input these values into equation1 to solve for C_{A}

\frac{C_{A}*20}{0.022*23.72}=\frac{1}{1}

C_{A}*20=0.022*23.72

C_{A}=0.026092M

7 0
3 years ago
What is the mass, in grams, of 1.33 mol of water, H2O? Express the mass in grams to three significant figures.
puteri [66]

Answer:

I need point

Explanation:

I need point

4 0
3 years ago
Read 2 more answers
Making the simplistic assumption that the dissolved NaCl(s) does not affect the volume of the solvent water, determine the const
shtirl [24]

Answer:

  • m = 1,000/58.5
  • b = - 1,000 / 58.5

1) Variables

  • molarity: M
  • density of the solution: d
  • moles of NaCl: n₁
  • mass of NaCl: m₁
  • molar mass of NaCl: MM₁
  • total volume in liters: Vt
  • Volume of water in mililiters: V₂
  • mass of water: m₂

2) Density of the solution: mass in grams / volume in mililiters

  • d = [m₁ + m₂] / (1000Vt)

3) Mass of NaCl: m₁

    Number of moles = mass in grams / molar mass

    ⇒ mass in grams = number of moles × molar mass

        m₁ = n₁ × MM₁


4) Number of moles of NaCl: n₁

   Molarity = number of moles / Volume of solution in liters

   M = n₁ / Vt

   ⇒ n₁ = M × Vt


5) Substitue in the equation of m₁:

   m₁ = M × Vt × MM₁


6) Substitute in the equation of density:

    d = [M × Vt × MM₁ + m₂] / (1000Vt)


7) Simplify and solve for M

  • d = M × Vt × MM₁ / (1000Vt) + m₂/ (1000Vt)
  • d = M × MM₁ / (1000) + m₂/ (1000Vt)

Making the simplistic assumption that the dissolved NaCl(s) does not affect the volume of the solvent water means 1000Vt = V₂  

  • d = M × MM₁ / (1000) + m₂/ V₂

        m₂/ V₂ is the density of water: 1.00 g/mL

  • d = M × MM₁ / (1000) + 1.00 g/mL
  • M × MM₁ / (1000) = d - 1.00 g/mL
  • M = [1,000/MM₁] d - 1,000/ MM₁

8) Substituting MM₁ = 58.5 g/mol

  • M = [1,000/58.5] d - [1,000/ 58.5]

Comparing with the equation Molarity = m×density + b, you obtain:

  • m = 1,000/58.5
  • b = - 1,000/58.5
7 0
3 years ago
Read 2 more answers
Consider the following reaction where Kc = 1.80×10-2 at 698 K:
Klio2033 [76]

Answer:

The system is not in equilibrium and the reaction must run in the forward direction to reach equilibrium.

Explanation:

The reaction quotient Qc is a measure of the relative amount of products and reagents present in a reaction at any given time, which is calculated in a reaction that may not yet have reached equilibrium.

For the reversible reaction aA + bB⇔ cC + dD, where a, b, c and d are the stoichiometric coefficients of the balanced equation, Qc is calculated by:

Qc=\frac{[C]^{c}*[D]^{d}  } {[A]^{a}*[B]^{b}}

In this case:

Qc=\frac{[H_{2} ]*[I_{2} ] } {[HI]^{2}}

Since molarity is the concentration of a solution expressed in the number of moles dissolved per liter of solution, you have:

  • [H_{2} ]=\frac{2.09*10^{-2} moles}{1 Liter}=2.09*10⁻² \frac{moles}{liter}
  • [I_{2} ]=\frac{4.14*10^{-2} moles}{1 Liter}=4.14*10⁻² \frac{moles}{liter}
  • [I_{2} ]=\frac{0.280 moles}{1 Liter}= 0.280 \frac{moles}{liter}

So,

Qc=\frac{2.09*10^{-2} *4.14*10^{-2}  } {0.280^{2} }

Qc= 0.011

Comparing Qc with Kc allows to find out the status and evolution of the system:

If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.

If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.

If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.

Being Qc=0.011 and Kc=1.80⁻²=0.018, then Qc<Kc. <u><em>The system is not in equilibrium and the reaction must run in the forward direction to reach equilibrium.</em></u>

8 0
3 years ago
Beaker A contains 2.06 mol of copper ,and Barker B contains 222 grams of silver.Which beaker the larger number of atom?
Dmitry [639]

Answer:

The number of copper atoms 12.405 ×10²³ atoms.  

The number of silver atoms  13.13 ×10²³ atoms.

Beaker B have large number of atoms.

Explanation:

Given data:

In beaker A

Number of moles of copper = 2.06 mol

Number of atoms of copper = ?

In beaker B

Mass of silver = 222 g

Number of atoms of silver = ?

Solution:

For beaker A.

we will solve this problem by using Avogadro number.

The number 6.022×10²³ is called Avogadro number and it is the number of atoms in one mole of substance.

While we have to find the copper atoms in 2.06 moles.

So,

63.546 g = 1 mole = 6.022×10²³ atoms

For 2.06 moles.

2.06 × 6.022×10²³ atoms

The number of copper atoms 12.405 ×10²³ atoms.  

For beaker B:

107.87 g = 1 mole = 6.022×10²³ atoms

For 222 g

222 g / 101.87 g/mol = 2.18 moles

2.18 mol × 6.022×10²³ atoms = 13.13 ×10²³ atoms

8 0
3 years ago
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