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Shkiper50 [21]
3 years ago
6

"A student takes 152 mL of a 3.0M HCl solution and places it into a beaker. The student then adds enough water to the beaker to

make 750 mL. Calculate the molarity of this new solution."
Chemistry
1 answer:
Jet001 [13]3 years ago
7 0

Answer: I think it's 0.608

Explanation: (152×3.0)÷750=0.608

Also the number of moles of the first solution is .456, so if we apply that to the new solution we get .456÷.750=.608

Sorry if wrong

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I need a answer quick. PLEASE<br><br>What is the volume of 2.43 x 1023 molecules of N2 gas at STP?
Readme [11.4K]
I think it is 2485.89.
6 0
3 years ago
In nature, one common strategy to make thermodynamically unfavorable reactions proceed is to couple them chemically to reactions
padilas [110]

Answer:

\triangle G= -6.7 KJ/mol

Explanation:

From the question we are told that:

Chemical Reactions:

X=A⇌B,ΔG= 14.8 kJ/mol

Y=B⇌C,ΔG= -29.7 kJ/mol

Z=C⇌D,ΔG= 8.10 kJ/mol

Since

Hess Law

The law states that the total enthalpy change during the complete course of a chemical reaction is independent of the number of steps taken.

Therefore

Generally the equation for the Reaction is mathematically given by

T = +1 * X +1 * Y +1 *Z

Therefore the free energy, ΔG is

\triangle G=1 * \triangle G*X +1 * \triangle G*Y +1 * \triangle G *Z

\triangle G= +1 * (14.9) +1 * (-29.7) +1 * (8.10)

\triangle G= -6.7 KJ/mol

5 0
3 years ago
How many moles of gold, Au, are in 3.60 x 10^-5 g of gold?
zlopas [31]
<h3>Answer:</h3>

1.83 × 10⁻⁷ mol Au

<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>Atomic Structure</u>

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

<u>Step 1: Define</u>

3.60 × 10⁻⁵ g Au (Gold)

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

Molar Mass of Au - 196.97 g/mol

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 3.60 \cdot 10^{-5} \ g \ Au(\frac{1 \ mol \ Au}{196.97 \ g \ Au})
  2. Multiply:                            \displaystyle 1.82769 \cdot 10^{-7} \ mol \ Au

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1.82769 × 10⁻⁷ mol Au ≈ 1.83 × 10⁻⁷ mol Au

4 0
3 years ago
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rjkz [21]
Your reaction
.. Fe + O2 ---> FexOy
for this reaction..
.. the Fe on the left is in the 0 oxidation state
.. the Fe on the right is in the +(2y/x) oxidation state
.. the O on the left is in the 0 oxidation state
.. the O on the right is in the -2 oxidation state
meaning
.. the O is reduced... . . (it's reduced in oxidation state)
.. the Fe is oxidized.. . .(oxidation state increased)
this is a REDOX reaction

*********
AND.. it's also a synthesis reaction.. (aka combination reaction)
4 0
3 years ago
Read 2 more answers
A. Calculate the empirical formula of a molecule with percent compositions: 55.3% potassium (K), 14.6% phosphorus (P), and 30.1%
Otrada [13]
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6 0
3 years ago
Read 2 more answers
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