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3 years ago

A reaction is spontaneous if delta G is 1)positive 2)negative 3)zero

Chemistry

2 answers:

Brrunno [24]3 years ago

7 0

Positive is the right answer

miskamm [114]3 years ago

4 0

delta g must be negative in order for the reaction to be spontaneous bb ♡

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Suppose you want to do an experiment to hear how three different objects with sound when dropped from the same distance. You dro

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4 years ago

How many moles of ammonium ions are in 125 mL of 1.40 M NH4NO3 solution? ________ moles (give answer with correct sig figs in un

Sholpan [36]

The number of mole of ammonium ion, NH₄⁺ in the solution is 0.175 mole

We'll begin by calculating the number of mole of NH₄NO₃ in the solution. This can be obtained as follow:

Volume = 125 mL = 125 / 1000 = 0.125 L

Molarity = 1.40 M

Mole = Molarity x Volume

Mole of NH₄NO₃ = 1.40 × 0.125

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NH₄NO₃(aq) —> NH₄⁺(aq) + NO₃¯(aq)

From the balanced equation above,

1 mole of NH₄NO₃ contains 1 mole of NH₄⁺

Therefore,

0.175 mole of NH₄NO₃ will also contain 0.175 mole of NH₄⁺

Thus, the number of mole of ammonium ion, NH₄⁺ in the solution is 0.175 mole

Learn more: brainly.com/question/25469095

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2 years ago

2. How does the way elements are grouped in the periodic table serve as an advantage to scientists?

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3 years ago

A student prepared a stock solution by dissolving 10.0 g of KOH in enough water to make 150. mL of solution. She then took 15.0

yanalaym [24]

Answer: The concentration of KOH for the final solution is 0.275 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in ml = 150 ml

moles of solute = $\frac{\text {given mass}}{\text {molar mass}}=\frac{10.0g}{56g/mol}=0.178moles$

Now put all the given values in the formula of molality, we get

$Molality=\frac{0.178\times 1000}{150}=1.19M$

According to the dilution law,

$C_1V_1=C_2V_2$

where,

$C_1$ = molarity of stock solution = 1.19 M

$V_1$ = volume of stock solution = 15.0 ml

$C_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 65.0 ml

Putting in the values we get:

$1.19\times 15.0=M_2\times 65.0$

$M_1=0.275M$

Therefore, the concentration of KOH for the final solution is 0.275 M

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3 years ago