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3 years ago

Any reaction that absorbs 150 kcal of energy can be classified as

1 answer:

6 0

The answer should be (D) endothermic reaction.

Reason : - Reactions in which energy is absorbed are called endothermic reactions. In your provided question, 150 kcal energy is being <u>absorbed</u>, and thus, we can say that it is an endothermic reaction.

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Determine the molality of ions in a solution formed by dissolving 0.187 moles of NaCl in 456 grams of water. The density of the

PSYCHO15rus [73]

Answer:

Molality Ions Na⁺ = 0.410 m

Molality Ions Cl⁻ = 0.410 m

Total ion molality = 0.820 m

Explanation:

Molality (m) is defined as moles of solute that are dissolved in 1000g (1kg) of solvent. In this case, the molality of the ions is requested, so it should be borne in mind that when the NaCl salt dissolves in the water, it dissociates forming ions as follows:

NaCl ⇒Na⁺ (aq) + Cl⁻ (aq)

Therefore, <em>for each mole of salt that dissolves, you will have 1 mole of Na⁺ and 1 mole of Cl⁻ </em>

Knowing the amount of initial solute moles and the mass of water, we proceed to perform the calculations for each ion:

456 g of H₂O ____ 0.187 mol Na⁺

1000 g of H₂O ____ X = 0.410 m

Calculation: 1000 g x 0.187 mol / 456 g = 0.410 mol

<em>Ions Cl⁻</em>

456 g of H₂O ____ 0.187 mol Cl⁻

1000 g of H₂O ____ X = 0.410 m

Calculation: 1000 g x 0.187 mol / 456 g = 0.410 m

If both results are added, the total molality of the ions will be obeyed:

0.410 m + 0.410 m = 0.820 m

This means that <em>for every kilogram of water, 0.820 moles of dissociated ions from NaCl are dissolved.</em>

8 0

2 years ago

The titration of 25.00 ml a 0.125 m hclo4 solution requires 27.07 ml of koh to reach the endpoint. what is the concentration of

agasfer [191]

Answer : The concentration of the $KOH$ is, 0.115 M

Explanation :

Using dilution law,

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1$ = basicity of an acid = 1

$n_2$ = acidity of a base = 1

$M_1$ = concentration of $HClO_4$ = 0.125 M

$M_2$ = concentration of KOH = ?

$V_1$ = volume of $HClO_4$ = 25 ml

$V_2$ = volume of NaOH = 27.07 ml

Now put all the given values in the above law, we get the concentration of the $KOH$ .

$1\times 0.125M\times 25ml=1\times M_2\times 27.07ml$

$M_2=0.115M$

Therefore, the concentration of the $KOH$ is, 0.115 M

5 0

3 years ago

A scientist fills a large, tightly sealed balloon with 75,000 mL of helium at STP.

pav-90 [236]

Considering the definition of STP conditions, 3.35 moles of helium are contained within the balloon.

<h3>Definition of STP condition</h3>

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

<h3>Amount of moles of helium within the balloon</h3>

In this case, you know that scientist fills a large, tightly sealed balloon with 75,000 mL of helium at STP.

So, you can apply the following rule of three: if by definition of STP conditions 22.4 liters are occupied by 1 mole of helium, 75 L (75 L= 75000 mL, being 1 L= 1000 L) are occupied by how many moles of helium?

$amount of moles of helium=\frac{75 Lx1 mole}{22.4 L}$