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2 years ago

_____ energy is energy available to do work.

2 answers:

7 0

The answer is C. Free Energy.

Activation energy is the energy that must be exceeded before a chemical reaction can occur. Free energy best describes as energy available to do work.

Masja [62]2 years ago

7 0

Answer:

When talking about the available energy within the cells, we're talking about "Free Energy", which is the kind of energy available to do work; and if we express that concempt into a thermal environment, that kind of energy is also known as potential energy; that energy in the cells is stored by ATP molecules and later on moved from one part of the cell to another, where it can be used for other biochemical purposes. Thereby (C.)

Explanation:

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Two billiard balls having masses of 0.2 kg and 0.15kg approach each other. The first ball having a velocity of 2m/s hit the seco

Gekata [30.6K]

Answer: The velocity of the first ball is 0.875 m/s if the second ball travels at 1.5 m/s after collision.

Explanation:

Given: $m_{1}$ = 0.2 kg, $m_{2}$ = 0.15 kg

$v_{1}$ = 2 m/s, $v_{2}$ = 0 m/s, $v'_{1}$ = ?, $v'_{2}$ = 1.5 m/s

Formula used is as follows.

$m_{1}v_{1} + m_{2}v_{2} = m_{1}v'_{1} + m_{2}v'_{2}$

where,

v = velocity before collision

v' = velocity after collision

Substitute the values into above formula as follows.

$m_{1}v_{1} + m_{2}v_{2} = m_{1}v'_{1} + m_{2}v'_{2}\\0.2 kg \times 2 m/s + 0.15 kg \times 0 m/s = 0.2 kg \times v'_{1} + 0.15 kg \times 1.5 m/s\\0.4 kg m/s + 0 = 0.2v'_{1} + 0.225 kg m/s\\0.2v'_{1} kg = 0.175 kg m/s\\v'_{1} = \frac{0.175 kg m/s}{0.2 kg}\\= 0.875 m/s$

Thus, we can conclude that the velocity of the first ball is 0.875 m/s if the second ball travels at 1.5 m/s after collision.

4 0

3 years ago

Acetonitrile (CH3CN) is a polar organic solvent that dissolves a wide range of solutes, including many salts. The density of a 1

erastovalidia [21]

Answer:

a. [LiBr] = 2.70 m

b. Xm for LiBr = 0.1

c. 81% by mass CH₃CN

Explanation:

Solvent → Acetonitrile (CH₃CN)

Solute → LiBr, lithium bromide

We convert the moles of solute to mass → 1.80 mol . 86.84 g/1 mol = 156.3 g

This mass of solute is contained in 1L of solution

1 L = 1000 mL → 1mL = 1cm³

We determine solution mass by density

Solution density = Solution mass / Solution volume

Solution density . Solution volume = solution mass

0.824 g/cm³ . 1000 cm³ = 824 g

Mass of solution = 824 g (solvent + solute)

Mass of solute = 156.3 g

Mass of solvent = 824 g - 156.3 g = 667.7 g

Molality → Moles of solute in 1kg of solvent

We convert the mass of solvent from g to kg → 667.7 g . 1kg /1000g = 0.667 kg

Mol/kg → 1.80 mol / 0.667 kg = 2.70 m → molality

Mole fraction → Mole of solute / Total moles (moles solute + moles solvent)

Moles of solvent → 667.7 g . 1mol/ 41g = 16.3 moles

Total moles = 16.3 + 1.8 = 18.1

Mole fraction Li Br → 1.80 moles / 18.1 moles = 0.1

Mass percentage → (Mass of solvent, in this case / Total mass) . 100

We were asked for the acetonitrile → (667.7 g / 824 g) . 100 = 81%

3 0

2 years ago

The flask contains 10.0 mL of HCl and a few drops of phenolphthalein indicator. The buret contains 0.160 M NaOH. It requires 18.

olchik [2.2K]

Answer:

Approximately $0.291\; \rm M$ (rounded to two significant figures.)

Explanation:

The unit of concentration $\rm M$ is the same as $\rm mol \cdot L^{-1}$ (moles per liter.) On the other hand, the volume of both the $\rm NaOH$ solution and the original $\rm HCl$ solution here are in milliliters. Convert these two volumes to liters:

$V(\mathrm{NaOH}) = 18.2\; \rm mL = 18.2 \times 10^{-3}\; \rm L = 0.0182\; \rm L$ .
$V(\text{$\mathrm{HCl}$, original}) = 10.0\; \rm mL = 10.0\times 10^{-3}\; \rm L = 0.0100\; \rm L$ .

Calculate the number of moles of $\rm NaOH$ in that $0.0182\; \rm L$ of $0.160\; \rm M$ solution:

$\begin{aligned} n(\mathrm{NaOH}) &= c(\mathrm{NaOH})\cdot V(\mathrm{NaOH})\\ &= 0.160\; \rm mol \cdot L^{-1} \times 0.0182\; \rm L \approx 0.00291\; \rm mol\end{aligned}$ .

$\rm HCl$ reacts with $\rm NaOH$ at a one-to-one ratio:

$\rm HCl\; (aq) + NaOH\; (aq) \to NaCl\; (aq) + H_2O\; (l)$ .

Coefficient ratio:

$\displaystyle \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} = 1$ .

In other words, one mole of $\rm NaOH$ would neutralize exactly one mole of $\rm HCl$ . In this titration, $0.291\; \rm mol$ of $\rm NaOH\!$ was required. Therefore, the same amount of $\rm HC$ should be present in the original solution:

$\begin{aligned}&n(\text{$\mathrm{HCl}$, original})\\ &= n(\mathrm{NaOH})\cdot \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} \\ &\approx 0.00291\; \rm mol \times 1 = 0.00291\; \rm mol\end{aligned}$ .

Calculate the concentration of the original $\rm HCl$ solution:

$\displaystyle c(\text{$\mathrm{HCl}$, original}) = \frac{n(\text{$\mathrm{HCl}$, original})}{V(\text{$\mathrm{HCl}$, original})} \approx \frac{0.00291\; \rm mol}{0.0100\; \rm L} \approx 0.291\; \rm M$ .

5 0

3 years ago

Use the data set to answer the question.

andrezito [222]

Answer:

It is both accurate and precise.

Explanation:

Precision and accuracy are two different terms used to describe data or measurements. Accuracy refers to how close a set of measurements/experimental values is to an accepted or correct value while Precision refers to how close a series of experimental values are to one another.

In the given set of data in the question below, the Correct Value is 59.2 while the experimental values are as follows;

Trial 1: 58.7

Trial 2: 59.3

Trial 3: 60.0

Trial 4: 58.9

Trial 5: 59.2

Based on comparison, it can be observed that these experimental values are close to the correct value (59.2). Hence, they are said to be ACCURATE. Also, the experimental values are close to one another, hence, they are said to be PRECISE.

Therefore, the data set is both accurate and precise.

7 0

2 years ago

Maya is playing the guitar. She strums the strings and the guitar produces noise. Maya is transforming _______ energy into _____

SOVA2 [1]

The answer would be mechanical; sound

6 0

2 years ago

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