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2 years ago

How many moles are present in 34 grams of cu(oh)2? *?

Chemistry

1 answer:

lakkis [162]2 years ago

3 0

Question Answer
How many moles are present in 34 grams of Cu(OH)2?
34 grams = 0.35 moles

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Why should scientists be able to replicate an investigation?

Ivenika [448]

Answer:

Explanation:

It is also better so scientists know that the data is accurate and that the results will be reliable. If an investigation is replicable, somebody else can use it. In order for others to achieve the same results scientists also have to be able to collect the same data ...

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3 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}$, original})} \approx \frac{0.00291\; \rm mol}{0.0100\; \rm L} \approx 0.291\; \rm M$ .

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2 years ago

Explain what would happen in Step 2 if the cylinder were heated with the plunger held steady.

Svetradugi [14.3K]

Answer:

be more descriptive

Explanation:

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2 years ago

The neutrons of an atom have ______ charges and are located ____ the nucleus

djyliett [7]

Neutral charges and are found in the center? I'm not sure for the last answer

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3 years ago

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Acetone is one of the most important solvents in organic chemistry. It is used to dissolve everything from fats and waxes to air

Phantasy [73]

Answer:

The half-life time, the team equired for a quantity to reduce to half of its initial value, is 79.67 seconds.

Explanation:

The half-life time = the time required for a quantity to reduce to half of its initial value. Half of it's value = 50%.

To calculate the half-life time we use the following equation:

[At]=[Ai]*e^(-kt)

with [At] = Concentration at time t

with [Ai] = initial concentration

with k = rate constant

with t = time

We want to know the half-life time = the time needed to have 50% of it's initial value

50 = 100 *e^(-8.7 *10^-3 s^- * t)

50/100 = e^(-8.7 *10^-3 s^-1 * t)

ln (0.5) = 8.7 *10^-3 s^-1 *t

t= ln (0.5) / -8.7 *10^-3 = 79.67 seconds

The half-life time, the team equired for a quantity to reduce to half of its initial value, is 79.67 seconds.

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3 years ago