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Soloha48 [4]
4 years ago
5

A droplet of pure mercury has a density of 13.6 g/cm^3. What is the density of a sample of pure mercury that is 10 times as larg

e as the droplet?
Chemistry
2 answers:
Vlad1618 [11]4 years ago
6 0

Answer: Answer

Explanation:

GG I BEAT YOU IN FORTNITE GET GOOD

andre [41]4 years ago
4 0
The only thing you need to solve it is to know that 
<span>density = mass or volumes
</span>It does't matter which mass do you have- density is always the same, therefore <span> 13.6 g/cm^3</span>×10
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Analyze the following reaction and the given scenarios to discuss the relationship between volume and reaction rate. Determine w
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The second reaction with the volume of 5L will occur faster as compared to the first reaction of volume 10L.

Volume is inversely proportional to the rate of reaction.

As volume increases rate of reaction decreases and as volume decreases rate of reaction increases.

Let's consider a reaction A   →  B.

r is rate of reaction and K is rate constant, A is the concentration of reaction.

r = k(A)

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2 years ago
Cu(NO3)2 + Zn (s) → Cu (s) + Zn(NO3)2 is an example of which type of reaction?
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Single replacements since cu is being replaced by zn
4 0
4 years ago
A student mixes in a test tube 3.00mL of 0.050M CuSO4with 7.00mL of 0.20M NH3/NH41 . The solution becomes a deep blue color. Ass
valkas [14]

Answer:

\large \boxed{\text{0.0035 mol/L}}

Explanation:

We are given the volumes and concentrations of two reactants, so this is a limiting reactant problem.

We know that we will need moles, so, lets assemble all the data in one place.

                   Cu²⁺ + 4NH₃ ⟶ Cu(NH₃)₄²⁺

    V/mL:   3.00      7.00

c/mol·L⁻¹:  0.050   0.20

1. Identify the limiting reactant

(a) Calculate the moles of each reactant  

\text{Moles of Cu}^{2+}= \text{3.00 mL solution} \times \dfrac{\text{0.050 mmol Cu}^{2+}}{\text{1 mL solution}} = \text{0.150 mmol Cu}^{2+}\\\\\text{Moles of NH}_{3} = \text{7.00 mL solution} \times \dfrac{\text{0.20 mmol NH}_{3}}{\text{1 mL solution}} = \text{0.140 mmol NH}_{3}

(b) Calculate the moles of Cu(NH₃)₄²⁺ that can be formed from each reactant

(i) From Cu²⁺

\text{Moles of Cu(NH$_{3}$)$_{4}$$^{2+}$} = \text{0.150 mmol Cu}^{2+} \times \dfrac{\text{1 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}}{\text{1 mmol Cu}^{2+}}\\\\= \text{0.150 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}

(ii) From NH₃

\text{Moles of Cu(NH$_{3}$)$_{4}$$^{2+}$} = \text{0.140 mmol NH}_{3} \times \dfrac{\text{1 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}}{\text{4 mmol NH}_{3}}\\\\= \text{0.0350 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}

NH₃ is the limiting reactant, because it forms fewer moles of the complex ion.

(c) Concentration of the complex ion

\text{The reaction forms 0.0350 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$ in a total volume of 10.00 mL.}\\c = \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{0.0350 mmol}}{\text{10.00 mL}} = \textbf{0.0035 mol/L}\\\\\text{The concentration of the complex ion is $\large \boxed{\textbf{0.0035 mol/L}}$}

7 0
3 years ago
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