This part of the plant cell turns sunlight into food. It is called a

In a student experiment, the empirical formula of a copper halide was found by adding aluminum metal to an aqueous solution of t

Radda [10]

Answer: The empirical formula for the given compound is $CuCl_3$

Explanation:

We are given:

Mass of copper chloride in 1 L or 1000 mL of solution = 42.62 grams

Taking Trial A:

Volume of solution = 49.6 mL

Applying unitary method:

In 1000 mL of solution, the mass of copper chloride present is 42.62 grams

So, in 49.6 mL of solution, the mass of copper chloride will be = $\frac{42.62}{1000}\times 49.6=2.114g$

We are given:

Mass of filter paper = 0.908 g

Mass of filter paper + copper = 1.694 g

Mass of copper = [1.694 - 0.908] g = 0.786 g

Mass of chlorine in the sample = [2.114 - 0.786]g = 1.328 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Copper = $\frac{\text{Given mass of Copper}}{\text{Molar mass of Copper}}=\frac{0.786g}{63.5g/mole}=0.0124moles$

Moles of Chlorine = $\frac{\text{Given mass of Chlorine}}{\text{Molar mass of Chlorine}}=\frac{1.328g}{35.5g/mole}=0.0374moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0124 moles.

For Copper = $\frac{0.0124}{0.0124}=1$

For Chlorine = $\frac{0.0374}{0.0124}=3.02\approx 3$

Step 3: Taking the mole ratio as their subscripts.

The ratio of Cu : Cl = 1 : 3

Hence, the empirical formula for the given compound is $CuCl_3$

3 0

3 years ago

The density of gold is 19.3 g/cm3. What is the volume of a 575 gram bar of pure gold?

Gnoma [55]

Answer:

The answer to your question is: Volume = 29.79 cm³

Explanation:

Data

Density of gold = 19.3 g/cm3

Mass = 575 grams

Formula

density = mass / volume

Volume = mass/density

Volume = 575 g / 19.3 g/cm3

Volume = 29.79 cm³

3 0

3 years ago

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1. john needs to create a buffered solution at a ph of 3.5 for his biomedical laboratory

Lunna [17]

Answer:

Use a ratio of 0.44 mol lactate to 1 mol of lactic acid

Explanation:

John could prepare a lactate buffer.

He can use the Henderson-Hasselbalch equation to find the acid/base ratio for the buffer.

$\text{pH} = \text{pK}_{\text{a}} + \log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}}\\\\3.5 = 3.86 + \log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}}\\\\\log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}} = 3.5 - 3.86 = -0.36\\\\\dfrac{\text{[A$^{-}$]}}{\text{[HA]}} = 10^{-0.36} = \mathbf{0.44}$

He should use a ratio of 0.44 mol lactate to 1 mol of lactic acid.

For example, he could mix equal volumes of 0.044 mol·L⁻¹ lactate and 0.1 mol·L⁻¹ lactic acid.

6 0

3 years ago

Compare and contrast all the periodic trends as you move down a group of metais. Organize

elena55 [62]

Answer:

The number of energy levels increases as you move down a group as the number of electrons increases. Each subsequent energy level is further from the nucleus than the last. Therefore, the atomic radius increases as the group and energy levels increase.

Explanation:

Dose this help? Tell me if it dose.

4 0

3 years ago

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Emission Spectrum Questions for Quiz Print - Quizizz

-Dominant- [34]

Isotopes of the same element

₃₅⁷⁷X and ₃₅⁸¹X

<h3>Further explanation</h3>

Given

Isotopes of element

Required

Isotopes of the same element

Solution

The elements in nature have several types of isotopes

Isotopes are elements that have the same Atomic Number (Proton) and different mass numbers

Element symbols that meet the 2 conditions above are :

₃₅⁷⁷X and ₃₅⁸¹X

3 0

2 years ago