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Sati [7]
2 years ago
8

A sample of aluminium has a mass of 54 gram. What is the mass of the same number of magnesium atom.​

Chemistry
1 answer:
Arada [10]2 years ago
5 0

Answer:

48g of Magnesium.

Explanation:

We know that:

Mole=given mass/molar mass

Then, moles of aluminum = 54/27 = 2.

A/c to question ,

For 2 moles of magnesium it's mass is 2×24=48g.

Hence, the required mass for same number of magnesium atoms is 48g.

(Hope this helps can I pls have brainlist (crown)☺️)

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The right answer for the question that is being asked and shown above is that: "C) carbon monoxide and carbon dioxide" hydrocarbons burn completely in an excess of oxygen, the products are <span>C) carbon monoxide and carbon dioxide</span>
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Which shows the correct order of increasing trophic level, from producer to tertiary consumer?
polet [3.4K]

Answer:

Grass, deer, wolf, fungi.

Explanation:

The tropical level of an organism is defined as the position of organism occupied in the food chain. The food chain is starts from tropical level one which is primary producers (such as plants), then move to herbivores (such as deer) in level two, carnivores, (such as wolf) at level three, apex predators in level four (such as lions), and it ends in decomposers such as fungi.

Fungi play important role in ecosystem to help in breakdown of dead organic matter and return nutrients to the soil. Without fungi nutrients cannot cycle in an ecosystem, and causing the breakdown of entire food web.

6 0
3 years ago
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If you had a 0.5 M KCl solution, how much solute would you have in moles, and what would the solute be?
Svetradugi [14.3K]

Answer:

37.25 grams/L.

Explanation:

  • Molarity (M) is defined as the no. of moles of solute dissolved per 1.0 L of the solution.

<em>M = (no. of moles of KCl)/(volume of the solution (L))</em>

<em></em>

∵ no. of moles of KCl = (mass of KCl)/(molar mass of KCl)

∴ M = [(mass of KCl)/(molar mass of KCl)]/(volume of the solution (L))

∴ (mass of KCl)/(volume of the solution (L)) = (M)*(molar mass of KCl) = (0.5 M)*(74.5 g/mol) = 37.25 g/L.

<em>So, the grams/L of KCl = 37.25 grams/L.</em>

6 0
3 years ago
A student conducts an experiment to determine how the temperature of water affects the time for sugar to dissolve. In each trial
V125BC [204]

The fact that the student used different amount of water (another independent variable) is wrong with the experimental design

WHAT ARE THE COMPONENTS OF AN EXPERIMENT?

  • An experiment aims at solving a scientific problem or answering a scientific question. An experiment should contain a variable being changed called INDEPENDENT VARIABLE and a variable being measured called DEPENDENT VARIABLE.

  • In an ideal experiment, only one independent variable should be used while every other variable should be kept constant. This is done so as not to affect the result of the experiment.

In the experiment conducted by the student in this question, two independent variables were used i.e. the different amount of water and the different temperatures. This is what is wrong about the experimental design.

  • In a nutshell, the fact that two independent variables were used by the student is what is wrong about the experimental design.

Learn more at: brainly.com/question/967776

8 0
3 years ago
An ideal gas at a given initial state expands to a fixed final volume. would the work be greater if the expansion occurs at cons
crimeas [40]

Answer:

Constant pressure

Explanation:

At constant pressure,

w = -p\Delta V = -p(V_{f} - V_{i})

At constant temperature,

w = -RT \ln \left(\dfrac{V_{f}}{V_{i}} \right)

1 mol of an ideal gas at STP has a volume of 22.71 L.

Let's compare the work done as it expands under each condition from an initial volume of 22.71 L.

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w = -100p(V_{2} - 22.71}); \text{(1 bar$\cdot$L = 100 J)}

A plot of -w vs V₂ gives a straight line (red) with a constant slope of 100 J/L as in the diagram below (Note that w is work done on the system, so -w is the work done by the system). \

Isothermal expansion

w= -8.314 \times 273.15 \ln \left(\dfrac{V_{f}}{22.71} \right)\\\\= -2271 \left( \ln V_{f} -\ln22.71 \right)\\= -2271 \left(\ln V_{f} - 3.123 \right)\\= 7092 - 2271\ln V_{f}

A plot of -w vs V₂ is a logarithmic curve. Its slope starts at 100 J/mol but decreases as the volume increases (the blue curve below).

Thus, the work done during an expansion at constant pressure is greater than if the system is at constant temperature.

4 0
3 years ago
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