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bulgar [2K]
3 years ago
6

997.62 mol = 56.367 L =

Chemistry
1 answer:
Galina-37 [17]3 years ago
4 0
14.626 Gallons because yeah dhhsjsjshshsh
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How many Al atoms are there in 40.5 g of Al foil? Al:27
andrew-mc [135]

Answer:

c) 9.03 x 10^23

Explanation:

find the molar mass of Al

Al is 27.0 grams

Then use that, to find the number of moles in Aluminum.

Then use Avogadro's number which is 6.02 * 10^23

After that, write all of that down with dimensional analysis.

40.5 g * 1 mol/ 27.0 g of Al  * 6.02 x 10^23 / 1mol

As your final answer, you will get 9.03 * 10^23 atoms with sig figs.

Hope it helped!

4 0
3 years ago
Which of the following best describe the chronological way of specimen preparation?
nata0808 [166]

Answer:

There are many reasons to examine human cells and tissues under the microscope. Medical and biological research is underpinned by knowledge of the normal structure and function of cells and tissues and the organs and structures that they make up. In the normal healthy state, the cells and other tissue elements are arranged in regular, recognizable patterns. Changes induced by a wide range of chemical and physical influences are reflected by alterations in the structure at a microscopic level, and many diseases are characterized by typical structural and chemical abnormalities that differ from the normal state. Identifying these changes and linking them to particular diseases is the basis of histopathology and cytopathology, important specializations of modern medicine. Microscopy plays an important part in haematology (the study of blood), microbiology (the study of microorganisms including parasites and viruses), and more broadly in the areas of biology, zoology, and botany. In all these disciplines, specimens are examined under a microscope.

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8 0
2 years ago
Read 2 more answers
How many grams of Sg is required to produce 83.10 g SF6? S: +24F--&gt;8SF
ozzi

Answer : The mass of S_8 required is 18.238 grams.

Explanation : Given,

Mass of SF_6 = 83.10 g

Molar mass of SF_6 = 146 g/mole

Molar mass of S_8 = 256.52 g/mole

The balanced chemical reaction is,

S_8+24F_2\rightarrow 8SF_6

First we have to determine the moles of SF_6.

\text{Moles of }SF_6=\frac{\text{Mass of }SF_6}{\text{Molar mass of }SF_6}=\frac{83.10g}{146g/mole}=0.569moles

Now we have to determine the moles of S_8.

From the balanced chemical reaction we conclude that,

As, 8 moles of SF_6 produced from 1 mole of S_8

So, 0.569 moles of SF_6 produced from \frac{0.569}{8}=0.0711 mole of S_8

Now we have to determine the mass of S_8.

\text{Mass of }S_8=\text{Moles of }S_8\times \text{Molar mass of }S_8

\text{Mass of }S_8=(0.0711mole)\times (256.52g/mole)=18.238g

Therefore, the mass of S_8 required is 18.238 grams.

7 0
3 years ago
At 20.°C, a 1.2-gram sample of Mg ribbon reacts rapidly with 10.0 milliliters of 1.0 M HCl(aq). Which change in conditions would
Pavlova-9 [17]

Answer: (2) decreasing the concentration of HCl(aq) to 0.1 M

Explanation: Rate of a reaction depends on following factors:

1. Size of the solute particles: If the reactant molecules are present in smaller size, surface of particles and decreasing the size increases the surface area of the solute particles. Hence, increasing the rate of a reaction.  

2. Reactant concentration: The rate of the reaction is directly proportional to the concentration of reactants.

3. Temperature: Increasing the temperature increases the energy of the molecules and thus more molecules can react to give products and rate increases.

(1) Increasing the initial temperature to 25°C will increase the reaction rate.

(2) Decreasing the concentration of HCl(aq) to 0.1 M will decrease the reaction rate due to lesser concentration.

(3) Using 1.2 g of powdered Mg will increase the reaction rate due to large surface area.

(4) Using 2.4 g of Mg ribbon will increase the reaction rate due to high concentration of reactants.

7 0
3 years ago
Read 2 more answers
The distance between two points on a map is 0.04 kilometers what is the distance in meters
ra1l [238]
The answer is 400 meters
8 0
2 years ago
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