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3 years ago

PLEASE HELP ITS VERY EASY FOR OTHER PLEASE ITS A TEST DUE IN AN HOUR PLEASEEEEEEEEEE

Chemistry

2 answers:

STatiana [176]3 years ago

5 0

Answer:

I think it is three times I'm not sure

AveGali [126]3 years ago

3 0

Answer:

once

Explanation:

You can tell each time there is a horizontal line. There is only one horizontal line that shows the object stayed 10 meters away for about 20 seconds.

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Which system in the body is able to fight off most infectious diseases, such as the common cold?

pashok25 [27]

<span>The immune system fights off infectious disease. This is the system which produces cells that fight off the common cold. This system is also very closely related to the lymphatic system. Some people have much stronger immune systems that others.</span>

7 0

3 years ago

How many moles in 28.0 grams of Oxygen

anastassius [24]

Answer:

1.) 28.0 grams of oxygen28 grams (1 mole/16 grams per mole)=1.75 moles oxygen2.)5.0 moles of Iron5 moles(55.845 grams/1 mole)=279.225 grams Iron3.) 452 grams Argon452 grams(1 mole/39.948 grams)=11.315 moles Argon4.) 16.5 moles Hydrogen16.5 moles(1.01 grams/1 mole)=16.665 grams Hydrogen

Explanation:

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2 years ago

The lewis representation above depicts a reaction between hydrogen (blue) and a main-group element from group (red). in this rep

Fed [463]

The question is incomplete, the complete question is;

The Lewis representation above depicts a reaction between hydrogen (blue) and a main-group element from group______ (red).

In this representation, each Y atom needs ______ electron(s) to complete its octet, and gains these electrons by forming______ bond(s) with atoms of H .

There are ______ unshared electron pair(s) and _______bonding electron pair(s) in the product molecule.

The bonds in the product are _________ (Ionic or Covalent)

Answer:

1) 16

2) 2 electrons

3) 2 bonds

4) 2 unshared pairs of electrons

5) 2 bonding pairs of electrons

6) The bonds in the product are covalent

Explanation:

Group sixteen elements have six electrons on their outermost shell. These include two unshared pairs of electrons and two unpaired electrons. These two unpaired electrons can now be covalently bonded to two hydrogen atoms to give H2Y. The compound H2Y has two lone pairs and two bond pairs of electrons.

H2Y can be a general formula for all hydrides of group 16. They are all very similar in structure but gradually differ in physical and chemical properties according to the graduated variation observed down the group.

6 0

3 years ago

(2 KClO3 (s) → 2 KCl (s) + 3 O2 (g) ) If 165 mL of oxygen is produced at 30.0 °C and 90.0 kPa, what mass of KClO3 was decomposed

soldier1979 [14.2K]

Taking into account the reaction stoichiometry and ideal gas law, 0.48144 grams of KClO₃ was decomposed.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 KClO₃ → 2 KCl + 3 O₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

KClO₃: 2 moles
KCl: 2 moles
O₂: 3 moles

The molar mass of the compounds is:

KClO₃: 122.45 g/mole
KCl: 74.45 g/mole
O₂: 32 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

KClO₃: 2 moles ×122.45 g/mole= 244.8 grams
KCl: 2 moles ×74.45 g/mole= 148.9 grams
O₂: 3 moles ×32 g/mole= 96 grams

<h3>Ideal gas law</h3>

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

P is the gas pressure.
V is the volume that the gas occupies.
T is the temperature of the gas.
R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
n is the number of moles of the gas.

<h3>Number of O₂ produced.</h3>

165 mL of oxygen is produced at 30.0 °C and 90.0 kPa. This is, you know:

P= 90 kPa= 0.888231 atm (being 101.325 kPa= 1 atm)
V= 165 mL= 0.165 L (being 1000 mL= 1 L)
n= ?
R= 0.082 $\frac{atmL}{molK}$
T= 30 C= 303 K (being 0 C= 273 K)

Replacing in the ideal gas law:

0.888231 atm× 0.165 L = n× 0.082 $\frac{atmL}{molK}$ × 303 K

Solving:

n= (0.888231 atm× 0.165 L)÷ (0.082 $\frac{atmL}{molK}$ × 303 K)

<u><em>n= 0.0059 moles</em></u>

Finally, 0.0059 moles of oxygen is produced at 30 °C and 90 kPa.

<h3>Mass of KClO₃ required</h3>

The following rule of three can be applied: If by stoichiometry of the reaction 3 moles of O₂ are produced by 244.8 grams of KClO₃, 0.0059 moles of O₂ are produced by how much mass of KClO₃?

$mass of KClO_{3}= \frac{0.0059 moles of O_{2}x 244.8 grams of KClO_{3}}{3 moles of O_{2}}$