2. zinc (III) chlorate →

Why are fixed resistors’ values indicated by color bands rather than printing the numeric value on their exterior?

gladu [14]

Because there are so many different values of numbers, it would be impractical to use 1Ω, 2Ω, 3Ω... etc... Using colored bands helps make reading it a little easier to the trained eye. There are hundreds of thousands, if not tens of millions of different resistors would need to exist to cover every value. So you just use something called "preferred values" with their resistance values posted on them instead.

6 0

3 years ago

A reaction produces 0.809 moles of h2o. how many molecules are produced?

PSYCHO15rus [73]

Use Avogadro's number which is numerically equal to ...

( 6.02 multiply by 10 to the power of negative 23)

Avocado's number is the number of particle(molecules, ions, atoms) present in one mole of substance.

number of molecule in 0.809 mole of h20 = ....

one mole of h20 contains (6.02* 10^-23)

0.809 mole contains....(6.02* 10^-23) (0.809)
which is equal to [4.951* 10 ^-23]

if I made no mistake in calculation...it must be correct

3 0

3 years ago

Read 2 more answers

How many grams of chlorine gas are present in a 150. liter cylinder of chlorine held at a pressure of 1.00 atm and 0. °C? Group

OlgaM077 [116]

Answer:

474 grams of chlorine gas are present in a 150 liter cylinder of chlorine held at a pressure of 1.00 atm and 0 °C

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

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 occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

In this case:

P= 1.00 atm
V= 150 L
n= ?
R= 0.082 $\frac{atm*L}{mol*K}$
T= 0 C= 273 K

Replacing:

1.00 atm* 150 L= n*0.08206 $\frac{atm*L}{mol*K}$ *273 K

Solving:

$n=\frac{1.00 atm* 150 L}{0.08206 \frac{atm*L}{mol*K}*273 K}$

n= 6.69 moles

Being Cl= 35.45 g/mole, the molar mass of chlorine gas is:

Cl₂=2*35.45 g/mole= 70.9 g/mole

So if 1 mole has 70.9 grams, 6.69 moles of the gas, how much mass does it have?

$mass=\frac{6.69 moles*70.9 grams}{1 mole}$

mass= 474.321 grams ≅ 474 grams

474 grams of chlorine gas are present in a 150 liter cylinder of chlorine held at a pressure of 1.00 atm and 0 °C

4 0

3 years ago

Circle the letter of each sentence that is part of John Dalton’s atomic theory.

satela [25.4K]

Answer:

A) All the elements are composed of atoms.

d) Atoms cannot be created or destroyed in any chemical changes.

c) An atom of one element cannot be changed into an atom of a different element.

Explanation:

The main postulates of Dalton atomic theory was:

1) The matter is composed of atoms and atom is indivisible particle which can not be destroyed.

2) The atoms of an element are identical.

3) The atoms of different elements are combining with definite proportion and form compounds.

4) The atoms of different elements are different from each other.

5) During chemical reaction rearrangement of atoms occur and form new compounds.

4 0

3 years ago

A compound is 21.6% Mg, 21.4% C, and 57.0% O. What is the empirical formula of the compound?

nignag [31]

Answer:

Compound x = $MgC_{2}O_{4}$

Explanation:

Let the compound be x

Assuming we have a 100g of compound x

Given the following data;

Magnesium, Mg = 21.6% = 21.6g

Carbon, C = 21.4% = 21.4g

Oxygen, O = 57.0% = 57.0g

Now, we would find the amount of moles for each element.

Atomic mass of Mg = 24.30g

Atomic mass of C = 12.01g

Atomic mass of O = 16.00g

Amount of moles for Mg;

21.6*(1/24.30) = 0.89mol

Amount of moles for C;

21.4*(1/12.01) = 1.78mol

Amount of moles for O;

57.0*(1/16.00) = 3.56mol

We then divide by the smallest to find the ratio;

0.89/0.89 = 1 Mol of Mg

1.78/0.89 = 2 Mol of C

3.56/0.89 = 4 Mol of O

Therefore, the ratio of Mg, C and O is 1:2:4.

Compound x = $MgC_{2}O_{4}$

Hence, the empirical formula of the compound is $MgC_{2}O_{4}$

7 0

3 years ago