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Vaselesa [24]
3 years ago
6

Element (X) forms different compounds with Element (Y)

Chemistry
1 answer:
qaws [65]3 years ago
8 0

Complete question:

Element (X) forms different compounds with Element (Y)

Ex:

Compound 1: XY6 (2.82g)

Compound 2: XY? (1.41g)

Compound 3: XY? (0.94g)

Compound 1: 2.82/6 = 0.47

Compound 2: 1.41/0.47 = 2

Compound 3: 0.94/0.47 = 3

What Law is this?

Answer:

Law of Multiple Proportions

Explanation:

The Law of Multiple Proportions states that when two elements combine to form more than one compound, they do so in simple whole number ratios of their masses/weights.  

You might be interested in
Now moles:<br> moles of H2+<br> A moles of O2<br> moles of H20
yan [13]

Answer:

H2+ + O2 —— H2O.

Explanation:

Hope it helps you.

4 0
3 years ago
Help(Must be done by 2/22/2018)
alekssr [168]
These are 6 questions and 6 answers.

Question 1:

Answer: 33.7 atm

Explanation:

1) Data:

p=?
m = 1360.0 g N2O
V = 25.0 liter
T = 59.0°C

2) Formulas:

Ideal gas law: p V = n R T
n = mass in grams / molar mass

3) Solution

n = mass of N2O in grams / molar mass of N2O

molar mass of N2O = 2 * 14 g/mol + 16 g/mol = 44 g/mol

n = 1360.0 g / 44 g/mol = 30.9 mol

T = 59.0 + 273.15 K = 332.15 K

R = 0.0821 atm*liter / K*mol

=> p = nRT / V = 30.9 mol * 0.0821 [atm*liter / K * mol] * 332.15K / 25.0 liter = 33.7 atm

Answer: 33.7 atm

Question 2:

Answer: 204.5 liter

Explanaton:

1) Data:

m = 11.7 g of He
V = ?
p = 0.262 atm
T = - 50.0 °C

2) Formulas:

pV = nRT

n = mass in grams / atomic mass

3) Solution:

atomic mass of He = 4.00 g/mol

n = 11.7 g / 4.00 g/mol = 2.925 mol

T = - 50.0 + 273.15 K = 223.15 K

pV = nRT => V = nRT / p

V = 2.925 mol * 0.0821 [* liter / K*mol] *223.15K / 0.262 atm = 204.5 liter

Answer: 204.5 liter

Question 3.

Answer: 97.8 mol

Explanation:

1) Data:

Ethane
T = 15.0 °C
p = 100.0 kPa
V = 245.0 ml
n = ?

2) Formula

pV = nRT

3) Solution

pV = nRT => n = RT / pV

T = 15.0 + 273.15K = 288.15K

R = 8.314 liter * kPa / (mol*K)

n = 8.314 liter * kPa / (mol*K) * 288.15K / [100.0 kPa * 0.245 liter] = 97.8 mol

Answer: 97.8 mol

Question 4:

Answer: 113.67 K = - 159.48 °C

Explanation:

1) Data:

V = 629 ml of O2
p = 0.500 atm
n = 0.0337 moles
T = ?

2) Formula:

pV = nRT

3) Solution:

pV = nRT => T = pV / (nR)

T = 0.500 atm * 0.629 liter / (0.0337 mol * 0.0821 atm*liter/K*mol ) = 113.67 K

°C = T - 273.15 = - 159.48 °C

Question 5.

Answer: 5.61 g

Explanation:

1) Data:

V = 3.75 liter of NO
T = 19.0 °C
p = 1.10 atm
m = ?

2) Formulas

pV = nRT

mass = number of moles * molar mass

3) Solution:

pV = nRT => n = pV / (RT)

T = 19.0 + 273.15 K = 292.15 K

n = 1.10 atm * 3.75 liter / [ (0.0821 atm*liter / K*mol) * 292.15 K ] = 0.17 mol

molar mass of NO = 17.0 g/mol + 16.0 g/mol = 33.0 g/mol

mass = 0.17 mol * 33.0 g/mol = 5.61 g

Question 6:

Answer: 22.4 liter

Explanation:

1) Data:

STP
n = 1.00 mol
V = ?

Solution:

1) It is a notable result that 1 mol of gas at STP occupies a volume of 22.4 liter, so that is the answer.

2) You can calculate that from the formula pV = nRT

3) STP stands for stantard pressure and temperature. That is p = 1 atm and T = 0°C = 273.15 K

4) Clear V from the formula:

V = nRT / p = 1.00 mol * 0.0821 atm*liter / (K*mol) * 273.15 K / 1.00 atm = 22.4 liter
7 0
3 years ago
The titration of 25.00 ml a 0.125 m hclo4 solution requires 27.07 ml of koh to reach the endpoint. what is the concentration of
agasfer [191]

Answer : The concentration of the KOH is, 0.115 M

Explanation :

Using dilution law,

n_1M_1V_1=n_2M_2V_2

where,

n_1 = basicity of an acid = 1

n_2 = acidity of a base = 1

M_1 = concentration of HClO_4 = 0.125 M

M_2 = concentration of KOH = ?

V_1 = volume of HClO_4 = 25 ml

V_2 = volume of NaOH = 27.07 ml

Now put all the given values in the above law, we get the concentration of the KOH.

1\times 0.125M\times 25ml=1\times M_2\times 27.07ml

M_2=0.115M

Therefore, the concentration of the KOH is, 0.115 M

5 0
4 years ago
In this experiment, the metal cations in the solutions were initially in the ground state. When placed in the flame, the metals
Andrej [43]
Initially, you should know that what the text describes is a test used to analyze and identify some metal ions in a compound. This is the principle behind it: first the ions are excited, by the heat of the flame, they absorb energy and the electrons are promoted to higher energy levels; after this, the electrons will fall back down to lower energy levels releasing energy as light. The flame color that is seen is related with the frequency of the light emitted.

Here is the text with the right words.

In this experiment, the metal cations in the solutions were initially in the ground state. When placed in the flame, the metals then (absorbed) energy as (heat). When this occurred, electrons made transitions from (low) energy levels to (high) energy levels. The metals were then in the (excited) state. The electrons in these metals then made transitions from (high) energy levels to (low) energy levels, resulting in the (emission) of energy as (EM radiation).

7 0
3 years ago
Read 2 more answers
31) The mass of an electron is approximately 1/1836
dolphi86 [110]

Answer:

(_{1} ^ {1}\textrm{H})  is the correct choice that is option A.

Explanation:

An electron has \frac{1}{1836} times the mass of a proton, but an equal and opposite negative charge.

Lets _{A}^{Z}\textrm{X} is any element represented here.

Where Z is the number of protons as well as its atomic number.

And

A is the mass number which is the number of protons and neutrons in any atom.

Note: We know that Hydrogen (_{1} ^ {1}\textrm{H}) is devoid of neutron,so number of neutrons in a Hydrogen atom is zero.

Then

(_{1} ^ {1}\textrm{H}) is also a representation of single proton.

So (_{1} ^ {1}\textrm{H}) is the right choice that is option A.

6 0
3 years ago
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