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3 years ago

!! Homogeneous or Heterogeneous mixture?

Chemistry

2 answers:

katrin2010 [14]3 years ago

6 0

A, Homogenous | b, heterogenous | c, homogenous

Liula [17]3 years ago

3 0

A) Homogeneous
b) Heterogeneous
c) Homogeneous

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Consider the reaction between HCl and O2: 4HCl(g)+O2(g)→2H2O(l)+2Cl2(g) When 63.1 g of HCl is allowed to react with 17.2 g of O2

alina1380 [7]

Answer:

The limiting reactant for this reaction is the HCl

Explanation:

This is my reaction:

4HCl(g)+O2(g)→2H2O(l)+2Cl2(g)

Molar mass O2 = 32 g/mol

Molar mass HCl = 36,45 g/mol

Mass / Molar mass = Moles

Moles HCl : 63,1 g / 36,45 g/m = 1,73 moles

Moles O2: 17,2 g / 32g/m = 0,54 moles

4 moles of HCl react with 1 mol of O2, according to reaction, so

1,73 moles of HCl, are going to react with, how many moles of O2.

4 moles HCl ___ 1 mol O2

1,73 moles HCl ___ (1,73 . 1)/ 4 = 0,43 moles of O2

O2 is my excess reagent because I need 0,43 moles and I have 0,54 so I have moles in excess.

1 mol of O2 are going to react with 4 moles of HCl

0,54 moles of O2 are going to react with, how many moles of HCl ?

1 mol O2 ____ 4 moles HCl

0,54 mol O2 ___ (0,54 . 4)/ 1 = 2,16 moles

I need 2,16 moles to consume my moles of O2, but I only have 1,73 moles, tha's why the HCl is mi limiting reactant.

3 0

3 years ago

To which factor can the principles of Nicholas Steno be applied?

Tcecarenko [31]

Answer:

the rock layers of the Grand Canyon

Explanation:

Steno's law are laws that applied to sedimentary rocks. These laws helps in understanding sedimentary sequences.

Sedimentary rocks are derived from the deposition of pre-existing rocks in basins. In order to understand some important relationships between these rock layers, Steno's law offer a good insight.

Steno's law are often applied when we want to do relative dating of rock layers. Some of the laws are:

Law of superposition of strata
Law of original horizontality
Law of lateral continuity
Law of inclusion
Law of fossil and fauna succession

These laws helps to interpret sedimentary rock sequences better.

7 0

4 years ago

Read 2 more answers

What is the vapor pressure of a solution that contains 70.0 mL of ether and 30.0 mL of benzene?

Lera25 [3.4K]

Answer:

it is 50

if I m right pls mark me brainliest

6 0

3 years ago

How many moles are in a 275g sample of K2CrO3

Marrrta [24]

Answer:

There are approximately 1.54 moles in a 275 g sample of $K_2CrO_3$ .

Explanation:

To find out number of moles, fistly we have to calculate molecular mass of $K_2CrO_3$ .

There are 2 atoms of Potassium 1 atom of Chromium and 3 atoms of oxygen in the given compound.

For molecular mass we have to add the value of mass of 2 atoms of Potassium with mass of 1 atom of Chromium and with mass of 3 atoms of oxygen.

Atomic mass of Potassium = 39

Atomic mass of Chromium = 52

Atomic mass of Oxygen = 16

Now,

Molecular mass of $K_2CrO_3$ = $2\times 39+52+3\times 16=78+52+48=178\ gm$

The molecular mass of a compound is the mass of compound in one mole.

To find out the number of moles, we have to divide given mass of compound by its molecular mass.

$Number\ of\ moles=\frac{given\ mass}{molecular\ mass}$

$Number\ of\ moles=\frac{275}{178}= 1.54$

Hence the number of moles in 275 gm of $K_2CrO_3$ is 1.54.

8 0

3 years ago

A gas exerts a pressure of 1.8 atm at a temperature of 60 degrees celsius. What is the new temperature when the pressure of the

Kamila [148]

The answer for the following problem is mentioned below.

Therefore the final temperature of the gas is 740 K

Explanation:

Given:

Initial pressure of the gas ( $P_{1}$ ) = 1.8 atm

Final pressure of the gas ( $P_{2}$ ) = 4 atm

Initial temperature of the gas ( $T_{1}$ ) = 60°C = 60 + 273 = 333 K

To solve:

Final temperature of the gas ( $T_{2}$ )

We know;

From the ideal gas equation;

we know;

P × V = n × R × T

So;

we can tell from the above equation;

P ∝ T

(i.e.)

$\frac{P}{T}$ = constant

$\frac{P_{1} }{P_{2} }$ = $\frac{T_{1} }{T_{2} }$

Where;

$P_{1}$ = initial pressure of a gas

$P_{2}$ = final pressure of a gas

$T_{1}$ = initial temperature of a gas

$T_{2}$ = final temperature of a gas

$\frac{1.8}{4}$ = $\frac{333}{T_{2} }$

$T_{2}$ = $\frac{333*4}{1.8}$

$T_{2}$ = 740 K

Therefore the final temperature of the gas is 740 K

8 0

3 years ago