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Eva8 [605]
3 years ago
7

Define a fluid very sort answer ​

Chemistry
1 answer:
Ganezh [65]3 years ago
4 0

Answer:

A fluid is a medium that has a defined mass and volume, but no fixed shape, at a constant temperature and pressure. This may include gases, liquids, plasmas, and to some extent plastic solids. A fluid can flow and deform, preventing it from carrying loads in a static equilibrium.  A fluid is always compressible and internal frictional forces always occur due to the viscosity of the fluid.

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A small diamond, which is made of pure carbon, contains too many carbon atoms to count individually. Which is closest to the num
8090 [49]

The correct answer is option C, 5.02 x 10²² carbon atoms

Atomic mass of C = 12 g/mol

According to Avogadro, 1 mole of C has 6.023 x 10²³C atoms

Now 1 mole of C is equal to 12 g

Therefore, 12 g of C = 6.023 x 10²³ C atoms

1 g of C = \frac{6.023 x 10^2^3}{12} C atoms = 5.02 x 10²² C atoms

7 0
3 years ago
LIMITING REACTANT!! Please help I’m very confused.
alexira [117]

Answer:

We'll have 1 mol Al2O3 and 3 moles H2

Explanation:

Step 1: data given

Numer of moles of aluminium = 2 moles

Number of moles of H2O = 6 moles

Step 2: The balanced equation

2Al + 3H2O → Al2O3 + 3H2

Step 3: Calculate the limiting reactant

For 2 moles Al we need 3 moles H2O to produce 1 mol Al2O3 and 3 moles H2

Aluminium is the limiting reactant. It will completely be consumed (2 moles).

H2O is in excess. There will react 3/2 * 2 = 3 moles

There will remain 6 - 3 = 3 moles

Step 4: Calculate moles products

For 2 moles Al we need 3 moles H2O to produce 1 mol Al2O3 and 3 moles H2

For 2 moles Al we'll have 2/1 = 1 mol Al2O3

For 2 moles Al We'll have 3/2 * 2 = 3 moles H2

We'll have 1 mol Al2O3 and 3 moles H2

8 0
3 years ago
The chemical combination of two or more different atoms in fixed amounts is called a(n)
valkas [14]
I think the correct answer from the list of choices above is option B. <span>The chemical combination of two or more different atoms in fixed amounts is called a compound. There are two type of compounds namely the ionic and covalent compounds.</span>
4 0
3 years ago
Read 2 more answers
Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp
gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200  

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in  j/mol R Ea is

Ea=35.5*1000 j/mol R

ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}

Read more about rate constant

brainly.com/question/20305871

#SPJ1

5 0
2 years ago
Kc for the reaction N2O4 &lt;=&gt; 2NO2 is 0.619 at 45 degrees C If 50.0g of N2O4 is introduced into an empty 2.10L container, w
Nadya [2.5K]

Answer:

p(N2O4) = 0.318 atm

p(NO2) = 7.17 atm

Explanation:

Step 1: Data given

Kc = 0.619

Temperature = 45.0 °C

Mass of N2O4 = 50.0 grams

Volume = 2.10 L

Molar mass N2O4 = 92.01 g/mol

Step 2: The balanced equation

N2O4 ⇔ 2NO2

Step 3: Calculate moles N2O4

Moles N2O4 = 50.0 grams / 92.01 g/mol

Moles N2O4 = 0.543 moles

Step 4: The initial concentration

[N2O4] = 0.543 moles/2.10 L = 0.259 M

[NO2]= 0 M

Step 5: Calculate concentration at the equilibrium

For 1 mol N2O4 we'll have 2 moles NO2

[N2O4] = (0.259 -x)M

[NO2]= 2x

Step 6: Calculate Kc

Kc = 0.619=  [NO2]² / [N2O4]

0.619 = (2x)² / (0.259-x)

0.619 = 4x² / (0.259 -x)

x = 0.1373  

Step 7: Calculate concentrations

[N2O4] = (0.259 -x)M = 0.1217 M

[NO2]= 2x = 0.2746 M

Step 8: The moles

Moles = molarity * volume

Moles N2O4 = 0.1217 M * 2.10  = 0.0256 moles

Moles NO2 = 0.2746 M * 2.10 = 0.577 moles

Step 9: Calculate partial pressure

p*V = n*R*T

⇒ with p = the partial pressure

⇒ with V = the volume = 2.10 L

⇒ with n = the number of moles

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 45 °C = 318 K

p = (nRT)/V

p(N2O4) = (0.0256 *0.08206 * 318)/ 2.10

p(N2O4) = 0.318 atm

p(NO2) = (0.577 *0.08206 * 318)/ 2.10

p(NO2) = 7.17 atm

6 0
3 years ago
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