All categories

3 years ago

Need help with 2 to 6 please :) Will give brainliest + extra points if you use words from the word search

Chemistry

2 answers:

inn [45]3 years ago

5 0

Answer:

6 collide .4 sesuoh 5 gravitational pull 3 sroetem

1seriorite collide and comment

Nookie1986 [14]3 years ago

4 0

Answer: .6 collide .4 sesuoh 5 gravitational pull 3 sroetem

1seriorite collide and comment

Explanation:

You might be interested in

A shopping cart rolling down a hill for 8 seconds has an acceleration of 6.0 meters/second?. If the cart

Komok [63]

Answer:

51.0 meters/second is the final speed

6 0

3 years ago

A 80.0 g piece of metal at 88.0°C is placed in 125 g of water at 20.0°C contained in a calorimeter. The metal and water come to

grandymaker [24]

Answer:

The specific heat of the metal is 0.485 J/g°C

Explanation:

<u>Step 1:</u> Data given

Mass of the piece of metal = 80.0 grams

Mass of the water = 125 grams

Initial temperature of the metal = 88.0 °C

Initial temperature of water =20.0 °C

Final temperature = 24.7 °C

pecific heat of water is 4.18 J/g*°C

<u>Step 2:</u> Calculate specific heat of the metal

Qgained = -Qlost

Q =m*c*ΔT

Qwater = - Qmetal

m(water)*c(water)*ΔT(water) = -m(metal)*c(metal)*ΔT(metal)

with mass of water = 125 grams

with c( water) = 4.18 J/g°C

with ΔT(water) = T2-T1 = 24.7 - 20 = 4.7°C

with mass of metal = 80.0 grams

with c(metal) = TO BE DETERMINED

with ΔT(metal) = 24.7 - 88.0 = -63.3 °C

125*4.18*4.7 = -80 * C(metal) * -63.3

2455.75 = -80 * C(metal) * -63.3

C(metal) = 2455.75 / (-80*-63.3)

C(metal) = 0.485 J/g°C

The specific heat of the metal is 0.485 J/g°C

3 0

4 years ago

Assume human blood is iso-osmotic or isotonic with a 0.16 mol/lit saline (NaCl) solution (the solution does not change the volum

const2013 [10]

Answer:

π = 4.1 atm

Explanation:

We can calculate the osmotic pressure exerted by a solution using the following expression.

π = M . R . T

where,

π is the osmotic pressure

M is the molar concentration of the solution

R is the ideal gas constant

T is the absolute temperature

The absolute temperature is 37 + 273 = 310 K

π = M . R . T

π = (0.16 mol/L) . (0.082atm.L/mol.K) . 310 K = 4.1 atm

8 0

4 years ago

Which component of an atom would you not expect to find in the nucleus, but might be orbiting around it?

cricket20 [7]

Answer: Electrons orbit the nucleus

4 0

3 years ago

Iron(III) oxide reacts with carbon monoxide to produce iron and carbon; Fe2O3(s)+3CO(g)-->2Fe(s)+3CO2(g). a) What is the perc

VMariaS [17]

Answer:

a) %yield= 33.00 %

b) %yield= 72.1 %

Explanation:

Step 1: Data given

Mass of iron(III) oxide = 65.0 grams

mass of iron produced = 15.0 grams

Molar mass of Fe2O3 = 159.69 g/mol

Molar mass of CO = 44.01 g/mol

Step 2: The balanced equation:

Fe2O3(s)+3CO(g) →2Fe(s)+3CO2(g)

Step 3: Calculate moles of Fe2O3

Moles Fe2O3 = 65.0 grams / 159.69 g/mol

Moles Fe2O3 = 0.407 moles

Step 4: Calculate moles Fe

For 1 mole Fe2O3 we'll have 2 moles Fe

For 0.407 moles Fe2O3 we'll have 2*0.407 = 0.814 molesFe

Step 5: Calculate mass Fe

Mass fe = 0.814*55.845 g/mol

Mass Fe = 45.46 grams = theoretical yield

Step 6: Calculate % yield

%yield = (actual yield/theoretical yield)*100%

%yield = (15.0/45.46)*100%

%yield= 33.00 %

b) What is the percent yield for the reaction if 75.0 g of carbon monoxide produces 85.0 g of carbon dioxide?

Step 1: Data given

Mass of CO = 75.0 grams

mass of CO2 produced = 85.0 grams

Molar mass of CO = 28.01 g/mol

Molar mass of CO2 = 44.01 g/mol

Step 2: The balanced equation:

Fe2O3(s)+3CO(g) →2Fe(s)+3CO2(g)

Step 3: Calculate moles of CO

Moles CO = 75.0 grams / 28.01 g/mol

Moles CO = 2.68 moles

Step 4: Calculate moles CO2

For 1 mole Fe2O3 we need 3 moles CO to produce 2 moles Fe and 3 moles CO2

For 2.68 moles CO we'll have 2.68 moles CO2

Step 5: Calculate mass CO2

Mass CO2= 2.68 * 44.01 g/mol

Mass CO2 = 117.95 grams = theoretical yield

Step 6: Calculate % yield

%yield = (actual yield/theoretical yield)*100%

%yield = (85.0/117.95)*100%

%yield= 72.1 %

7 0

4 years ago