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barxatty [35]
3 years ago
6

Temperate grasslands have trees and shrubs True or false

Chemistry
2 answers:
laila [671]3 years ago
5 0
it should be True unless if I’m wrong
Volgvan3 years ago
3 0
The answer is False.
Temperate grasslands do not have trees and shrubs.
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List 3 common house hold items and 2 physical properties for each
BARSIC [14]

physical properties for a bar of soap: hardness and color

physical properties for a book: texture and size

physical properties for tooth past: cohesiveness and taste

7 0
3 years ago
Read 2 more answers
You placed 5.5 grams of salt in 10 ml of water. What is the w/v%
ra1l [238]

Answer:

By definition, a percent w/v solution is the measure of weight per 100 mL. 7.5 g/100 mL = 7.5%

Explanation:

6 0
3 years ago
To study a key fuel-cell reaction, a chemical engineer has 20.0-L tanks of H₂ and of O₂ and wants to use up both tanks to form 2
Lapatulllka [165]

the pressure needed on H2 tank is 34.9477319atm and the pressure needed in O2 tanks is 16.7690007atm .

Given , to study a key fuel-cell reaction , a chemical engineer has 20.0L tanks of H2 and O2 and wants to use up both tanks to form 28.0mol of water at 23.8°C .

the reaction of the fuel cell is given by ,

H2(g) +1/2 O2 ( g) →H2O

Moles of H2 required = 28 mol

moles of O2 required = 14mol

Now according to van der waals equation ,

p= (nRT/ V-nb)  - an^2 / V^2

for H2 , a = 0.2453L^2bar /mol^2 , b= 0.02651L/mol

P for H2

= (28×0.0821×296.8/20-28×0.02651 ) -(0.2456 ×28×28/20×20)

P for H2 = 35.4291079-0.481376 = 34.9477319 atm

for O2 ,  a = 1.38L^2bar/mol^2 , b = 0.03186L/mol

P for O2 = (14×0.0821×296.8 /20-14×0.03186) - (1.382×14×14/20×20)

P for O2 = 17.4461807 - 0.67718 =16.7690007 atm

Hence , the pressure needed on H2 tank is 34.9477319atm and the pressure needed in O2 tanks is 16.7690007atm .

Learn more about pressure here :

brainly.com/question/25965960

#SPJ4

6 0
1 year ago
You sealed an Erlenmeyer flask that was determined to have a volume of 272.2 mL with a stopper. The density of air at 25°C is 0.
musickatia [10]

Answer:

The amount of air is 11.1 mmol

Explanation:

Density shows the relation between mass and volume, so you can know the mass which has been sealed in the Erlenmayer.

Density = mass/ volume

Density . volume = mass

0.001185 g/ml . 272.2 ml = 0.322 g

We have the mass now, so the molar weight determinates the mols

Mass/molar weight = mol

0.322g/ 28.96 g/m = 0.0111 mol

To get a better number we can inform on milimols

Mols . 1000 = milimols

8 0
4 years ago
Can some help me do this we have a final tomorrow and i have now idea how to do it. Step by Step explanation and tips If you kno
Romashka-Z-Leto [24]

Answer:

87.27 grams

Explanation:

The mole ratio of nitrogen to hydrogen is 1:3; while that one of hydrogen to the products (ammonia) is 3:2

Thus if 3 moles of hydrogen gas produce 2 moles of ammonia gas

7.7 moles of hydrogen will produce:

(7.7moles×2)/3

77/15 moles

1 mole of ammonia gas has a mass of  14+3=17

since the mass of an atom of nitrogen is 14 while that of hydrogen atom is 1.

Therefore 77/15 moles will have a mass of

77/15 moles × 17=87.27 grams

3 0
3 years ago
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