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Marianna [84]
3 years ago
11

Mannitol salt agar is defined as both a selective and a differential agar since it selects against non-halophiles and allows the

researcher to distinguish between organisms that vary in their ability to ferment mannitol. True or false?
Chemistry
1 answer:
viktelen [127]3 years ago
8 0

Answer: The above statement is true

Explanation:

Mannitol salt agar is a medium which contain high concentration of salt,about 7.5%-10%. It is commonly used as a selective and differential growth medium.Because of its high concentration of salt,it inhibits growth if non halophiles(as they only grow best in media containing less tha 0.2M salts). If the bacteria can ferment mannitol,an acidic byproduct is formed which turns phenol red agar into yellow.So only those bacteria grow well which can ferment mannitol.

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40.0L of N₂ gas are in a sealed container at STP.How many moles of N₂ are present?9 mol
Vinvika [58]

Explanation:

We have to find the number of moles of N₂ that are present in a sample that has a volume of 40.0 L at STP.

STP means Standard Conditions of Temperature and Pressure. These conditions are 273.15 K and 1 atm. We know that 1 mol of N₂ will occupy 22.4 L. We can use that ratio to find the answer to our problem.

1 mol of N₂ = 22.4 L

moles of N₂ = 40.0 L * 1 mol/(22.4 L)

moles of N₂ = 1.79 mol

Answer: 1.79 moles of nitrogen are present.

8 0
1 year ago
The number of grams of H2 in 1470 mL of H2 gas. ​
Lorico [155]

Answer:

0.1313 g.

Explanation:

  • It is known that at STP, 1.0 mole of ideal gas occupies 22.4 L.
  • Suppose that hydrogen behaves ideally and at STP conditions.

<u><em>Using cross multiplication:</em></u>

1.0 mol of hydrogen occupies → 22.4 L.

??? mol of hydrogen occupies → 1.47 L.

∴ The no. of moles of hydrogen that occupies 1.47 L = (1.0 mol)(1.47 L)/(22.4 L) = 6.563 x 10⁻² mol.

  • Now, we can get the no. of grams of hydrogen in 6.563 x 10⁻² mol:

<em>The no. of grams of hydrogen = no. of hydrogen moles x molar mass of hydrogen</em> = (6.563 x 10⁻² mol)(2.0 g/mol) = <em>0.1313 g.</em>

5 0
3 years ago
The equilibrium constant has been estimated to be 0.12 at 25 °C. If you had originally placed 0.069 mol of cyclohexane in a 2.8
scZoUnD [109]

Answer: Concentrations of cyclohexane and methylcyclopentane at equilibrium are 0.0223 M and 0.0027 M respectively

Explanation:

Moles of cyclohexane = 0.069 mole

Volume of solution = 2.8 L

Initial concentration of cyclohexane =\frac{moles}{Volume}=\frac{0.069}{2.8}=0.025M

The given balanced equilibrium reaction is,

                            cyclohexane  ⇔  methylcyclopentane

Initial conc.                 0.025 M           0

At eqm. conc.       (0.025-x)M       (x) M

The expression for equilibrium constant for this reaction will be,

K= methylcyclopentane / cyclohexane

Now put all the given values in this expression, we get :

0.12=\frac{(x)}{(0.025-x)}

By solving the term 'x', we get :

x =  0.0027

Concentration of cyclohexane at equilibrium = (0.025-x ) M = (0.025-0.0027) M = 0.0223 M

Concentration of methylcyclopentane at equilibrium = (x ) M = (0.0027) M

4 0
4 years ago
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I think it’s a carrot . pretty sure
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