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3 years ago

What is the mass of a bacterial cell?

2 answers:

6 0

Thus, it was found that the intestinal microbiota has approximately 100 trillion bacterial cells - about ten times more than the total number of cells in the human body.

Ray Of Light [21]3 years ago

4 0

100 trillion bacteria cells

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The energy from nuclear fission is potentially limitless but production is restricted by the limited availability of

sineoko [7]

Answer: uranium

Explanation: It’s hard to find when mining and is required for nuclear fission

6 0

3 years ago

What is the volume of 2.00 moles of ideal gas at 25'c and 121.59 kpa of pressure

katrin [286]

Answer:

40.73 L.

Explanation:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm (P = 121.59 kPa/101.325 = 1.2 atm).

V is the volume of the gas in L (V = ??? L).

n is the no. of moles of the gas in mol (n = 2.0 mol).

R is the general gas constant (R = 0.082 L.atm/mol.K),

T is the temperature of the gas in K (T = 25°C + 273 = 298 K).

∴ V = nRT/P = (2.0 mol)(0.082 L.atm/mol.K)(298 K)/(1.2 atm) = 40.73 L.

6 0

3 years ago

The solubility of nitrogen gas at 25 ◦C and 1 atm is 6.8×10−4 mol/L. If the partial pressure of nitrogen gas in air is 0.76 atm,

iris [78.8K]

Answer:

Concentration of dissolved nitrogen = 5.2 × 10⁻⁴ mol/L

Explanation:

More the pressure of the gas, more will be its solubility.

So, for two different pressure, the relation between them is shown below as:-

$\frac {P_1}{C_1}=\frac {P_2}{C_2}$

Given ,

P₁ = 1 atm

P₂ = 0.76 atm

C₁ = 6.8 × 10⁻⁴ mol/L

C₂ = ?

Using above equation as:

$\frac{1\ atm}{6.8\times 10^{-4}\ mol/L}=\frac{0.76\ atm}{C_2}$

$C_2=\frac{0.76\times 6.8\times 10^{-4}}{1}\ mol/L$

Concentration of dissolved nitrogen = 5.2 × 10⁻⁴ mol/L

5 0

3 years ago

How many grams of sodium nitrate (NaNO2) will dissolve in 100g of water at 20°C?

inn [45]

88g ( so sorry if this isn’t correct )

8 0

3 years ago

At a certain temperature the rate of this reaction is first order in HI with a rate constant of :0.0632s

Shalnov [3]

Answer : The time taken for the reaction is, 28 s.

Explanation :

Expression for rate law for first order kinetics is given by :

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = 0.0632

t = time taken for the process = ?

$[A_o]$ = initial amount or concentration of the reactant = 1.28 M

$[A]$ = amount or concentration left time 't' = $1.28\times \frac{17}{100}=0.2176M$

Now put all the given values in above equation, we get:

$0.0632=\frac{2.303}{t}\log\frac{1.28}{0.2176}$

$t=28s$

Therefore, the time taken for the reaction is, 28 s.

8 0

3 years ago