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3 years ago

Calculate number of grams in 1.15*10^21 molecules of P2O5

1 answer:

6 0

To go from molecules, we first need to convert to moles and then convert to grams.

To convert from molecules to moles, we need to divide by Avogadro's constant.

1.15*10^21 molecules * (1 mole/6.022*10^23 molecules) = 0.0019097 moles

To convert from moles to grams, we need to use the molar mass.

The molar mass of P₂O₅ is (2*30.98)+(5*16.00) = 141.96
You can find the molar mass using the periodic table.

0.0019097 moles * (141.96 grams/1 mole) = 0.2711 grams.

Since we have 3 significant digits in 1.15*10^21, that means our final answer is limited to 3 significant digits.

Your final answer is 0.271 grams.

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How many subatomic particles are in the NUCLEUS of a neutral atom of Beryllium?

ser-zykov [4K]

Answer:

Explanation:

There are 4 protons, 5 neutrons, and 4 electrons. This is a neutral beryllium atom.

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2 years ago

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A small sack of sand has a density of 1.5 g/cm3 and a mass of 1500 g. How much space (volume) does the sand occupy? 750 cm3 2250

Cloud [144]

1.5 = 1500g/Volume

Rearrange the equation:

Volume = 1500/1.5 = 1000

So your answer is 3. 1000 cm3

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2 years ago

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You are managing a city that needs to upgrade its disinfection basin at your 40 MGD surface water drinking water treatment plant

nignag [31]

Solution :

According to Chick's law

$\frac{N_t}{N_0}=e^{-k'C^n t}$

where, t = contact time

c = concentration of disinfectant

k' = lethality coefficient = 4.71

n = dilution coefficient = 1

4 log removal = % removal = 99.99

$\frac{N_t}{N_0}=\frac{\text{bacteria remaining}}{\text{bacteria initailly present}}$

= 1 - R

= 1 - 0.9999

Now for plug flow reactor contact time,

$\tau =\frac{V}{Q} =\frac{75000}{40 \times 10^6}$

= 0.01875 days

= 27 minutes

For CSTR, $\tau =\frac{V}{Q} =\frac{150000}{40 \times 10^6}$

$=3.75 \times 10^{-3}$ days

= 5.4 minute

There are 3 reactors, hence total contact time = 3 x 5.4

= 16.2 minute

Or $\frac{N_t}{N_0}=e^{-k'C^n t}$

or $(1-0.9999)=e^{-4.71 \times C \times t}$

∴ C x t = 1.955

For PFR, $t_1 = 27$ min

∴ C $=\frac{1.955}{27}$ = 0.072 mg/L

For CSIR, $t_2=16.2$ min

$C=\frac{1.955}{16.2} = 0.1206$ mg/L

∴ Chlorine required for PFR in kg/day

$=\frac{0.072 \times 40 \times 10^6 \times 3.785}{10^6}$ (1 gallon = 3.785 L)

= 18.25 kg/day

Therefore we should go for PFR system.

3 0

2 years ago

The limiting reactant, O2, can form up to 2.7 mol Al2O3. What mass of Al2O3 forms?

Margaret [11]

Answer:

280 g Al₂O₃

Explanation:

To find the mass, you need to multiply the given value by the molar mass. This will cause the conversion because the molar mass exists as a ratio; technically, the ratio states that there are 101.96 grams per every 1 mole Al₂O₃. It is important to arrange the ratio in a way that allows for the cancellation of units. In this case, the desired unit (grams) should be in the numerator. The final answer should have 2 sig figs to reflect the given value (2.7 mol).

Molar Mass (Al₂O₃): 101.96 g/mol

2.7 moles Al₂O₃ 101.96 g
------------------------ x ------------------- = 275 g Al₂O₃ = 280 g Al₂O₃
1 mole

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1 year ago

How would you classify raisin bran?

Nostrana [21]

Pure substance element I think

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3 years ago