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4 years ago

What did Mendel do that made his predictions so accurate

Chemistry

2 answers:

Hoochie [10]4 years ago

8 0

He worked with large numbers of plants.

murzikaleks [220]4 years ago

6 0

Answer:

he worked wit a large number of plants

Explanation: the other guy got it right i just got it done

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Jill and Susan were absent when the instructions were given for the lab investigation. They gathered their materials and watched

MrRissso [65]

Jill and Susan violated safety procedures by not properly listening and/or reading over the instructions to know all the materials, steps, and equipment they need for the lab. Hope this helps!

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3 years ago

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Elena makes the table below to determine the number of atoms of each element in the chemical formula 3(NH4)2SO4

frozen [14]

Answer:

she should not have multiplied the sulfur atoms by the subscript 4

Explanation:

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3 years ago

Read 2 more answers

Please help I don’t know if my answer is right!!

Temka [501]

Answer:

neutralization reaction! Aka: option C!

HOPE THIS HELPS! :)

Explanation:

8 0

3 years ago

20 poll

aleksley [76]

Answer: n=15.56moles

Explanation:

PV = nRT

where

P is pressure in atmospheres

V is volume in Liters

n is the number of moles of the gas

R is the ideal gas constant = given as (0.0821L -atm/k-mol

PV = nRT

n= PV/RT

n= (1.5 X 230)/ (0.0821 X 270)

n= 15.56 moles

5 0

4 years ago

Where the oxygen comes from the air (21% O2 and 79% N2). If oxygen is fed from air in excess of the stoichiometric amount requir

guajiro [1.7K]

Answer:

$y_{O2} =4.3$ %

Explanation:

The ethanol combustion reaction is:

$C_{2}H_{5} OH+3O_{2}$ → $2CO_{2}+3H_{2}O$

If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

$x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}$

Dividing the previous equation by x:

$1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}$

We would need 3.30 oxygen moles per ethanol mole.

Then we apply the composition relation between O2 and N2 in the feed air:

$3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )$

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

$3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles$

Calculate the number of moles of CO2 and water considering the same:

$0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)$

$0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)$

The total number of moles at the reactor output would be:

$N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)$

So, the oxygen mole fraction would be:

$y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3$ %

6 0

3 years ago