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3 years ago

What animals are listed in the picture?

Chemistry

2 answers:

Lapatulllka [165]3 years ago

8 0

star fish, sea shell,sea turtle,sea merchants, squid, and i dont know the other one

Dmitriy789 [7]3 years ago

7 0

starfish , squid , sea turtle are the only ones i know

You might be interested in

Earth is unique because it is the _____.

astraxan [27]

Answer:

B. The only planet to support life.

Explanation:

5 0

3 years ago

Read 2 more answers

0.467 mol NaCl are needed to

Alisiya [41]

Answer:

27.4 gram is the solution it's simple dude...

Explanation:

don't be afraid of huge question they confuse you you need not to be confused

now see simple solution

molality is denoted by m

m= moles of solute / mass of solvent in kg.

i hope your know the meaning of solute and solvent....

so moles are given 0.467

and molar mass is given 58.44

so just take out the gram means

by applying formula

58.44×0.467

it will give 27.4 grams simple.....

8 0

2 years ago

Using the thermodynamic information in the ALEKS Data tab, calculate the standard reaction entropy of the following chemical rea

Aleksandr [31]

Answer:

ΔS° = - 47.2 J/mol.K

Explanation:

P4O10 + 6H2O → 4H3PO4

ΔS°= 4(S°mH3PO4) - 6(S°mH2O) - S°mP4O10

∴ S°mH2O(l) = 69.9 J/mol.K

∴ S°mP4O10 = 231 J/mol.K

∴ S°mH3PO4 = 150.8 J/mol.K

⇒ ΔS° = 4*(150.8) - 6*(69.9) - 231

⇒ ΔS° = - 47.2 J/mol.K

5 0

3 years ago

Read 2 more answers

It is proposed to use Liquid Petroleum Gas (LPG) to fuel spark-ignition engines. A typical sample of the fuel on a volume basis

Norma-Jean [14]

Answer:

The overall balanced combustion reaction is written as :

$0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> 3.05CO_2 \ + \ 3.8H_2O \ + \ 18.612N_2$

$(F/A)_{stoichiometric} = 0.0424$

$(A/F)_{stoichiometric} = 23.562$

the higher heating values $(HHV)_f$ per unit mass of LPG = 49.9876 MJ/kg

the lower heating values $(LHV)_f$ per unit mass of LPG = 46.4933 MJ/kg

Explanation:

The stoichiometric equation can be expressed as :

$0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> aCO_2 \ + \ bH_2O \ + \ cN_2$

Now, equating the coefficient of carbon; we have:

(0.7×3)+(0.05×4)+(0.25×3) = a

a = 3.05

Also, Equating the coefficient of hydrogen : we have:

(0.7 × 8) +(0.05 × 10) + ( 0.25 × 6) = 2 b

2b = 7.6

b = 3.8

Equating the coefficient of oxygen

2x = 2a + b

$x = \frac{2a+b}{2} \\ \\ x = \frac{2(3.05)+3.8}{2} \\ \\ x = 4.95$

Equating the coefficient of Nitrogen

$c = 3.76x \\ \\ c = 3.76 *4.95 \\ \\ c = 18.612$

Therefore, The overall balanced combustion reaction can now be written as :

$0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> 3.05CO_2 \ + \ 3.8H_2O \ + \ 18.612N_2$

Now; To determine the stoichiometric F/A and A/F ratios; we have:

$(F/A)_{stoichiometric} = \frac{n_f}{n_a } \\ \\ (F/A)_{stoichiometric} = \frac{1}{4.95*(1+3.76)} \\ \\ (F/A)_{stoichiometric} = 0.0424$

$(A/F)_{stoichiometric} = \frac{n_a}{n_f } \\ \\ (A/F)_{stoichiometric} = \frac{4.95*(1+3.76)}{1} \\ \\ (A/F)_{stoichiometric} = 23.562$

What are the higher and lower heating values per unit mass of LPG?

Let calculate the molecular mass of the fuel in order to determine their mass fraction of the fuel components.

Molecular mass of the fuel $M_f = (0.7*M_{C_3H_5} ) + (0.05 *M_{C_4H_{10}}) + (0.25*M _{C_3H_6})$

= 30.8 + 2.9 + 10.5

= 44.2 kg/mol

Mass fraction of the fuel components can now be calculated as :

$m_{C_3H_8} = \frac{30.8}{44.2} \\ \\ m_{C_3H_8} = 0.7 \\ \\ \\ m_{C_4H_{10}} = \frac[2.9}{44.2} \\ \\ m_{C_4H_{10}} = 0.06 \\ \\ \\ m_{C_3H_6} = \frac{10.5}{44.2} \\ \\ m_{C_3H_6} = 0.24$

Finally; calculating the higher heating values $(HHV)_f$ per unit mass of LPG; we have:

$(HHV)_f=(0.7 * HHV_{C_3H_8}) + (0.06 *HHV_{C_4H_{10}})+(0.24*HHV_{C_3H_6} \\ \\ (HHV)_f=(0.7*50.38)+(0.06*49.56)+(0.24*48.95) \\ \\ (HHV)_f=49.9876 \ MJ/kg$

calculating the lower heating values $(LHV)_f$ per unit mass of LPG; we have:

$(LHV)_f = (HHV)_f - \delta H_w \\ \\ (LHV)_f = (HHV)_f - [\frac{m_w}{m_f}h_{vap}] \\ \\ (LHV)_f = 49.9876 \ MJ/kg - [\frac{3.8*18}{44.2}*2.258 \ MJ/kg] \\ \\ (LHV)_f = 46.4933 \ M/kg$

7 0

3 years ago

the enthalpy of vaporization of an roganic alchol is 35.3 kJ/mol at the boiling pointof 64.2 C calculate the entropy change for

bonufazy [111]

Answer:

Explanation:

s=h/t

=35.3/64.2(c)

=35.3/337.2(k)

=0.104

3 0

2 years ago