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Ostrovityanka [42]

4 years ago

11

Galileo discovered four of Jupiter's moons. They are Io, Europa, Callisto, and _____, which is the largest moon in the solar sys

tem. A. Titan B. Triton C. Phobos D. Ganymede

1 answer:

Ganezh [65]4 years ago

7 0

Answer:

D. Ganymede

I hope this helps!

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3 Cu + SHNO3 — 3 Cu(NO3)2 + 2 NO + 4 H20

LenaWriter [7]

Considering the reaction stoichiometry, the mass of H₂O that is produced when 11.9 moles of HNO₃ react is 107.1 grams.

<h3>Reaction stoichiometry</h3>

The balanced reaction is:

3 Cu + 8 HNO₃ → 3 Cu(NO₃)₂ + 2 NO + 4 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Cu: 3 moles
HNO₃: 8 moles
Cu(NO₃)₂: 3 moles
NO: 2 moles
H₂O: 4 moles

The molar mass of the compounds present in the reaction is:

Cu: 63.54 g/mole
HNO₃: 63 g/mole
Cu(NO₃)₂: 187.54 g/mole
NO: 30 g/mole
H₂O: 18 g/mole

Then, by reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of mass of each compound participate in the reaction:

Cu: 3 moles ×63.54 g/mole= 190.62 grams

HNO₃: 8 moles ×63 g/mole= 504 grams

Cu(NO₃)₂: 3 moles ×187.54 g/mole= 562.62 grams

NO: 2 moles ×30 g/mole= 60 grams

H₂O: 4 moles ×18 g/mole= 72 grams

<h3>Mass of H₂O produced</h3>

It is possible to determine the the amount of mass of H₂O produced by a rule of three: if by stoichiometry 8 moles of HNO₃ produce 72 grams of H₂O, if 11.9 moles of HNO₃ react how much mass of H₂O will be formed?

$mass of H_{2}O=\frac{11.9 moles of HNO_{3} x72 grams of H_{2}O}{8 moles of HNO_{3}}$

<u><em>mass of H₂O= 107.1 grams</em></u>

In summary, the mass of H₂O that is produced when 11.9 moles of HNO₃ react is 107.1 grams.

Learn more about reaction stoichiometry:

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3 0

2 years ago

The tabulated data were collected for this reaction:

erastovalidia [21]

Answer:

ai) Rate law, $Rate = k [CH_3 Cl] [Cl_2]^{0.5}$

aii) Rate constant, k = 1.25

b) Overall order of reaction = 1.5

Explanation:

Equation of Reaction:

$CH_{3} Cl (g) + 3 Cl_2 (g) \rightarrow CCl_4 (g) + 3 HCl (g)$

If $A + B \rightarrow C + D$ , the rate of backward reaction is given by:

$Rate = k [A]^{a} [B]^{b}\\k = \frac{Rate}{ [A]^{a} [B]^{b}}\\k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}$

k is constant for all the stages

Using the information provided in lines 1 and 2 of the table:

$0.014 / [0.05]^a [0.05]^b = 00.029/ [0.100]^a [0.05]^b\\0.014 / [0.05]^a [0.05]^b = 00.029/ [2*0.05]^a [0.05]^b\\0.014 / = 0.029/ 2^a\\2^a = 2.07\\a = 1$

Using the information provided in lines 3 and 4 of the table and insering the value of a:

$0.041 / [0.100]^a [0.100]^b = 0.115 / [0.200]^a [0.200]^b\\0.041 / [0.100]^a [0.100]^b = 0.115 / [2 * 0.100]^a [2 * 0.100]^b\\$

$0.041 = 0.115 / [2 ]^a [2]^b\\ \[[2 ]^a [2]^b = 0.115/0.041\\ \[[2 ]^a [2]^b = 2.80\\\[[2 ]^1 [2]^b = 2.80\\\[[2]^b = 1.40\\b = \frac{ln 1.4}{ln 2} \\b = 0.5$

The rate law is: $Rate = k [CH_3 Cl] [Cl_2]^{0.5}$

The rate constant $k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}$ then becomes:

$k = 0.014 / ( [0.050] [0.050]^(0.5) )\\k = 1.25$

b) Overall order of reaction = a + b

Overall order of reaction = 1 + 0.5

Overall order of reaction = 1.5

3 0

3 years ago

What is the entropy change for freezing 2.71 g of C2H5OH at 158.7 K? ∆H = −4600 J/mol. Answer in units of J/K.

Ira Lisetskai [31]

Answer:

-1.71 J/K

Explanation:

To solve this problem we use the formula

ΔS = n*ΔH/T

Where n is mol, ΔH is enthalpy and T is temperature.

ΔH and T are already given by the problem, so now we calculate n:

Molar Mass C₂H₅OH = 46 g/mol

2.71 g C₂H₅OH ÷ 46g/mol = 0.0589 mol

Now we calculate ΔS:

ΔS = 0.0589 mol * −4600 J/mol / 158.7 K

ΔS = -1.71 J/K

4 0

3 years ago

The chemical formula of aspirin is C9H8O4. What is the mass of 0.40 mol<br> of aspirin?

dybincka [34]

Answer:

$\boxed {\boxed {\sf About \ 72 \ grams }}$

Explanation:

To convert from moles to grams, we must use the molar mass (also known as the gram formula mass).

First, look up the molar masses of the elements in the formula.

C: 12.011 g/mol
H: 1.008 g/mol
O: 15.999 g/mol

Next, multiply by the subscript, because it tells us the number of atoms of each element in the formula.

C₉: 4(12.011 g/mol)= 108.099 g/mol
H₈: 8(1.008 g/mol)= 8.064 g/mol
O₄: 4(15.999 g/mol)= 63.996 g/mol

Add the values.

108.099 + 8.064+63.996=180.159 g/mol

Use this molar mass as a ratio.

$\frac {180.159 \ g \ C_9H_8O_4}{1 \ mol \ C_9H_8O_4}$

Multiply by the given number of moles, 0.40

$0.40 \ mol \ C_9H_8O_4 *\frac {180.159 \ g \ C_9H_8O_4}{1 \ mol \ C_9H_8O_4}$

The units moles of aspirin cancel.

$0.40 *\frac {180.159 \ g \ C_9H_8O_4}{1 }$

$72.0636 \ g \ C_9H_8O_4$

The original number of moles has 2 sig figs (4 and 0), so answer must have the same. For the number we calculated, it is the ones place. The 0 in the tenth place tells us to leave the 2.

$72 \ g$

5 0

3 years ago

The eu(iii) standards added to the groundwater sample contained 0.168 ng/ml (ppb) of eu(iii). calculate the concentration of eu(

Andre45 [30]

The standard addition equation is as followsI_(S+X) (V/V_O )=I_X+I_X/[X]_i [S]_4 (V_S/V_0 ) Here, [X]_i is the initial concentration of analyte, [S]_i is the initial concentration of standard, I_X is signal for analyte, I_(S+X) is signal for standard and analyte, V_0 is the initial volume, V_S is the added standard volume, and V is the total volume.Added volume of standard V_S is-23.3 mL. Initial volume of the sample V_0 is 10.00 mL. Initial concentration of standard ([S]_i) is 0.156 ng/mL.[X]_i= -[S]_i (V_S/V_0 )〖[X]〗_(i )= -(0.156 ng/mL)((-23.3 mL)/(10.00 mL))=0.363 ng/mL
Concentration of U(III) in ground sample is 0.363 ng/mL

6 0

3 years ago