When solving mass to mass stoichiometry problems, where do you get the molar ratio for the first and second chemicals?

You measure salt water in a tank to have a density of 1.02 g/mL. A balloon weighs 2.0 g and you weights have a mass of 30.0 g ea

Elden [556K]

Weight of the balloon = 2.0 g

Six weights each of mass 30.0 g is added to the balloon.

Total mass of the balloon = 2.0 g + 6*30.0 g = 182 g

Density of salt water = 1.02 g/mL

Calculating the volume from mass and density:

$182g*\frac{mL}{1.02g} =178mL$

Converting the volume from mL to cubic cm:

$178 mL * \frac{1cm^{3} }{1mL} =178cm^{3}$

Assuming the balloon to be a sphere,

Volume of the sphere = $\frac{4}{3}$ π $r^{3}$

$178 cm^{3} = \frac{4}{3}(\frac{22}{7})r^{3}$

r = 3.49 cm

Radius of the balloon = 3.49 cm

Diameter of the balloon = 2 r = 2*3.49 cm = 6.98cm

4 0

3 years ago

A 44.0 g sample of an unknown metal at 99.0 oC was placed in a constant-pressure calorimeter of negligible heat capacity contain

tatiyna

Answer:

$C_m=0.474\frac{J}{g\°C}$

Explanation:

Hello.

In this case, since this is a system in which the water is heated up and the metal is cooled down in a calorimeter which is not affected by the heat lose-gain process, we can infer that the heat lost by the metal is gained be water, it means that we can write:

$Q_m=-Q_w$

Thus, in terms of masses, specific heats and temperatures we can write:

$m_mC_m(T_{eq}-T_m)=-m_wC_w(T_{eq}-T_w)$

Whereas the equilibrium temperature is the given final temperature of 28.4 °C and we can compute the specific heat of the metal as shown below:

$C_m=\frac{-m_wC_w(T_{eq}-T_w)}{m_m(T_{eq}-T_m)}$

Plugging the values in and since the density of water is 1.00 g/mL so the mass is 80.0g, we obtain:

$C_m=\frac{-80.0g*4.184\frac{J}{g\°C} (28.4\°C-24.0\°C)}{44.0g(28.4\°C-99.0\°C)}\\\\C_m=0.474\frac{J}{g\°C}$

Best regards!

6 0

3 years ago

How Many Moles Of HCl Need To Be Added To 150.0 ML Of 0.50 M NaZ To Have A Solution With A PH Of 6.50

Aleks04 [339]

The number of mole of HCl needed for the solution is 1.035×10¯³ mole

<h3>How to determine the pKa</h3>

We'll begin by calculating the pKa of the solution. This can be obtained as follow:

Equilibrium constant (Ka) = 2.3×10¯⁵
pKa =?

pKa = –Log Ka

pKa = –Log 2.3×10¯⁵

pKa = 4.64

<h3>How to determine the molarity of HCl </h3>

pKa = 4.64
pH = 6.5
Molarity of salt [NaZ] = 0.5 M
Molarity of HCl [HCl] =?

pH = pKa + Log[salt]/[acid]

6.5 = 4.64 + Log[0.5]/[HCl]

Collect like terms

6.5 – 4.64 = Log[0.5]/[HCl]

1.86 = Log[0.5]/[HCl]

Take the anti-log

0.5 / [HCl] = anti-log 1.86

0.5 / [HCl] = 72.44

Cross multiply

0.5 = [HCl] × 72.44

Divide both side by 72.44

[HCl] = 0.5 / 72.4

[HCl] = 0.0069 M

<h3>How to determine the mole of HCl </h3>

Molarity of HCl = 0.0069 M
Volume = 150 mL = 150 / 1000 = 0.15 L

Mole of HCl =?

Mole = Molarity x Volume

Mole of HCl = 0.0069 × 0.15

Mole of HCl = 1.035×10¯³ mole

<h3>Complete question</h3>

How many moles of HCl need to be added to 150.0 mL of 0.50 M NaZ to have a solution with a pH of 6.50? (Ka of HZ is 2.3 x 10 -5 .) Assume negligible volume of the HCl

Learn more about pH of buffer:

brainly.com/question/21881762

7 0

2 years ago

KFell Fe"(CN), + e + Nat → KNaFe'Fe(CN)6

Alinara [238K]

Answer:

Most common oxidation states: +2, +3

M.P. 1535º

B.P. 2750º

Density 7.87 g/cm3

Characteristics: Iron is a gray, moderately active metal.

Characteristic reactions of Fe²⁺ and Fe³⁺

The [Fe(H2O)6]3+ ion is colorless (or pale pink), but many solutions containing this ion are yellow or amber-colored because of hydrolysis. Iron in both oxidation states forms many complex ions.

Aqueous Ammonia

Aqueous ammonia reacts with Fe(II) ions to produce white gelatinous Fe(OH)2, which oxidizes to form red-brown Fe(OH)3:

Fe2+(aq)+2NH3(aq)+3H2O(l)↽−−⇀Fe(OH)2(s)+2NH+4(aq)(1)

Fe3appt.gif

Aqueous ammonia reacts with Fe(III) ions to produce red-brown Fe(OH)3:

Fe3+(aq)+3NH3(aq)+3H2O(l)↽−−⇀Fe(OH)3(s)+3NH+4(aq)(2)

Fe3bppt.gif

Both precipitates are insoluble in excess aqueous ammonia. Iron(II) hydroxide quickly oxidizes to Fe(OH)3 in the presence of air or other oxidizing agents.

Sodium Hydroxide

Sodium hydroxide also produces Fe(OH)2 and Fe(OH)3 from the corresponding oxidation states of iron in aqueous solution.

Fe2+(aq)+2OH−(aq)↽−−⇀Fe(OH)2(s)(3)

Fe4appt.gif

Fe3+(aq)+3OH−(aq)↽−−⇀Fe(OH)3(s)(4)

Fe4bppt.gif

Neither hydroxide precipitate dissolves in excess sodium hydroxide.

Potassium Ferrocyanide

Potassium ferrocyanide will react with Fe3+ solution to produce a dark blue precipitate called Prussian blue:

K+(aq)+Fe3+(aq)+[Fe(CN)6]4−(aq)↽−−⇀KFe[Fe(CN)6](s)(5)

Fe5a1ppt.gif

With Fe2+ solution, a white precipitate will be formed that will be converted to blue due to the oxidation by oxygen in air:

2Fe2+(aq)+[Fe(CN)6]4−(aq)↽−−⇀Fe2[Fe(CN)6](s)(6)

Fe5a2ppt.gif

Many metal ions form ferrocyanide precipitates, so potassium ferrocyanide is not a good reagent for separating metal ions. It is used more commonly as a confirmatory test.

Potassium Ferricyanide

Potassium ferricyanide will give a brown coloration but no precipitate with Fe3+. With Fe2+, a dark blue precipitate is formed. Although this precipitate is known as Turnbull's blue, it is identical with Prussian blue (from Equation 5).

K+(aq)+Fe+2(aq)+[Fe(CN)6]3−(aq)↽−−⇀KFe[Fe(CN)6](s)(7)

Fe5b.gif

Potassium Thiocyanate

KSCN will give a deep red coloration to solutions containing Fe3+:

Fe+3(aq)+NCS−(aq)↽−−⇀[FeNCS]+2(aq)(8)

Fe5cppt.gif

No Reaction

Cl−, SO2−4

7 0

3 years ago

A sample of methane gas has a volume of 355 mL at 25.0°C and 2414 mm Hg. If the temperature is raised to 100.°C and the pressure

Drupady [299]

Answer:

23019

Explanation:

5 0

3 years ago

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