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ss7ja [257]
2 years ago
6

A geochemist in the field takes a 36.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X.

He notes the temperature of the pool, 170 C, and caps the sample carefully. Back in the lab, the geochemist first dilutes the sample with distilled water to 500. mL. Then he filters it and evaporates all the water under vacuum. Crystals of X are left behind. The researcher washes, dries and weighs the crystals. They weigh 3.96 g Using only the information above, can you calculate yes the solubility of X in water at 17.0 C? If you said yes, calculate it. Be sure your answer has a unit symbol and 3 O no significant digits.
Chemistry
1 answer:
Ostrovityanka [42]2 years ago
4 0

Answer:

The solubility of X in water at 17°C is 0.110 g/mL.

Explanation:

The water of a rock pool lined with mineral crystals is a <em>saturated solution</em> of said mineral, this means the concentration of X in those 36 mL is the solubility of compound X in water at 17 °C.

  • This means<u> it is possible to calculate said solubility</u>.

The dilution of the sample is not relevant, nor is that 500 mL volume. What's important is that 3.96 g of X form a saturated solution with 36.0 mL of water, so the solubility is:

  • 3.96 g / 36.0 mL = 0.110 g/mL
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When 50 ml (50 g) of 1.00 m hcl at 22oc is added to 50 ml (50 g) of 1.00 m naoh at 22oc in a coffee cup calorimeter, the tempera
vitfil [10]

Answer:

\boxed{\text{2700 J}}

Explanation:

HCl + NaOH ⟶ NaCl + H₂O

There are two energy flows in this reaction.  

Heat of reaction + heat to warm water = 0  

           q₁             +              q₂                 = 0  

           q₁             +          mCΔT              = 0  

Data:

    m(HCl) = 50 g

m(NaOH) = 50 g

           T₁ = 22       °C

          T₂ = 28.87 °C

           C = 4.18 J·°C⁻¹g⁻¹

Calculations:

 m = 50 + 50 = 100 g

ΔT = 28.87 – 22 = 6.9 °C

 q₂ = 100 × 4.18 × 6.9 = 2900 J

q₁ + 2900 = 0

q₁ = -2900 J

The negative sign tells us that the reaction produced heat.

The reaction produced \boxed{\textbf{2900 J}}.

7 0
2 years ago
calculate the specific heat capacity for gold n 105 joules are required to heat 30.0 grams of gold from 27.7c to 54.9c
Pepsi [2]

<u>Answer:</u>

<em>The specific heat capacity for gold in 105 joules which are required to heat 30.0 grams of gold is 0.129 J/(g℃)</em>

<u>Explanation:</u>

We make use of the formula

Q=m \times c \times \Delta T

where

∆T = final T - initial T

= 54.9℃ - 27.7℃ = 27.2℃

Q is the heat energy in Joules = 105J

c is the specific heat capacity = ?

m is the mass of Gold = 30.0g

Q=m \times c \times \Delta T

Rearranging the formula

c= \frac {Q}{(m\times \Delta T)}

= \frac {105J}{(30.0g \times 27.2 ^\circ{C})}\\\\= \frac {105J}{(816g^\circ{C})}

So,

c = 0.129 J/(g℃)

(Answer)

7 0
2 years ago
Calculate the wavelength of the photon emitted when an electron makes a transition from n=6 to n=3. You can make use of the foll
Angelina_Jolie [31]

<u>Answer:</u> The wavelength of light is 1.094\times 10^{-6}m

<u>Explanation:</u>

To calculate the wavelength of light, we use Rydberg's Equation:

\frac{1}{\lambda}=R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2} \right )

Where,

\lambda = Wavelength of radiation

R_H = Rydberg's Constant  = 1.097\times 10^7m^{-1}

n_f = Final energy level = 3

n_i = Initial energy level = 6

Putting the values in above equation, we get:

\frac{1}{\lambda }=1.097\times 10^7m^{-1}\left(\frac{1}{3^2}-\frac{1}{6^2} \right )\\\\\lambda =\frac{1}{914617m^{-1}}=1.094\times 10^{-6}m

Hence, the wavelength of light is 1.094\times 10^{-6}m

6 0
2 years ago
In a car piston shown above, the pressure of the compressed gas (red) is 5.00 atm. If the area of the piston is 0.0760 m^2. What
Shkiper50 [21]

Answer:

38503.5N

Explanation:

Data obtained from the question include:

P (pressure) = 5.00 atm

Now, we need to convert 5atm to a number in N/m2 in order to obtain the desired result of force in Newton (N). This is illustrated below:

1 atm = 101325N/m2

5 atm = 5 x 101325 = 506625N/m^2

A (area of piston) = 0.0760 m^2

Pressure is force per unit area. Mathematically it is written as

P = F/A

F = P x A

F = 506625 x 0.0760

F = 38503.5N

Therefore, the force exerted on the piston is 38503.5N

6 0
2 years ago
C6H12O6+6O2→6CO2+6H2O
babunello [35]

Answer:

<em>The answer that is true is </em><em>A</em>

Explanation:

The first thing to do is to perform stoichiometry relationships and calculations.

<em>First relationship</em><em>: 0.3 moles of CO₂ are produced </em>

6 mol CO₂ _____ 1 mol C₆H₁₂O₆

0.3 mol CO₂ _____ X = <em>0.05 mol of C₆H₁₂O₆</em>

Calculation: 0.3 mol x 1 mol / 6 mol = 0.05 mol of C₆H₁₂O₆

<em>Second relationship:</em><em> 0.60 moles of CO₂ are produced </em>

6 mol CO₂ _____ 1 mol C₆H₁₂O₆

0.6 mol CO₂_____ X = <em>0.1 mol of C₆H₁₂O₆ </em>

Calculation: 0.6 mol x 1 mol / 6 mol = 0.1 mol of C₆H₁₂O₆

6 mol CO₂ _____ 6 mol O₂

0.6 mol CO₂ _____ X =<em> 0.6 mol of O₂</em>

Calculation: 0.6 mol x 6 mol / 6 mol = 0.6 mol of O₂

Then, the answer that is true is A, because in order for 0.6 moles of CO₂ to be produced, there must initially be 0.1 mol of C6H12O6 as calculated.

The B is not correct because for more CO₂ to be produced, more C₆H₁₂O₆ must react, this is defined according to <em>Proust's law</em> or<em> law of defined proportions</em>, stated as: “<u><em>When the combination of two or more elements occurs to form a new compound, they always do it in a mass relationship that is completely constant</em></u>”. It is by this law that no more CO₂ can be produced with the same amount of C₆H₁₂O₆ reagent.

The C is not correct either, although the temperature and volume of the container are the same, the O₂ is in excess to react, so 0.6 mol of it is needed to produce 0.6 mol of CO₂, according to the calculations made .

The D is not correct because it would also not respect the <em>law of defined proportions</em> or <em>Proust's law.</em> Although O₂ is in excess, it cannot react more than the stoichiometric ratio. Simply, if it is in excess, the rest of the O₂ will remain unreacted.

6 0
2 years ago
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