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mote1985 [20]
3 years ago
8

A chef uses a large aluminum spoon with a mass of 38.0 g to stir a pot of boiling water. As a result, the spoon’s temperature in

creases by 3.60°C. Consider that the specific heat capacity of aluminum is 0.90 J/g–°C. How much heat has the spoon absorbed? For credit, you must show all steps of your calculations.
Chemistry
1 answer:
Setler [38]3 years ago
7 0

Answer:

12,312 J or 0.12312 kJ

Explanation:

q = mCT

q = J of heat  

m = mass

C = heat capacity

T = temp change

q = 38 x 0.90 x 3.6

q = 12312 J

q = 0.12312 kJ

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Now that Snape and Dumbledore has taught you the finer points of hydration calculations they have a slightly more challenging pr
Mrrafil [7]

Answer:

The value of an integer x in the hydrate is 10.

Explanation:

Molarity=\frac{Moles}{Volume(L)}

Molarity of the solution = 0.0366 M

Volume of the solution = 5.00 L

Moles of hydrated sodium carbonate = n

0.0366 M=\frac{n}{5.00 L}

n=0.0366 M\times 5 mol=0.183 mol

Mass of hydrated sodium carbonate = n= 52.2 g

Molar mass of hydrated sodium carbonate = 106 g/mol+x18 g/mol

n=\frac{\text{mass of Compound}}{\text{molar mass of compound}}

0.183 mol=\frac{52.2 g}{106 g/mol+x\times 18 g/mol}

106 g/mol+x\times 18 g/mol=\frac{52.2 g}{0.183 mol}

Solving for x, we get:

x = 9.95 ≈ 10

The value of an integer x in the hydrate is 10.

6 0
3 years ago
200 grams of an organic sample which contains only carbon, hydrogen, and oxygen is analyzed and found to contain 97.30 grams of
ipn [44]
The empirical formula for 200 grams of an organic sample which contains carbon,hydrogen, and oxygen is C3H6O

3 0
3 years ago
Read 2 more answers
While heating up a 25 gram sample of concrete (specific heat = 0.210-cal/g°C), your initial tempărature is room temperature (25°
Lana71 [14]

Answer:

Final temperature  = 83.1 °C

Explanation:

Given data:

Mass of concrete = 25 g

Specific heat capacity = 0.210 cal/g. °C

Initial temperature = 25°C

Calories gain = 305 cal

Final temperature = ?

Solution:

Q = m. c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

305 cal = 25 g ×0.210 cal/g.°C × T2 -  25°C

305 cal = 5.25cal/°C × T2 -  25°C

305 cal / 5.25cal/°C = T2 -  25°C

58.1 °C = T2 -  25°C

T2 = 58.1 °C + 25°C

T2 = 83.1 °C

7 0
3 years ago
What is the product of the reaction of hydrobromic acid and 2-bromo-1-butene in the presence of acid and ether?
sergejj [24]

Answer: The major product of the reaction between Hydrobromic Acid and 2-bromo-1-butene in the presence of ether and acid is 2,2-dibromobutane.

Explanation:

The mechanism of the reaction is supported by the Markovnikov's rule which explains that in the addition reaction of alkenes by hydrogen-halogen compounds, the incoming halogen substituent goes to the more substituted Carbon. It can also be stated that incoming hydrogen atom goes to the Carbon with more Hydrogen atoms.

The only case when the reverse of Markovnikov's rule takes place is when Hydrogen peroxide is present and the addition reagent is HBr.

This case is not like that and it simply follows the Markovnikov's rule.

I'll add an attachment of the reaction to this now.

6 0
4 years ago
Read 2 more answers
opper(I) ions in aqueous solution react with NH 3 ( aq ) according to Cu + ( aq ) + 2 NH 3 ( aq ) ⟶ Cu ( NH 3 ) + 2 ( aq ) K f =
Y_Kistochka [10]

Answer:

55.373g/l

Explanation:

The dissolved amount of sparingly soluble salts is interlinked with a unitlesss quantity called as solubility product. It is a fixed quantity that only increases with the rise in temperatures and is used to predict the salting out of compounds. If the value of ionic product (Q) is larger than (Ksp), precipitation of compound occurs.

Given:

The solubility product of CuBr is 6.3×10−9.

The concentration of NH3 is 0.10 M.

Formula and Calculations:

The dissolution reaction (I) of CuBr is shown below.

The reaction showing dissolution of CuBr in NH3 is shown below.

The above reaction can be obtained by adding reaction (I) and (II) as shown below.

The equilibrium constants will get multiplied.

Suppose the solubility of CuBr is “s”.

It is given that concentration of NH3 is 0.10 M.

The equilibrium constant expression for the above reaction is as follows,

Here,

The concentration of pure solids is 1 M. Thus, the concentration of CuBr is 1 M.

As calculated, the value of Ksp is 396.9.

Substitute all the required values in above formula.  

On further solving above equation,

Therefore, the solubility of CuBr in ammonia is 0.386 M.

The formula to calculate solubility

Solubuility (g/l)= Molarity(M) x Molarmass

Chemistry homework question answer, step 2, image 10

The molar mass of CuBr is 143.45 g/mol.

The formula to calculate solubility in g/L is given below.

The molar mass of CuBr is 143.45 g/mol.

therefore,

solubility = 0.386M x 143.45g/mol

where (M = mol/l)

solubility = 55.375g/l

5 0
3 years ago
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