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Lelu [443]
2 years ago
11

A substance has a specific heat capacity of 2.0 J/g °C. How many joules are needed to raise the temperature of 30.0 g of this su

bstance from 0°C to15°C?
a. 1 J

b. -1 J

c. -900 J

d. 900 J
Chemistry
2 answers:
Vesnalui [34]2 years ago
7 0

Answer:

\Large \boxed{\sf 900 \  J}

Explanation:

Use formula

\displaystyle \sf Energy\ (J)=mass \ (g) \times specific \ heat \ capacity \ (Jg^{-1}\°C^{-1}) \times change \ in \ temperature \ (\°C)

Substitute the values in formula and evaluate

\displaystyle \sf Energy\ (J)=30.0\ g \times 2.0 \ Jg^{-1}\°C^{-1} \times (15\°C-0\°C)

\displaystyle Q=30.0 \times 2.0 \times (15-0  )=900

3241004551 [841]2 years ago
5 0

Answer:

g

Explanation:

know what you think about th can get it

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The reaction between alcohols and carboxylic acids is called
aleksandrvk [35]

Explanation:

When a carboxylic acid is treated with an alcohol and an acid catalyst, an ester is formed (along with water). This reaction is called the Fischer esterification.

4 0
2 years ago
When salt is dissolved in water, the salt particles are referred to as the _____.
Sergio039 [100]
The salt is a solid compound and is considered the "Solute" of the solution.
5 0
3 years ago
Read 2 more answers
C3H8(g) + 5O2(g) ⟶ 3CO2(g) + 4H2O(g) H = -2220 kJ If 865.9 g of H2O is produced during this combustion, how much heat is generat
dem82 [27]

Answer:

3 × 10⁴ kJ

Explanation:

Step 1: Write the balanced thermochemical equation

C₃H₈(g) + 5 O₂(g) ⟶ 3 CO₂(g) + 4 H₂O(g) ΔH = -2220 kJ

Step 2: Calculate the moles corresponding to 865.9 g of H₂O

The molar mass of H₂O is 18.02 g/mol.

865.9 g × 1 mol/18.02 g = 48.05 mol

Step 3: Calculate the heat produced when 48.05 moles of H₂O are produced

According to the thermochemical equation, 2220 kJ of heat are evolved when 4 moles of H₂O are produced.

48.05 mol × 2220 kJ/4 mol = 2.667 × 10⁴ kJ ≈ 3 × 10⁴ kJ

5 0
2 years ago
onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency
mestny [16]

Answer : The excess reactant is, H_2O

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of CaCN_2 = 105.0 g

Mass of H_2O = 78.0 g

Molar mass of CaCN_2 = 80.11 g/mole

Molar mass of H_2O = 18 g/mole

Molar mass of CaCO_3 = 100.09 g/mole

First we have to calculate the moles of CaCN_2 and H_2O.

\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles

\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3

From the balanced reaction we conclude that

As, 1 mole of CaCN_2 react with 3 mole of H_2O

So, 1.31 moles of CaCN_2 react with 1.31\times 3=3.93 moles of H_2O

From this we conclude that, H_2O is an excess reagent because the given moles are greater than the required moles and CaCN_2 is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)

\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0
3 years ago
Calculate e°cell for a silver-aluminum cell in which the cell reaction is al(s) + 3ag+(aq) → al3+(aq) + 3ag(s) –2.46 v 0.86 v –0
tekilochka [14]
When E° cell is an electrochemical cell which comprises of two half cells.
 
So,

when we have the balanced equation of this half cell :

Al3+(aq) + 3e- → Al(s)   and E°1 = -1.66 V 

and we have  also this balanced equation of this half cell :

Ag+(aq)  + e- → Ag(s)  and E°2 = 0.8 V 

so, we can get E° in Al(s) + 3Ag (aq) → Al3+(aq) + 3Ag(s)

when E° = E°2 - E°1

∴E° =0.8 - (-1.66)

      = 2.46 V

∴ the correct answer is 2.46 V




6 0
3 years ago
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