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77julia77 [94]
3 years ago
11

1. How many joules must be added to 10.0 g of water to raise its temperature from 10°C to 15°C?

Chemistry
1 answer:
weqwewe [10]3 years ago
6 0

Answer:

209.3 Joules require to raise the temperature from 10 °C to 15 °C.

Explanation:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m × c × ΔT

Given data:

mass of water = 10 g

initial temperature T1= 10 °C

final temperature T2=  15 °C

temperature change =ΔT= T2-T1 = 15°C - 10°C = 5 °C

Energy or joules added to increase the temperature Q = ?

Solution:

We know that specific heat of water is 4.186 J/g .°C

Q = m × c × ΔT

Q = 10 g × 4.186 J/g .°C × 5 °C

Q = 209.3 J

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*A & B*

Answers A & B are not possible, as Hydrogen “bonds” are intermolecular forces and do not actually involve transfer or sharing of electrons.

*C & D*

Viscosity and surface tension are not the answer as they are not specific enough to the question.

*E*

Polarity of water molecules is the correct answer, as water molecules are highly polar. The partial positive of the Hydrogen on one water molecule is highly attracted to the partial negative of the Oxygen (due to its lone pairs) on another water molecule.
4 0
3 years ago
3. What methods are you using to test this hypothesis? Outline the steps of the procedure in full sentences.
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Answer:

AHYPOTHISES IS A GUESS

Explanation:

4 0
3 years ago
If 18.7ml of 0.01500M aqueous HCl is required to titrâtes 15.00ml of an aqueous solution of NaOH to the equivalence point, what
sweet [91]

Answer:

0.0187 M

Explanation:

Step 1: Write the balanced neutralization reaction

NaOH + HCl ⇒ NaCl + H₂O

Step 2: Calculate the reacting moles of HCl

18.7 mL of 0.01500 M HCl react.

0.0187 L × 0.01500 mol/L = 2.81 × 10⁻⁴ mol

Step 3: Calculate the reacting moles of NaOH

The molar ratio of HCl to NaOH is 1:1. The reacting moles of NaOH are 1/1 × 2.81 × 10⁻⁴ mol = 2.81 × 10⁻⁴ mol.

Step 4: Calculate the molarity of NaOH

2.81 × 10⁻⁴ moles are in 15.00 mL of NaOH.

[NaOH] = 2.81 × 10⁻⁴ mol/0.01500 L = 0.0187 M

6 0
3 years ago
The standard free-energy changes for the reactions below are given. Phosphocreatine → creatine + Pi ∆ G'° = –43.0 kJ/mol ATP → A
Jet001 [13]

Answer:

-12.5 kJ/mol

Explanation:

The free-energy predicts if a reaction is spontaneous or not. If it is, ΔG < 0. When a reaction happens by steps, the free-energy of the global reaction can be calculated by the sum of the free-energy of the steps (Hess law). If it's needed to operations at the reaction the same operation must be done in the value of ΔG (if the reaction is inverted, the signal of ΔG must be inverted).

Phosphocreatine → creatine + Pi ∆G'° = –43.0 kJ/mol

ATP → ADP + Pi                             ∆G'° = –30.5 kJ/mol (x-1)

--------------------------------------------------------------------------------------

Phosphocreatine → creatine + Pi ∆G'° = –43.0 kJ/mol

Pi + ADP → ATP                             ∆G'° = 30.5 kJ/mol

The bold compounds are in opposite sides, so they'll be canceled in the sum of the reactions:

Phosphocreatine + ADP → creatine + ATP

∆G'° = -43.0 + 30.5

∆G'° = -12.5 kJ/mol

5 0
3 years ago
What is the next atomic orbital in the series 1s 2p 3s 3p
liq [111]

Answer:

s an example, the ground state configuration of the sodium atom is 1s22s22p63s1, as deduced from the Aufbau principle (see below). The first excited state is obtained by promoting a 3s electron to the 3p orbital, to obtain the 1s22s22p63p1 configuration, abbreviated as the 3p level.

Explanation:

3 0
3 years ago
Read 2 more answers
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