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ololo11 [35]
3 years ago
8

A student attempts to measure the specific heat capacity of an unknown liquid through repeated trials. She measures its specific

heat capacity, in J/g•°C, as 2.14, 2.11, 2.13, 2.12, and 2.11. The specific heat capacity of the liquid should be recorded as -
Chemistry
1 answer:
Goryan [66]3 years ago
4 0

Answer:

2.13

Explanation:

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The scientific method is one application of critical thinking.true or false
kumpel [21]
The best answer between the two choices would be the first option TRUE because the scientific method is used to do more advance research and investigation on things.
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3 years ago
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Help please! I'd appreciate it 
babymother [125]

Answer:

16.56 g

Explanation:

Mass is the production of Volume and Density.

m = V. d = 6 × 2.76 = 16.56 g

6 0
3 years ago
A 25.0 mL sample of sulfuric acid is completely neutralized by adding 32.8 mL of 0.116 mol/L ammonia solution. Ammonium sulfate
Paul [167]

Answer:

0.08 mol L-1

Explanation:

Sulfuric acid Formula: H2SO4

Ammonia Formula: NH3

Ammonium sulfate Formula: (NH₄)₂SO₄

H2SO4 + 2NH3 = 2NH4+ + SO4 2-

H2SO4 + 2NH3 = (NH₄)₂SO₄

H2SO4 = (1/2)x (32.8 x 10^-3 L x 0.116 mol L-1)/25 x 10^-3 L

= 0.08 mol L-1

7 0
3 years ago
What do melted ice cooling lava and boiling water have in common?
KIM [24]
They are all transioning in states of matter
7 0
3 years ago
Calculate the standard reaction enthalpy for the reaction NO2(g) → NO(g) + O(g) given +142.7 kJ/mol for the standard enthalpy of
bulgar [2K]

Answer:

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

Explanation:

O_2(g) \rightarrow \frac{2}{3}O_3(g),\Delta H^o_{1}=142.7 kJ/mol..[1]

O_2(g) \rightarrow 2 O(g),\Delta H^o_{2}=498.4 kJ/mol..[2]

NO(g) + O_3(g)\rightarrow NO_2(g) + O_2(g) ,\Delta H^o_{3} = -200 kJ/mol..[3]

NO_2(g)\rightarrow NO(g) + O(g),\Delta H^o_{4}=?..[4]

Using Hess's law:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

2 × [4] = [2]- (3 ) × [1] - (2) × [3]

2\times \Delta H^o_{4}=\Delta H^o_{2} -3\times \Delta H^o_{1}-2\times \Delta H^o_{3}

2\times \Delta H^o_{4}=498.4 kJ/mol-3\times 142.7 kJ/mol-2\times -200 kJ/mol

2\times \Delta H^o_{4}=470.3 kJ/mol

\Delta H^o_{4}=\frac{470.3 kJ/mol}{2}=235.15 kJ/mol

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

7 0
3 years ago
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