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qaws [65]
4 years ago
7

How is hydrogen in heavy water different from hydrogen in normal water?

Chemistry
1 answer:
Neko [114]4 years ago
3 0
Hydrogen in heavy water is an isotope of Hydrogen (Deuterium)
so, it's different
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Casey remembers that molecular motion increases as temperature increases. Identify what effect increased molecular motion.
Anastaziya [24]

D

Molecular motion increases as temperature increases and therefore increasing the molecular motion increases collision frequency.

Explanation:

When a substance is heated, its molecules gain increased kinetic energy even as the substance increases in temperatures. With higher kinetic energy, this means the molecules are moving more rapidly and hence collision between the molecules will be more frequent and of higher energy. It is these high energy collisions that cause the molecules to move further away from each other with increased temperatures (why substances expand).  

Learn More:

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#LearnWithBrainly

3 0
3 years ago
Read 2 more answers
How would you define Acid
Zepler [3.9K]

Answer:

a chemical substance that neutralizes alkalis, dissolves some metals, and turns litmus red; typically, a corrosive or sour-tasting liquid of this kind.

Explanation:

8 0
3 years ago
A 5 g sample of lead (specific heat 0.129 /g˚C) is heated, then put in a calorimeter with 50 mL of water (specific heat 4.184 J/
Svetach [21]

Answer:

670.68°C

Explanation:

Given that:

volume of water = 50 ml but 1 g = 1 ml. Therefore the mass of water (m) = 50 ml × 1 g / ml = 50 g

specific heat (C) = 4.184 J/g˚C

Initial temperature = 20°C, final temperature = 22°C. Therefore the temperature change ΔT = final temperature - initial temperature = 22 - 20 = 2°C

The quantity of heat (Q) used to raise the temperature of a body is given by the equation:

Q = mCΔT

Substituting values:

Q = 50 g × 4.184 J/g˚C × 2°C = 418.4 J

Since the mass of lead = 5 g and specific heat = 0.129 J/g˚C. The heat used to raise the temperature of water is the same heat used to raise the temperature of lead.

-Q = mCΔT

-418.4 J = 5 g × 0.129 J/g˚C × ΔT

ΔT = -418.4 J / ( 5 g × 0.129 J/g˚C) = -648 .68°C

temperature change ΔT = final temperature - initial temperature

- 648 .68°C = 22°C - Initial Temperature

Initial Temperature = 22 + 648.68 = 670.68°C

4 0
3 years ago
A+common+iv+solution+is+0.9%+saline+(nacl+solution).+what+is+the+osmolarity+of+0.9%+saline+mosmoles/l?+the+molecular+weight+of+n
Vedmedyk [2.9K]

An osmolarity of saline solution is 308 mosmol/L.

m(NaCl) = 9 g; the mass of sodium chloride

V(solution) = 1 L; the volume of the saline solution

n(NaCl) = 9 g ÷ 58.44 g/mol

n(NaCl) = 0.155 mol; the amount of sodium chloride

number of ions = 2

Osmotic concentration (osmolarity) is a measure of how many osmoles of particles of solute it contains per liter.

The osmolarity = n(NaCl) ÷ V(solution)  × 2

The osmolarity = 0.154 mol ÷ 1 L × 2

The osmolarity = 0.154 mol/L × 1000 mmol/m × 2

The osmolarity of the saline solution = 308 mosm/L.

More about osmolarity: brainly.com/question/13258879

#SPJ4

8 0
1 year ago
When 40.0 mL of 0.200 M HCl at 21.5°C is added to 40.0 mL of 0.200 M NaOH also at 21.5°C in a coffee-cup calorimeter, the temper
MakcuM [25]

Answer:

The enthalpy change for the reaction is ΔH = - 54.3 kJ/mol

Explanation:

The reaction between HCl and NaOH is a neutralization reaction:

HCl + NaOH \rightarrow NaCl + H2O

Heat released during neutralization = Heat gained by water

i.e. \qrxn = -\q(solution)-----(1)

where:

\q(solution) = mc\Delta T-----(2)

m = total mass of solution

m = density*total\ volume = 1.00g/ml*(40.00+40.00)ml = 80.00\ g

ΔT = change in temperature = 22.8 - 21.5 = 1.3 C

c = specific heat = 4.18 J/g C

\q(solution) = 80.00g*4.18J/gC*1.3C = 434.7 J

As per equation (1): qrxn = -434.7 J

The reaction enthalpy ΔH is the heat released per mole of acid (or base)

Moles\ of\ HCl = V(HCl) * M(HCl) = 0.040 L*0.200moles/L = 0.008\ moles

\Delta Hrxn = \frac{q}{mole}=\frac{-434.7J}{0.008mole}=-54337 J/mol=-54.3 kJ/mol

7 0
3 years ago
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