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natulia [17]
4 years ago
11

The density of pure water is 1.0 g/ml. Four students measured and calculated the density of pure water samples. The results are

as follows: student 1 got 0.85
g/ml, student 2 got 0.94 g/ml, student 3 got 0.95 g/ml, and student 4 got 1.3 g/ml. Which student had the smallest error in their measurement?

A. Student 3
B. Student 2
C. Student 4
D. Student 1
Chemistry
1 answer:
balandron [24]4 years ago
5 0

Answer:

Student 4

Explanation:

Student 1: 15

student 2 : 6

Student 3: 5

Student 4: 3

P.S: Just guessed

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When 100 mL of 0.200 M NaCl(aq) and 100 mL of 0.200 M AgNO3(aq), both at 21.9 °C, are mixed in a coffee cup calorimeter, the tem
masya89 [10]

Answer:

There is 1.3 kJ heat produced(released)

Explanation:

<u>Step 1:</u> Data given

Volume of a 0.200 M Nacl solution = 100 mL = 0.1 L

Volume of a 0.200 M AgNO3 solution = 100 mL = 0.1 L

Initial temperature = 21.9 °C

Final temperature = 23.5 °C

Solid AgCl will be formed

<u>Step 2</u>: The balanced equation:

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

AgCl(s) + NaNO3(aq) → Na+(aq) + NO3-(aq) + AgCl(s)

<u>Step 3:</u> Define the formula

Pressure is constant.  → the heat evolved from the reaction is equivalent to the enthalpy of reaction.  

Q=m*c*ΔT

⇒ Q = the heat transfer (in joule)

⇒ m =the mass (in grams)

⇒ c= the heat capacity (J/g°C)

⇒ ΔT = Change in temperature = T2- T1

Step 4: Calculate heat

Let's vonsider the density the same as the density of water (1g/mL)

Mass = volume * density

Mass = 200 mL * 1g/mL

Mass = 200 grams

Q= m*c*ΔT

⇒ m = 200 grams

⇒ c = the heat capacity (let's consider the heat capacity of water) = 4.184 J/g°C

⇒ ΔT = 23.5 -21.9 = 1.6°C

Q = 200 * 4.184 * 1.6 = 1338 .9 J = 1.3 kJ

There is 1.3 kJ heat produced(released)

Therefore, we assumed no heat is absorbed by the calorimeter, no heat is exchanged between the  calorimeter and its surroundings, and the specific heat and mass of the solution are the same as those for  water (1g/mL and 4.184 J/g°C)

7 0
3 years ago
Please help, I really don’t understand this!!!
kap26 [50]

<u>Analysing the Question:</u>

We are given the balanced equation:

C₆H₁₂O₆ + 6O₂→ 6CO₂ + 6H₂O

from this equation, we can say that: <em>for every 1 mole of Glucose, we need 6 moles of Oxygen</em>

<u>Moles of Glucose used in the reaction:</u>

Molar mass of Glucose = 180 grams / mol

Given mass of Glucose = 1 gram

Mole of Glucose = Given mass / Molar mass

Moles of Glucose = 1 / 180 moles

<u>Mass of Oxygen required:</u>

We know that for every mole of Glucose, we need 6 moles of Oxygen

So, for 1/180 moles of Glucose, we need 6 / 180 = 1 / 30 moles of Oxygen

Mass of 1 / 30 moles of Oxygen:

Mass = Molar mass * number of moles

Mass of Oxygen = 32 * 1/30

Mass of Oxygen  = 32 / 30

Mass of Oxygen = 1.06 grams

5 0
3 years ago
What are some facts about Radon something that is unique and relatively unknown by the general population????
viktelen [127]

Answer:

The word radon is derived from radium, of which radon is gas. Early in its discovery it was also called radium emanation and niton, which comes from the Latin nitens, Since 1923, however, it has been called radon.

8 0
3 years ago
A balloon is filled to a volume of 1.6 L at 278 K. The balloon is left outside overnight and the temperature has dropped to 253
german

Explanation:

Charles' law gives the relationship between the volume and the temperature of the gas. Mathematically,

Volume ∝ Temperature

i.e. \dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}

We have, V₁ = 1.6 L, T₁ = 278 K, T₂ = 253, V₂=?

V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{1.6\times 253}{278}\\\\V_2=1.45\ L

So, the new volume is 1.45 L.

6 0
3 years ago
Which correctly describes the reaction between potassium and excess water?
nika2105 [10]
Balanced chemical reaction: 2K(s) + 2H₂O(l) → 2KOH(aq) + H₂(g).

KOH is inorganic compound p<span>otassium hydroxide, a strong base.
H</span>₂ is hydrogen gas.
In balanced chemical reaction number of atoms on both side of chemical reaction must be same. There are two potassium atoms, four hydrogen atoms and two oxygen atoms on both side of reaction.
7 0
3 years ago
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