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DedPeter [7]
3 years ago
6

Describe how you created the 0.075 M glucose solution from the 1 M stock solution. Show your calculations and give details about

the equipment used, which solutions you mixed together, and amounts of each. Show your calculations
Chemistry
1 answer:
soldier1979 [14.2K]3 years ago
8 0

Answer:

by addind

Explanation:idk how to do this

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Which is the best term to use when describing the energy of position?
Minchanka [31]
Answer: potential.

Chemical energy is the energy provided by a chemical reaction.

Kinetic energy is the energy due to the speed.

Potential energy is the energy due to the position. For example, an object on the top of a mountain, has the possibility to perform work if it falls.

Electromagnetic energy. is propagated by waves: radio waves, infrared radiation, microwaves, etc.
4 0
3 years ago
Read 2 more answers
Convert 3.82 inches to km
weeeeeb [17]

the answer is 0.000097 KM

6 0
3 years ago
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3. If all the waves were measured for a period of 5 seconds, which wave had the highest frequency?
Vanyuwa [196]

Answer:

Correct answer is A.

Explanation:

Frequency is the number of oscillations that a wave have per unit time. Since time is measured in seconds, the wave with the highest frequency must register the highest number of oscillation per second. Hence, correct answer is A.

8 0
3 years ago
f the Ksp for HgBr2 is 2.8×10−14, and the mercury ion concentration in solution is 0.085 M, what does the bromide concentration
goldfiish [28.3K]

Answer:

0.057 M

Explanation:

Step 1: Given data

Solubility product constant (Ksp) for HgBr₂: 2.8 × 10⁻⁴

Concentration of mercury (II) ion: 0.085 M

Step 2: Write the reaction for the solution of HgBr₂

HgBr₂(s) ⇄ Hg²⁺(aq) + 2 Br⁻

Step 3: Calculate the bromide concentration needed for a precipitate to occur

The Ksp is:

Ksp = 2.8 × 10⁻⁴ = [Hg²⁺] × [Br⁻]²

[Br⁻] = √(2.8 × 10⁻⁴/0.085) = 0.057 M

7 0
3 years ago
We can also perform a similar calculation for the mass defect and binding energy for nuclear reactions using the masses of the a
sukhopar [10]

Answer:

See Explanation

Explanation:

\frac{235}{92} U + \frac{1}{0} n ---->\frac{137}{52} Te + \frac{97}{40} Zr +2\frac{1}{0} n

Hence the mass defect is;

[235.04393 + 1.00867] - [ 136.92532 + 96.91095 + 2(1.00867)]

=  236.0526 - 235.85361

= 0.19899 amu

Since 1 amu = 1.66 * 10^-27 Kg

0.19899 amu = 0.19899 * 1.66 * 10^-27 = 3.3 * 10^-28 Kg

Binding energy = Δmc^2

Binding energy = 3.3 * 10^-28 Kg * (3 * 10^8)^2 = 2.97 * 10^-11 J

ii) \frac{10}{5}B + \frac{1}{0}n-----> \frac{7}{3} Li + \frac{4}{2} He + Energy

Hence the mass defect is;

[10.01294 + 1.00867] - [7.01600 + 4.00260]

= 11.02161 - 11.0186

= 0.00301 amu

Since 1 amu = 1.66 * 10^-27 Kg

0.00301 amu = 0.00301 * 1.66 * 10^-27 = 4.997 * 10^-30 Kg

Binding energy = Δmc^2

Binding energy = 4.997 * 10^-30 Kg * (3 * 10^8)^2 = 4.5 * 10^-13 J

7 0
3 years ago
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