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Neko
23 days ago
13

A kitchen worker at a local hospital was filling salt shakers. For those patients on a sodium restricted diet due to high blood

pressure, the hospital provided a salt substitute containing potassium chloride instead of sodium chloride. Unfortunately, the hospital worker mixed some of the containers up. How could the contents of the containers be indefited?
Chemistry
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An atom of beryllium (m = 8.00 u) splits into two atoms of helium (m = 4.00 u) with the release of 92.2 kev of energy. if the or
VMariaS [2998]
The energy released results in a kinetic energy of 92.2 keV for the products. We should convert keV into Joules, noting that 1 keV equals a kiloelectron volt. The required conversion is: 1.602×10⁻¹⁹ <span>joule = 1 eV

Kinetic energy = 92.2 keV * (1,000 eV/1 keV) * (</span>1.602×10⁻¹⁹ joule/1 eV) = 5.76×10²³ Joules

Next, we can determine the velocity of each He atom from the kinetic energy:
KE = 1/2*mv²
5.76×10²³ Joules = 1/2*(4)(v²)
This solves to give us: v = 5.367×10¹¹ m/s
4 0
2 months ago
Read 2 more answers
A 0.784 g sample of magnesium is added to a 250 ml flask and dissolved in 150 ml of water. magnesium hydroxide obtained from the
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Although multiple values are given, our focus is on HCl.

<span>We have 215 mL (0.215 L) of 0.300 M HCl fully consumed in the reaction. It's important to recall that the number of moles is found by multiplying volume by molarity:</span>

 

moles = 0.215 L × 0.300 M

<span>moles = 0.0645 moles of HCl</span>

4 0
4 months ago
How to correctly solve this problem : 4.05Kg+567.95g+100.1g correct and best way
castortr0y [3046]
Refer to the attached document for the solution.

7 0
3 months ago
Suppose a 20.0 g gold bar at 35.0°C absorbs 70.0 calories of heat energy. Given that the specific heat of gold is 0.0310 cal/g °
VMariaS [2998]

The change in temperature can be expressed as:

T_2-T_1=\dfrac{q}{mC_p(Gold)}

By substituting in the known values, we arrive at:

T_2-T_1=\dfrac{70\ cal}{20\ g\times 0.0310\ cal/g^o\ C}\\\\T_2-T_1=112.90^oC\\\\T_2-35^oC=112.90^oC\\\\T_2=(112.90+35)^oC\\\\T_2=147.9^oC

Thus, we obtain the required answer.

6 0
3 months ago
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