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grandymaker [24]
3 years ago
7

You are working in a lab when radiation alarms go off. You are able to hide inside a steel cabinet, whose sides are about 1.5 in

ches thick, until the alarm goes off. Preliminary reports show that the radiation was weakly ionizing and negatively charged. Were you safe in the cabinet?
A. Yes, it was beta radiation and the steel was enough to block it.
B. Yes, it was alpha radiation and not harmful.
C. No, it was beta radiation and very dangerous.
D. No, it was gamma radiation and can only be blocked by a thick wall of lead.
Chemistry
1 answer:
dusya [7]3 years ago
6 0

Answer:

Its either B or C (in my opinion)

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As pH increases, what happens to the hydrogen ion concentration?
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Answer:

According to libretexts the answer would be B. decreases.

Explanation:

If the hydrogen concentration increases, the pH decreases, causing the solution to become more acidic. This happens when an acid is introduced. ... If the hydrogen concentration decreases, the pH increases, resulting in a solution that is less acidic and more basic

6 0
3 years ago
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As temperature increases, the amount of solute that a solvent can dissolve increases. True False
ra1l [238]

<span>As temperature increases, the amount of solute that a solvent can dissolve increases.</span>



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3 years ago
Calculate the unit cell edge length for an 78 wt% Ag- 22 wt% Pd alloy. All of the palladium is in solid solution, and the crysta
BaLLatris [955]

Answer:

The unit cell edge length for the alloy is 0.405 nm

Explanation:

Given;

concentration of Ag, C_{Ag} = 78 wt%

concentration of Pd, C_P_d = 22 wt%

density of Ag = 10.49 g/cm³

density of Pd = 12.02 g/cm³

atomic weight of Ag, A_A_g = 107.87 g/mol

atomic weight of iron, A_P_d = 106.4 g/mol

Step 1: determine the average density of the alloy

\rho _{Avg.} = \frac{100}{\frac{C_A_g}{\rho _A_g} + \frac{C__{Pd}}{\rho _{Pd}} }

\rho _{Avg.} = \frac{100}{\frac{78}{10.49} + \frac{22}{12.02} } = 10.79 \ g/cm^3

Step 2: determine the average atomic weight of the alloy

A _{Avg.} = \frac{100}{\frac{C_v}{A _A_g} + \frac{C__{Pd}}{A _{Pd}} }

A _{Avg.} = \frac{100}{\frac{78}{107.87} + \frac{22}{106.4} } = 107.54 \ g/mole

Step 3: determine unit cell volume

V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}

for a FCC crystal structure, there are 4 atoms per unit cell; n = 4

V_c=\frac{4*107.54}{ 10.79*6.022*10^{23}} = 6.620*10^{-23} \ cm^3/cell

Step 4: determine the unit cell edge length

Vc = a³

a = V_c{^\frac{1}{3} }\\\\a = (6.620*10^{-23}}){^\frac{1}{3}}\\\\a = 4.05 *10^{-8} \ cm= 0.405 nm

Therefore, the unit cell edge length for the alloy is 0.405 nm

7 0
3 years ago
The substance which does the dissolving in a solution is called the
iVinArrow [24]
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Anvisha [2.4K]

Explanation:

This is correct!

Ions that exist in both the reactant and product side of the equation are referred to as spectator ions. Overall, they do not partake in the reaction. If they are present on both sides of the equation, you can cancel them out.

An example is;

Na+(aq) + Cl​−​​(aq) + Ag​+(aq) + NO​3​−​​(aq) → Na​+​​(aq) + NO​3​−​​(aq) + AgCl(s)

The ions; Na+, NO​3​−​​(aq) would be cancelled out to give;

Cl​−​​(aq) + Ag​+(aq)  → AgCl(s)

7 0
3 years ago
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