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1 year ago

In which scenario does radiation occur? A B C

Chemistry

2 answers:

goldenfox [79]1 year ago

6 0

Answer:

B: drying towels at the beach

Explanation:

I had this on my EDG assignment for radiation and I got it correct.

I hope this is what you were looking for

jarptica [38.1K]1 year ago

3 0

Answer:

if it has anything to do with plutonuim, uranium , or burilium its radiation

Explanation:

those elements emit radiation

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The chemical reaction involving an acid and a base or alkali is know as a

saul85 [17]

Ans

Neutralization reaction

Explanation:

The meaning is

acid + base = salt + water

5 0

2 years ago

Read 2 more answers

For the reaction 2 H2S(g) D 2 H2 (g) + S2 (g), Kp = 1.5 × 10−5 at 800.0°C. If the initial partial pressures of H2 and S2 in a cl

Usimov [2.4K]

Answer: The approximate equilibrium partial pressure of $H_2S$ is 3.92 atm

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

The given balanced equilibrium reaction is,

$2H_2S(g)\rightleftharpoons 2H_2(g)+S_2(g)$

$K_p=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}$

$1.5\times 10^{-5}=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}$

On reversing the reaction:

$2H_2(g)+S_2(g)\rightleftharpoons 2H_2S(g)$

initial pressure 4.00atm 2.00 atm 0

eqm (4.00-2x)atm (2.00-x) atm 2x atm

$K_p=\frac{[H_2S]^2}{[H_2]^2\times [S_2]}$

$K_p'=\frac{1}{K_p}=0.67\times 10^5$

$2H_2(g)+S_2(g)\rightleftharpoons 2H_2S(g)$

$0.67\times 10^5=\frac{2x]^2}{[4.00-2x]^2\times [2.00-x]}$

$x=1.96$

$[H_2S]=2x=2\times 1.96=3.92 atm$

Thus approximate equilibrium partial pressure of $H_2S$ is 3.92 atm

3 0

2 years ago

Based on their electrons dot diagrams, what is the formula for the covalently bonded compound

Luda [366]

Answer:

Nitrogen, the next nonmetal, has 5 electrons in the valence shell, so it needs to combine with 3 hydrogen atoms to fulfill the octet rule and form a stable compound called ammonia (NH3).

5 0

1 year ago

**25 POINTS**. I doubt anyone will do this but it's worth a try to ask. To do this exercise, you will need fifteen containers, f

Travka [436]

Answer:

done not that hard

Explanation:

6 0

2 years ago

Calculate the pH of a buffer solution prepared by dissolving 0.20 mole of cyanic acid (HCNO) and 0.80 mole of sodium cyanate (Na

Afina-wow [57]

The pH of a buffer solution : 4.3

<h3>Further explanation</h3>

Given

0.2 mole HCNO

0.8 mole NaCNO

1 L solution

Required

pH buffer

Solution

Acid buffer solutions consist of weak acids HCNO and their salts NaCNO.

$\tt \displaystyle [H^+]=Ka\times\frac{mole\:weak\:acid}{mole\:salt\times valence}$

valence according to the amount of salt anion

Input the value :

$\tt \displaystyle [H^+]=2.10^{-4}\times\frac{0.2}{0.8\times 1}\\\\(H^+]=5\times 10^{-5}\\\\pH=5-log~5\\\\pH=4.3$

7 0

2 years ago