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Kaylis [27]
3 years ago
6

Select the correct answer. Which notation is used to represent gamma decay?​

Chemistry
1 answer:
Marrrta [24]3 years ago
8 0

Answer:

B.

Explanation:

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Agustina was standing on the edge of a cliff of unknown height holding a bowling ball. She wanted to find the height of the clif
nlexa [21]

Answer:

433 m

Explanation:

Since the fall represents motion under gravity, we use the equation

s = ut - 1/2gt² where s = height of cliff or distance bowling ball falls through, u = initial velocity of bowling ball = 0 m/s(since it starts from rest), t = time = 9.4 s and g = acceleration due to gravity = -9.8 m/s².

So, substituting the values of the variables into the equation, we have

s = 0 m/s × 9.4 s - 1/2 × 9.8 m/s² × (9.4 s)²

s = 0 m - 1/2 × -9.8 m/s² × 88.36 s²

s = 1/2(865.928 m)

s = 432.964

s ≅ 433 m

5 0
3 years ago
Which of the following ions is formed when a base is dissolved in a solution?
Aneli [31]

Answer:

\huge\boxed{\sf OH-}

Explanation:

<u>According to Arrhenius concept of acid and base:</u>

"When a base in a solution, produces/yields OH- (Hydroxide) ions."

So, when a base is dissolved in a solution, it produces OH- ions.

<u>For example:</u>

NaOH ⇄ Na⁺ + OH⁻ (So, it is a base)

\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3>
3 0
2 years ago
The normal boiling point of bromine is 58.8°C, and its enthalpy of vaporization is 30.91 kJ/mol. What is the approximate vapor p
saul85 [17]

Answer : The vapor pressure of bromine at 10.0^oC is 0.1448 atm.

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of bromine at 10.0^oC = ?

P_2 = vapor pressure of propane at normal boiling point = 1 atm

T_1 = temperature of propane = 10.0^oC=273+10.0=283.0K

T_2 = normal boiling point of bromine = 58.8^oC=273+58.8=331.8K

\Delta H_{vap} = heat of vaporization = 30.91 kJ/mole = 30910 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{1atm}{P_1})=\frac{30910J/mole}{8.314J/K.mole}\times (\frac{1}{283.0K}-\frac{1}{331.8K})

P_1=0.1448atm

Hence, the vapor pressure of bromine at 10.0^oC is 0.1448 atm.

4 0
3 years ago
0.05 grams od cobalt ii oxalate will dissolve in one liter of solution at 25 degree celcius. Calculate the ksp for cobalt ii oxa
Veseljchak [2.6K]

Answer:

The equilibrium expression is:

CoC2O4(s)⇌Co2+(aq)+C2O2−4(aq)

For this reaction:

Ksp = [Co2+][C2O2−4]=1.96×10−8

Explanation:

Batteries will not clot if cobalt ions are removed from its cells. Some blood collection tubes contain salts of the oxalate ion,

C2O2−4

, for this purpose. At sufficiently high concentrations, the calcium

and oxalate ions form solid, CoC2O4·H2O (which also contains water bound in the solid). The concentration of Co2+ in a sample of blood serum is 2.2 × 10–3M. What concentration of

C2O2−4

ion must be established before CoC2O4·H2O begins to precipitate.

CoC2O4 does not appear in this expression because it is a solid. Water does not appear because it is the solvent.

Solid CoC2O4 does not begin to form until Q equals Ksp. Because we know Ksp and [Co2+], we can solve for the concentration of

C2O2−4

that is necessary to produce the first trace of solid:

5 0
3 years ago
How many moles of gold are there in 3.73 x 10^24 atoms of gold?
BARSIC [14]

Answer:

6.2moles of Gold

Explanation:

To solve this problem, we are going to use the mole concept approach.

 Given that;

  Number of atoms of gold is 3.73 x 10²⁴ atoms

Now;

  In 1 mole of any substance, we have 6.02 x 10²³ atoms;

   So;

  If there  6.02 x 10²³ atoms in  1 mole of any substance;

                 3.73 x 10²⁴ atoms will contain  \frac{3.73 x 10^{24} }{6.02 x 10^{23} }    = 6.2moles of Gold

4 0
3 years ago
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