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Pie
2 years ago
10

What kind of radioactive element is useful for dating an object?

Chemistry
1 answer:
OlgaM077 [116]2 years ago
4 0

The kind of radioactive element is useful for dating an object is one with a half-life close to the age of the object.

<h3>What are radioactive elements?</h3>

Radioactive elements are elements that involved in radioactivity and this comprises of atoms or particles in their nuclei whose atomic nuclei are not stable and they emit radiations to maintain stability.

Therefore, The kind of radioactive element is useful for dating an object is one with a half-life close to the age of the object.

Learn more about radioactive elements below.

brainly.com/question/10257016

#SPJ1

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Under what pressure will a scientist need to store 0.400 miles of gas if the container has a volume of 200.0 mL and the temperat
mihalych1998 [28]
Hello!

Ok so for this problem we use the ideal gas law of PV=nRT and I take it that the scientist needs to store 0.400 moles of gas and not miles.

So if we have
n=0.400mol
V=0.200L
T= 23degC= 273k+23c=296k
R=ideal gas constant= 0.0821 L*atm/mol*k

So now we rearrange equation for pressure(P)

P=nRT/V
P=((0.400mol)*(0.0821 L*atm/mol*k)*(296k))/(0.200L) = 48.6 atm of pressure

Hope this helps you understand the concept and how to solve yourself in the future!! Any questions, please feel free to ask!! Thank you kindly!!!
4 0
3 years ago
Which type of telescope would perform better on a mountaintop than at a lower elevation?
marishachu [46]

when an optical telescope located on the ground is used to view the sky the image has to surpass through the earth's atmosphere which is polluted. so the resolution wont be clear. When viewed on a mountain top both optical and infrared telescopes would perform better as the air above the atmosphere is less depth and cleaner.

5 0
3 years ago
A certain weak acid, ha, has a ka value of 6.0×10−7. calculate the percent ionization of ha in a 0.10 m solution.
joja [24]
For the purpose, we will use the equation for determining the dissociation constant from concentration and <span>percent of ionization:

Kd = c </span>× α²

α = √(Kd/c) × 100%

Kd = 6.0×10⁻⁷

c(HA) = 0.1M

α = √(6.0×10⁻⁷/0.1)  × 100% =  0.23%

So, in the solution, the acid <span>percent of ionization will be just 0.23%.</span>

5 0
3 years ago
My swimming pool is rectangular (16 feet by 34 feet) and has a depth of 6 feet. Lets imagine that my pool water is full to the t
Reil [10]

Answer:

Number of moles of photons required = 5.04 × 10⁴ moles

Explanation:

The energy of a photon can be calculated from Planck's equation E = hc/λ

Where h = 6.63 × 10-³⁴ Js and c, the velocity of light = 3.0 × 10⁸ m/s

Energy of one mole of photons = N₀ × hc/λ

wavelength of photon, λ = 520 nm = 5.20 × 10-⁷ m

Energy of one mole of photons = 6.02 × 10²³ × 6.63 × 10−³⁴ × 3 × 10⁸/5.20 × 10-⁷

Energy of one mole of photons = 2.30 × 10⁵ J/mol

Energy required to raise the temperature of a given mass of a substance, E = mcΔT

Where m is mass of substance,  c is specific heat capacity,  ΔT is temperature difference

Mass ofnwternin the pool = volume × density

Volume of water = Volume of swimming pool

Volume of water = 16 × 34 × 6 ft³ = 3264 ft³

1 ft³ = 28316.8 cm³; 3264 ft³ = 28316.8 × 3264 = 92426035.2 cm³

Density of water = 1 g/cm³

Mass of water = 92426035.2 cm³ × 1 g/cm³ = 92426035.2g

ΔT = 80°C - 50°C = 30°C, c = 4.18 J/g/K

Energy required to raise 92426035.2 g water by 30° C = 92426035.2 × 4.18 × 30

Energy required = 1.16 × 10¹⁰ J

Hence, number of moles of photons required = 1.16 × 10¹⁰ J/2.30 × 10⁵ J/mol

Number of moles of photons required = 5.04 × 10⁴ moles

5 0
3 years ago
44.8% of a 250. mL acid solution is used in an experiment, what volume of acid (in mL) was used?
postnew [5]

Answer:

112 mL  

Explanation:

The formula for percent by volume is

\text{Percent by volume} = \dfrac{\text{Volume of solute}}{\text{Volume of solution}}\times 100 \, \%

If you have 250 mL of a solution that is 44.8 % v/v,

\begin{array}{rcl}44.8\, \% & = & \dfrac{\text{Volume of solute}}{\text{250 mL}}\times 100 \, \%\\\\44.8 \times \text{ 250 mL} & = & \text{Volume of solute} \times 100\\\text{Volume of solute} & = & \dfrac{44.8 \times 250\text{ mL}}{100}\\\\ & = & \textbf{112 mL}\\\end{array}

7 0
3 years ago
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