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3 years ago

Where would a plant that lives on land get nutrients from?

Chemistry

2 answers:

Masja [62]3 years ago

8 0

Answer:

The sun, the soil. also water

klemol [59]3 years ago

5 0

Answer:

they absorb water and nutrients through the roots

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Charcoal from the dwelling level of the Lascaux Cave in France gives an average count of 0.97 disintegrations of ^14 C per minut

Verdich [7]

Answer:

Explanation:

count given by old sample = .97 disintegrations per minute per gram

count given by fresh sample = 6.68 disintegrations per minute per gram

Half life of radioactive carbon = 5568 years

rate of disintegration

dN / dt = λ N

In other words rate of disintegration is proportional to no of radioactive atoms present . As number reduces rate also reduces .

Let initial no of radioactive be N₀ and after time t , number reduces to N

N₀ / N = 6.68 / .97

Now

$N=N_0e^{-\lambda t}$

$\frac{N}{N_0} =e^{-\lambda t}$

$\frac{6.68}{.97} = e^{\lambda t}$

λ is disintegration constant

λ = .693 / half life

= .693 / 5568

= .00012446 year⁻¹

Putting the values in the equation above

$\frac{6.68}{.97} = e^{.00012446\times t}$

$6.8866 = e^{.00012446\times t}$

1.929577 = .00012446 t

t = 15503.6 years .

4 0

3 years ago

HELP ASAP!

Ilia_Sergeevich [38]

Answer:

All decrease

Explanation:

Trust, and I just got this question and got it right

8 0

3 years ago

One liter of oxygen gas at standard temperature and pressure has a mass of 1.43 g. The same volume of hydrogen gas under these c

Alchen [17]

Answer:

Indeed, the two samples should contain about the same number of gas particles. However, the molar mass of $\rm O_2\; (g)$ is larger than that of $\rm H_2\; (g)$ (by a factor of about $16$ .) Therefore, the mass of the $\rm O_2\; (g)$ sample is significantly larger than that of the $\rm H_2\; (g)$ sample.

Explanation:

The $\rm O_2\; (g)$ and the $\rm H_2\; (g)$ sample here are under the same pressure and temperature, and have the same volume. Indeed, if both gases are ideal, then by Avogadro's Law, the two samples would contain the same number of gas particles ( $\rm O_2\; (g)$ and $\rm H_2\; (g)$ molecules, respectively.) That is:

$n(\mathrm{O_2}) = n(\mathrm{H}_2)$ .

Note that the mass of a gas $m$ is different from the number of gas particles $n$ in it. In particular, if all particles in this gas have a molar mass of $M$ , then:

$m = n \cdot M$ .

In other words,

$m(\mathrm{O_2}) = n(\mathrm{O_2}) \cdot M(\mathrm{O_2})$ .
$m(\mathrm{H_2}) = n(\mathrm{H_2}) \cdot M(\mathrm{H_2})$ .

The ratio between the mass of the $\rm O_2\; (g)$ and that of the $\rm H_2\; (g)$ sample would be:

$\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})}\end{aligned}$ .

Since $n(\mathrm{O_2}) = n(\mathrm{H}_2)$ by Avogadro's Law:

$\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})} = \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}$ .

Look up relative atomic mass data on a modern periodic table:

$\rm O$ : $15.999$ .
$\rm H$ : $1.008$ .

Therefore:

$M(\mathrm{O_2}) = 2 \times 15.999 \approx 31.998\; \rm g \cdot mol^{-1}$ .
$M(\mathrm{H_2}) = 2 \times 1.008 \approx 2.016\; \rm g \cdot mol^{-1}$ .

Verify whether $\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}$ :

Left-hand side: $\displaystyle \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{1.43\; \rm g}{0.089\; \rm g} \approx 16.1$ .
Right-hand side: $\displaystyle \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}= \frac{31.998\; \rm g \cdot mol^{-1}}{2.016\; \rm g \cdot mol^{-1}} \approx 15.9$ .

Note that the mass of the $\rm H_2\; (g)$ sample comes with only two significant figures. The two sides of this equations would indeed be equal if both values are rounded to two significant figures.

7 0

4 years ago

You have knowledge of the outermost electron states of the atoms, trends in the periodic table, and the patterns of chemical pro

olganol [36]

Answer: ok pardon me but i think it's b

Explanation:

8 0

3 years ago

Does any solid Ag₂CrO₄ form when 2.7x10⁻⁵g of AgNO₃ is dissolved in 15.0 mL of 4.0x10⁻⁴MK₂CrO₄?

e-lub [12.9K]

Molarity of Ag+ is less than the molar solubility thus ppt will not occur.

Balanced reaction-:

Moles of AgNO3=mass(g)molar mass (g/mol) =2.7×10−5g / 169.86 gmol

=1.589⋅10^−7 mol

Molarity of Ag+=moles of solute(L)=1.589⋅10−7 mol0.015 L=1.059⋅10−5M

Ksp of Ag2CrO4

=[Ag+]2[CrO42−]

1.2⋅10−12=[2s]2[s]

4s3=1.2⋅10−12

s=6.69⋅10−5 M

Molarity of Ag+ is less than the molar solubility thus ppt will not occur.

<h3>What is the molarity calculation formula?</h3>

The volume of solvent required to dissolve the provided solute is multiplied by the ratio of the moles of the solute whose molarity has to be computed. (M=frac{n}{V}) The molality of the solution that needs to be computed in this case is M. n is the solute's molecular weight in moles.

Learn more about Molarity:

brainly.com/question/8732513

#SPJ4

3 0

1 year ago