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3 years ago

9

What are some questions you would ask to learn more about the toxic algal bloom in Lake Temescal? Include at least two questions

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1 answer:

Darina [25.2K]3 years ago

5 0

Answer:

How can we prevent harmful algae blooms?

How do you know if algal blooms are harmful?

Explanation:

I hope this helps

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Can anyone please help?

Marrrta [24]

is it me bc i cant even see the page:?

5 0

3 years ago

How many moles of glucose are in 19.1g of glucose?

Crank

Answer:

0.106 mol (3s.f.)

Explanation:

To find the number of moles, divide the mass of glucose (in grams) by its Mr. Glucose has a chemical formula of C6H12O6. To find the Mr, add all the Ar of all the atoms in C6H12O6.

Ar of C= 12, Ar of H= 1, Mr of O= 16

These Ar values can be found on the periodic table.

Mr of glucose= 6(12)+ 12(1) + 6(16)= 180

Moles of glucose

= mass ÷ mr

= 19.1 ÷ 180

= 0.106 mol (3 s.f.)

3 0

3 years ago

25.0cm3 of s saturated potassium hydroxide is neutralized by 35.0cm3 of hydrogen chloride acid of concentration 0.75 mol/dm3. Ca

Kamila [148]

Answer:

Concentration of the original $\rm KOH$ solution: approximately $1.05\; \rm mol \cdot dm^{-3}$ .

Explanation:

Notice that the concentration of the $\rm HCl$ solution is in the unit $\rm mol\cdot dm^{-3}$ . However, the unit of the two volumes is $\rm cm^{3}$ . Convert the unit of the two volumes to $\rm dm^{3}$ to match the unit of concentration.

$\begin{aligned} V(\mathrm{NaOH}) &= 25.0\; \rm cm^{3} \\ &= 25.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0250\; \rm dm^{3} \end{aligned}$ .

$\begin{aligned} V(\mathrm{HCl}) &= 35.0\; \rm cm^{3} \\ &= 35.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0350\; \rm dm^{3} \end{aligned}$ .

Calculate the number of moles of $\rm HCl$ molecules in that $0.0350\; \rm dm^{3}$ of $0.75\; \rm mol \cdot dm^{3}$ $\rm HCl\!$ solution:

$\begin{aligned}n(\mathrm{HCl}) &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.00350\; \rm dm^{3} \times 0.75\; \rm mol \cdot dm^{3} \\ &\approx 0.02625\; \rm mol\end{aligned}$ .

$\rm HCl$ is a monoprotic acid. In other words, each $\rm HCl\!$ would release up to one proton $\rm H^{+}$ .

On the other hand, $\rm KOH$ is a monoprotic base. Each $\rm KOH\!$ formula unit would react with up to one $\rm H^{+}$ .

Hence, $\rm HCl$ molecules and $\rm KOH\!$ formula units would react at a one-to-one ratio.

${\rm HCl}\, (aq) + {\rm NaOH}\, (aq) \to {\rm NaCl}\, (aq) + {\rm H_2O}\, (l)$ .

Therefore, that $0.02625\; \rm mol$ of $\rm HCl$ molecules would neutralize exactly the same number of $\rm NaOH$ formula units. That is: $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ .

Calculate the concentration of a $\rm NaOH$ solution where $V(\mathrm{NaOH}) = 0.0250\; \rm dm^{3}$ and $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ :

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} \\ &= \frac{0.02625\; \rm mol}{0.0250\; \rm dm^{3}}\approx 1.05\; \rm mol \cdot dm^{-3}\end{aligned}$ .

7 0

3 years ago

Which isotope in each pair is more stable? Why?(d) ¹⁴₇N or ¹⁸₇N

quester [9]

<u>¹⁴₇N</u><u> </u>is the more stable isotope

<h3>Briefly explained</h3>

We have ¹⁴₇N which has a neutron to proton ratio of one, and we look at ¹⁸₇N which has a neutron to proton ratio of 1.57 Again, you look at table 24 to and you see the atomic number of seven and there is really no stable isotope. It has any more than 10 neutrons.

When we have eight, protons will go down seven protons. There's really nothing stable that has more than maybe eight neutrons. So the fact that we have 11 neutrons with ¹⁸₇N suggests that this is very unstable and

¹⁴₇N is the stable isotope of the pair.

<h3>Stable and Unstable Nuclei</h3>

An atom is electrically neutral. It contains an equal number of positively charged protons and negatively charged electrons and their charges balance. The nucleus however contains only positively charged protons which are closely packed together in a very small volume (remember neutrons have no charge).

From the laws of physics (Coulomb’s Law) one would expect that the protons being of the same charge and so close together would exert strong repulsive forces on each other. The combined gravitational force from the protons and neutrons in a nucleus is insignificant as an attractive force because their masses are so tiny.

This implies there must be an additional attractive force similar in size to the electrostatic repulsion which holds the nucleus together.

Learn more about stable and unstable nuclei

brainly.com/question/24748035

#SPJ4

6 0

1 year ago

A sample of an ideal gas in a cylinder of volume 2.67 L at 298 K and 2.81 atm expands to 8.34 L by two different pathways. Path

Igoryamba

Explanation:

For path A, the calculation will be as follows.

As, for reversible isothermal expansion the formula is as follows.

W = $-2.303 nRT log(\frac{V_2}{V_1})$

Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

R = 8.314 J/(K mol), T = 298 K ,

$V_{2}$ = 8.34 L, $V_{1}$ = 2.67 L

Now, putting the given values into the above formula as follows.

W = $-2.303 nRT log(\frac{V_2}{V_1})$

= $-2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})$

= $-2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494$

= -2818.68 J

Hence, work for path A is -2818.68 J.

For path B, the calculation will be as follows.

Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure $P_{external}$ = 1.00 atm.

So, W = $-P_{external} \times \Delta V$

Now, putting the given values into the above formula as follows.

W = $-P_{external} \times \Delta V$

= $-1 atm \times (8.34 L - 2.67 L)$

= -5.67 atm L

As we known that, 1 atm L = 101.33 J

Hence, work will be calculated as follows.

W = $-\frac{101.33 J}{1 atm L} \times 5.67 atm L$

= -574.54 J

Therefore, total work done by path B = 0 + (-574.54 J)

W = -574.54 J

Hence, work for path B is -574.54 J.

3 0

4 years ago