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3 years ago

Why is a control needed in a valid experiment?

1 answer:

5 0

You need a control so you have something to compare the experimental group on-- If you don't have a control group, how will you know what's changed?

You might be interested in

Which way will the equilibrium shift if you..... cool it to a low temperature?

igor_vitrenko [27]

If you can provide the reaction you are looking at, then we can provide a more satisfactory answer.

If the forward reaction is exothermic, then reducing the temperature where the reaction occurs will shift the equilibrium towards the right. This is because exothermic reactions release heat, and this will counteract the change as stated in Le Chatelier's Principle.

If the forward reaction is endothermic, then reducing the temperature will shift to the left. This occurs as the backward reaction is the exothermic reaction, and by Le Chatelier's Principle, the reaction will favor the reaction that produces more in to counter a reduction in temperature, in this case the backward direction reaction.

5 0

3 years ago

5.98 g of K Al(SO4)2 is equivalent to how many moles?

Rasek [7]

Answer:

Molecular Weight Of KAl(SO4)2 12H2O = 475 G / Mol Molecular Weight Of Al = 27 G / Moles.

Explanation:

HOPE THIS HELPS YOU OUT AND IF IT DID PLS MARK ME AS BRAINIEST

4 0

3 years ago

A solid ball has a density of 2.5 g/cm and a volume of 20 cm3. What is the

Svetach [21]

Answer:

50 g

Explanation:

d= m/v

rearranging the above equation

m = d x v

m = 2.5 g x 20 g/cm3

m = 50 g

7 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

student has calibrated his/her calorimeter and finds the heat capacity to be 14.2 J/°C. S/he then determines the molar heat capa

nika2105 [10]

Answer:

24.03 J/mol.ºC

Explanation:

For a calorimeter, the heat lost must be equal to the heat gained from water plus the heat gained from calorimeter, which has the same initial temperature as the water.

-Qal = Qw + Qc (minus signal represents that the heat is lost)

-mal*Cal*ΔTal = mw*Cw*ΔTw + Cc*ΔTc

Where m is the mass, C is the specific heat, ΔT is the temperature variation, al is from aluminum. w from water and c from the calorimeter. Cw = 4.186 J/gºC

-25.5*Cal*(22.7 - 100) = 99.0*4.186*(22.7 - 18.6) + 14.2*(22.7 - 18.6)

1971.15Cal = 1699.10 + 58.22

1971.15Cal = 1757.32

Cal = 0.89 J/g.ºC

The molar mass of Al is 27 g/mol

Cal = 0.89 J/g.ºC * 27 g/mol

Cal = 24.03 J/mol.ºC

6 0

3 years ago