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3 years ago

The variable that a scientist observes to change while conducting an experiment is called the ___ variable.

Chemistry

2 answers:

QveST [7]3 years ago

5 0

The variable that a scientist observes to change while conducting an experiment is called the responding variable.

Nimfa-mama [501]3 years ago

4 0

Great Question!

The Answer Would Be "B" The "RESPONDING" Variable

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Consider two solutions made using water as the solvent: 1.0 m CaCl2 and 3.0 m sucrose. Describe how the osmotic pressures of the

mash [69]

Osmotic pressure is calculated by the product of the concentration in molarity, the temperature, the vant Hoff factor (3 for CaCl2 and 1 for sucrose) and R, universal gas constant. At the same temperature, the osmotic pressures of both solutions are equal.

π = CRTi
For CaCl2,
π = (1)RT(3) = 3RT

For sucrose,
π = (3)RT(1) = 3RT

7 0

3 years ago

Read 2 more answers

For the equilibrium

Mamont248 [21]

Answer:

Equilibrium concentrations of the gases are

$H_2S=0.596M$

$H_2=0.004 M$

$S_2=0.002 M$

Explanation:

We are given that for the equilibrium

$2H_2S\rightleftharpoons 2H_2(g)+S_2(g)$

$k_c=9.0\times 10^{-8}$

Temperature, $T=700^{\circ}C$

Initial concentration of

$H_2S=0.30M$

$H_2=0.30 M$

$S_2=0.150 M$

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles of reactant reduced and form product

Concentration of

$H_2S=0.30+2x$

$H_2=0.30-2x$

$S_2=0.150-x$

At equilibrium

Equilibrium constant

$K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}$

Substitute the values

$9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}$

By solving we get

$x\approx 0.148$

Now, equilibrium concentration of gases

$H_2S=0.30+2(0.148)=0.596M$

$H_2=0.30-2(0.148)=0.004 M$

$S_2=0.150-0.148=0.002 M$

3 0

3 years ago

Soda pop is carbonated with CO2. Mark puts one bottle of soda pop in the refrigerator and leaves the other out in the hot sunlig

fomenos

Answer:

The one left in the hot sunlight.

Explanation:

The solubility of gases decreases when temperature increases. The gas in the soda pop (CO2) left in the sun will not stay dissolved as much as the on left in the refrigerator.

3 0

2 years ago

How many particles are in a 151 g sample of Li2O?

neonofarm [45]

Answer:

3.052 × 10^24 particles

Explanation:

To get the number of particles (nA) in a substance, we multiply the number of moles of the substance by Avogadro's number (6.02 × 10^23)

The mass of Li2O given in this question is as follows: 151grams.

To convert this mass value to moles, we use;

moles = mass/molar mass

Molar mass of Li2O = 6.9(2) + 16

= 13.8 + 16

= 29.8g/mol

Mole = 151/29.8g

mole = 5.07moles

number of particles (nA) of Li2O = 5.07 × 6.02 × 10^23

= 30.52 × 10^23

= 3.052 × 10^24 particles.

4 0

3 years ago

Andrews [41]

Answer:

similarities: they both carry a charge

differences:

polyatomic ion- composed of more than one atom

monatomic ion- composed in a single atom.

6 0

3 years ago