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Alja [10]
2 years ago
5

Mrs. Nogaki tested four black markers in rubbing alcohol and in water. Each filter paper had four evenly spaced dots. She found

that all four markers had ink that separated into two or more different pigments when placed in the rubbing alcohol. What was her dependent variable?
Chemistry
1 answer:
ExtremeBDS [4]2 years ago
8 0

Answer:

Lost pigment of marker when dipped in alcohol

Explanation:

dependent viable = output

so it's the output of what happens after the input.

  • she put the marker in the water which is the independent variable, that's the input
  • the output or the result of that decision is having lost pigment in the marker

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Chemically combined substances, such as CO2, NH3, and H20 are called.....
Alex787 [66]
It is d. compounds because they have more than one element.
8 0
3 years ago
Read 2 more answers
Suppose 215 g of NO3- flows into a swamp each day. What volume of CO2 would be produced each day at 17.0°C and 1.00 atm?
charle [14.2K]

Answer:

The answer is "41.23 \ L\  N_2"

Explanation:

2 NO_3^{-} + 10 e^{-} + 12 H^{+} \longrightarrow N_2 + 6 H_2O\\\\= \frac{( 215 \ g \ NO_3^{-})}{(62.0049  \frac{\ g NO_3^{-}}{mol})} \times  \frac{(1 \ mol \ N_2}{ 2 \ mol \ NO_3^{-})}\\\\

=3.46746789 \times 0.5\\\\= 1.733 \ mol \ N_2 \\\\\to V = \frac{nRT}{P} \\\\= (1.733 \ mol) \times (0.08205746 \frac{L\ atm}{Kmol}) \times \frac{ (17 + 273) K}{(1.00 atm)}\\\\= 41.23

8 0
3 years ago
The ideal gas heat capacity of nitrogen varies with temperature. It is given by:
hammer [34]

Answer:

A)  1059 J/mol

B)  17,920 J/mol

Explanation:

Given that:

Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4

R (constant) = 8.314

We know that:

C_p=C_v+R

We can determine C_v from above if we make C_v the subject of the formula as:

C_v=C_p-R

C_V = 29.42-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4-8.314

C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4

A).

The formula for calculating change in internal energy is given as:

dU=C_vdT

If we integrate above data into the equation; it implies that:

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 1059J/mol

Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.

B).

If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.

then T = 273 K & T2 = 1073 K

∴

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 17,920 J/mol

3 0
3 years ago
B. The highly toxic gas phosgene is produced by the reaction CO(g) + Cl2(g) COCl2(g), ΔH = –108 kJ/mol. Use this reaction to ans
galben [10]

Answer:

I. Increasing pressure will allow more frequent successful collision between particles due to the particles being closer together.

II. Rate of reaction increases due to more products being made; as increased pressure favours the exothermic side of the equilibrium.

III. Increasing temperature provides particles lots of (Kinetic) energy, for more frequent successful collision due to the particles moving at a faster rate than before. However, favouring the endothermic side of the equilibrium due to lots of energy required to break and form new bonds.

IV. Rate of reaction increases due to increase temperature favouring both directions of the equilibrium - causing products to form faster.

Hope this helps!

6 0
2 years ago
Nonpolar substances do not dissolve in polar solvents like water. If you're conducting an investigation to test the solubility o
shusha [124]
I would expect silane because all the rest have an overall dipole movement
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