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max2010maxim [7]
3 years ago
9

Part b - the effect of an electric field (voltage) on a negatively charged oil droplet in the millikan oil droplet experiment, t

he oil is sprayed from an atomizer into a chamber. the droplets are allowed to pass through the hole into the chamber so that their fall can be observed. the top and bottom of the chamber consist of electrically charged plates. the upper plate is positively charged, and the lower plate is negatively charged. x rays are introduced into the chamber so that when they strike the oil droplets, the droplets will acquire one or more negative charges. the electric field (voltage) is applied to the metal plates. watch the animation and identify the effects of an electric field on the motion of a negatively charged oil droplet. consider the gravitational force as fg and the electric force as fe. all the other forces acting on the oil droplet can be ignored as their effect on the motion of the oil droplet is negligible. check all that apply.
Chemistry
1 answer:
AnnZ [28]3 years ago
6 0
There will be two dominant forces acting on the oil droplets. The first is the weight of the oil droplets, which will cause them to move downwards in the chamber. The other is the force of attraction the droplets will experience due to the positively charged plate and their own negative charge. The magnitude of the resultant force, which will be equivalent to the difference of these forces, will dictate whether the net movement of the droplet is downwards or upwards.
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Rank the solutions below in order of increasing acidity. (Drag and drop into the appropriate area)
Vladimir79 [104]

Answer:

0.1 M NaOH, 3 M NH3, 0.01 M CH3COOH, 0.01 M H2SO4, 0.1 M HCl

Explanation:

Strong acids are more acids than weak acids. In the same way, strong bases are more basic than weak bases that are in the same concentration.

Then, the more concentrated acid or base will be more acidic or basic.

CH3COOH. Weak acid

NaOH. Strong base

H2SO4. Strong acid

NH3. Weak base.

HCl. Strong acid

The less acid (More basic):

<h3>0.1 M NaOH, 3 M NH3, 0.01 M CH3COOH, 0.01 M H2SO4, 0.1 M HCl</h3>

Strong base, weak base, weak acid, diluted strong acid, undiluted strong acid

7 0
3 years ago
People take antacids, such as milk of magnesia, to reduce the discomfort of acid stomach or heartburn. The recommended dose of m
Llana [10]

Answer:

V_{HCl}=0.208L=208mL

Explanation:

Hello,

In this case, since the chemical reaction is:

2HCl+Mg(OH)_2\rightarrow MgCl_2+2H_2O

We can see that hydrochloric acid and magnesium hydroxide are in a 2:1 mole ratio, which means that the neutralization point, we can write:

n_{HCl}=2*n_{Mg(OH)_2}

In such a way, the moles of magnesium hydroxide (molar mass 58.3 g/mol) in 500 mg are:

n_{Mg(OH)_2}=500mg*\frac{1g}{1000mg}*\frac{1mol}{58.3g}  =0.00858mol

Next, since the pH of hydrochloric acid is 1.25, the concentration of H⁺ as well as the acid (strong acid) is:

[H^+]=[HCl]=10^{-pH}=10^{-1.25}=0.0562M

Then, since the concentration and the volume define the moles, we can write:

[HCl]*V_{HCl}=2*n_{Mg(OH)_2}

Therefore, the neutralized volume turns out:

V_{HCl}=\frac{2*0.00858mol}{0.0562\frac{mol}{L} }\\ \\V_{HCl}=0.208L=208mL

Best regards.

3 0
3 years ago
In a classroom which comparison would a teacher most likely use for describing a mole
irina [24]
Mole - one of the most important concepts in chemistry - is a kind of link to go from the microworld of atoms and molecules in a normal macrocosm grams and kilograms.
In chemistry often have to consider large numbers of atoms and molecules. For fast and efficient calculation made using the weighing method. But it is necessary to know the weight of individual atoms and molecules. In order to identify the molecular weight must be added the weight of all atoms in the compound.
3 0
3 years ago
Consider the following reaction 2 N2O(g) =&gt; 2 N2(g) + O2(g) rate = k[N2O]. For an initial concentration of N2O of 0.50 M, cal
den301095 [7]

Answer:

After 2.0 minutes the concentration of N2O is 0.3325 M

Explanation:

Step 1: Data given

rate = k[N2O]

initial concentration of N2O of 0.50 M

k = 3.4 * 10^-3/s

Step 2: The balanced equation

2N2O(g) → 2 N2(g) + O2(g)  

Step 3: Calculate the concentration of N2O after 2.0 minutes

We use the rate law to derive a time dependent equation.

-d[N2O]/dt = k[N2O]

ln[N2O] = -kt + ln[N2O]i

 ⇒ with k = 3.4 *10^-3 /s

⇒ with t = 2.0 minutes = 120s

⇒ with [N2O]i = initial conc of N2O = 0.50 M

ln[N2O] = -(3.4*10^-3/s)*(120s) + ln(0.5)

ln[N2O] = -1.101

e^(ln[N2O]) = e^(-1.1011)

[N2O} = 0.3325 M

After 2.0 minutes the concentration of N2O is 0.3325 M

3 0
3 years ago
URGENT!!!What volume of a 0.88 M solution can be made using 130. Grams of FeCI^2?
Likurg_2 [28]

Answer:

1.16L can be made

Explanation:

Molarity = Mol / Volume

Volume = Mol / Molarity

Let's determine the moles of salt, with that mass:

130 g FeCl₂ . 1mol / 126.75 g = 1.02 moles of FeCl₂

Volume = 1.02 mol / 0.88 mol/L → 1.16L

5 0
3 years ago
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