You are given a glass rod that is negatively charged and a pinwheel that has a negative charge. What will happen when the glass rod is brought close to the pinwheel? Why?

Answers

Answer 1

Answer:

Answer:

they will repel each other

Explanation:

When these two are brought close to one another they will repel each other. This is similar to what happens with magnets, when two objects share the same polarity one object will create a repulsive force upon the second object and push it away. This repellent force is caused by an electric field from the same charged electrons in the atoms of the object. Since in this case both the glass rod and the pinwheel have a negative charge they will repel each other when they come into proximity of one another.

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A mixture contains N a H C O 3 together with unreactive components. A 1.68 g sample of the mixture reacts with H A to produce 0.561 g of C O 2 . What is the percent by mass of N a H C O 3 in the original mixture

Answers

There is 65% of NaHCO3 in the sample.

The equation of the reaction is;

HA + NaHCO3 -----> NaA + CO2 + H2O

Amount of CO2 formed = mass/molar mass

mass of CO2 = 0.561 g/44 g/mol = 0.013 moles

From the balanced reaction equation;

1 mole of NaHCO3 yields 1 mole of CO2

0.013 moles of Na2CO3 yields 0.013 moles of CO2

Hence, mass of NaHCO3 in the sample = 0.013 moles × 84 g/mol = 1.092 g of NaHCO3

Percentage by mass of NaHCO3 = 1.092 g/1.68 g ×100/1

= 65%

Learn more: brainly.com/question/25150590

Answer:

63.75%.

Explanation:

The first step here is to write out the reaction showing the chemical reaction between the two chemical species. Thus, we have;

HA(aq) + NaHCO3 --------------> CO2(g) + H20(l) + NaA(aq).

Therefore, the mole ratio is 1 : 1 : 1 : 1 that is go say one mole of HA reacted with one mole of NaHCO3 to give one mole of CO2 and one .ole of NaA.

Hence, the number of moles of CO2 = mass/molar mass = 0.561/44 = 0.01275 moles.

Thus, the number of moles of NaHCO3 = number of moles of CO2 = 0.01275 moles.

Therefore, we have ( 0.01275 moles × 84 g/mol) grams = 1.071 g NaHCO3 in the mixture.

Therefore, the percent by mass of N a H C O 3 in the original mixture = 1.071/1.68 × 100 = 63.75%.

Ammonium nitrate dissociates in water according to the following equation:43() = 4+()+03−()

When a student mixes 5.00 g of NH4NO3 with 50.0 mL of water in a coffee-cup calorimeter, the temperature of the resultant solution decreases from 22.0 °C to 16.5 °C. Assume the density of water is 1.00 g/ml and the specific heat capacity of the resultant solution is 4.18 J/g·°C.

1) Calculate q for the reaction. You must show your work.

2) Calculate the number of moles of NH4NO3(s) which reacted. You must show your work.

3) Calculate ΔH for the reaction in kJ/mol. You must show your work.

Answers

Answer:

Explanation:

NH₄NO₃ = NH₄⁺ +NO₃⁻

heat released by water = msΔ T

m is mass , s is specific heat and ΔT is fall in temperature

= 50 x 4.18 x ( 22 - 16.5 ) ( mass of 50 mL is 50 g )

= 1149.5 J .

This heat will be absorbed by the reaction above .

q for the reaction = + 1149.5 J

2 )

molecular weight of NH₄NO₃ = 80

No of moles reacted = 5/80 = 1 / 16 moles.

3 )

5 g absorbs 1149.5 J

80 g absorbs 1149.5 x 16 J

= 18392 J

= 18.392 kJ.

= + 18.392 kJ

ΔH = 18.392 kJ / mol

How to prepare ethanoic acid from ethane

Answers

anonymous
anonymous 5 years ago
First you chlorinate it in presence of light.
C2H6 + Cl2 ---hv -> C2H5Cl + HCl
Then you add aqeuos KOH to get C2H5OH
C2H5Cl+KOH-> C2H5OH+ KCl.
Then you add KMnO4 to get the rquired compound.
C2H5OH ----KMnO4 ---> CH3COOH.

C2h6 + O2 ----> CH3CooH

Identify the true statements about introns.a- they code for polypeptide proteinsb- they have a branch site located 20 to 50 nucleotides upstream of the 3' splice sitec- they end with the nucleotides AG at the 3' endd- they begin with the nucleotides GU at the 5' ende- they tend to be common in bacterial genes

Answers

Answer:

The answer is "Option b, c, and d".

Explanation:

In such a gene, Autosomes are also the sequence for code and transposable elements, not the series of encoding. Through the expression of genes, such fragments of its introns are split through protein complexes throughout the translation process. There has been no kenaf fiber in the genomes of prokaryotic cells.

The following data is given to you about a reaction you are studying: Overall reaction: 2A  D Proposed mechanism: Step 1 A + B  C (slow) Step 2 C + A  D + B (fast) [A]o = 0.500 M [B]o = 0.0500 M [C]o = 0.500 M [D]o = 1.50 M This reaction was run at a series of temperatures and it was found that a plot of ln(k) vs 1/T (K) gives a straight line with a slope of -982.7 and a Y intercept of -0.0726. What is the initial rate of the reaction at 298K?

Answers

Answer : The initial rate of the reaction at 298 K is, $8.6* 10^(-4)M/s$

Explanation :

The Arrhenius equation is written as:

$K=A* e^{(-Ea)/(RT)}$

Taking logarithm on both the sides, we get:

$\ln k=-(Ea)/(RT)+\ln A$ ............(1)

where,

k = rate constant

Ea = activation energy

T = temperature

R = gas constant = 8.314 J/K.mole

A = pre-exponential factor

The equation (1) is of the form of, y = mx + c i.e, the equation of a straight line.

Thus, if we plot a graph of $\ln k$ vs $(1)/(T)$ then the graph shows a straight line with negative slope. That means,

Slope of the line = $-(Ea)/(R)$

And,

Intercept = $\ln A$

As we are given that:

Slope of the line = -982.7 = $-(Ea)/(R)$

Intercept = -0.0726 = $\ln A$

Now we have to calculate the value of rate constant by putting the value of slope, intercept and temperature (298K) in equation 1, we get:

$\ln k=-(982.7)/(298)+(-0.0726)$

$\ln k=-3.37$

$k=0.0344s^(-1)$

The value of rate constant is, $0.0344s^(-1)$

Now we have to calculate the initial rate of the reaction at 298 K.

As we know that the slow step is the rate determining step. So,

The slow step reaction is,

$A+B\rightarrow C$

The expression of rate law for this reaction will be,

$Rate=k[A][B]$

As we are given that:

[A] = 0.500 M

[B] = 0.0500 M

k = $0.0344s^(-1)$

Now put all the given values in the rate law expression, we get:

$Rate=(0.0344)* (0.500)* (0.0500)$

$Rate= 8.6* 10^(-4)M/s$

Therefore, the initial rate of the reaction at 298 K is, $8.6* 10^(-4)M/s$

50 mL of 0.1 M acetic acid is mixed with 50 mL of 0.1 M sodium acetate (the conjugate base). The Ka of acetic acid is approximately 1. 74 X 10 -5. What is the pH of the resulting solution?