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Georgia [21]
2 years ago
4

Yosef is playing with different kinds of rubber bands. Some are very narrow, and some are quite wide. Yosef is curious about the

rubber bands and develops this scientific question: Does the width of a rubber band affect how easily it can be stretched? He decides to develop a hypothesis to test this scientific question. What could Yosef’s hypothesis be?
Chemistry
2 answers:
Digiron [165]2 years ago
5 0
Yosef's hypothesis could be:
The larger the width of a rubber band, the harder it is to stretch it. 
umka2103 [35]2 years ago
3 0

Answer:

The correct answer will be-The narrow bands are easier to stretch compared to wider bands is the correct answer.

Explanation:

A scientific hypothesis is a prediction or an idea based on the previous knowledge which could be tested through experiments. The scientific hypothesis represents the limited explanation of the phenomenon.

In the given question, Yosef was playing with the rubber bands of various width from narrow to wide which lead to the observation that width of rubber bands affects the stretching ability or elasticity of the band.

After asking the question, Yosef could have done background research on the question and framed the hypothesis which could be the "The narrow bands are easier to stretch compared to wider bands".

Thus, The narrow bands are easier to stretch compared to wider bands" is the correct answer.

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Evaluate the following conversion. Will the answer be correct? Explain. If incorrect, how could you adjust one of the factors to
elena55 [62]

Given :

Rate = 75 m 1 s×60 s 1 min×1 h 60 m.

To Find :

Correct answer after conversion.

Solution :

We know, 1 min = 60 sec.

1 hour = 60 min = 60×60 sec = 3600 sec.

Putting value of min and hour in seconds , we get :

R=((75\times 60) + 1 )\times ( (60 + 1)\times 60 ) \times ( 3600+3600)\ s^3\\\\R=1.186\times 10^{11} \ s^3

Hence, this is the required solution.

6 0
2 years ago
How is data not actually obtained from the experiment represented in a line graph? with a double line with only dots with a colo
Fittoniya [83]
The way how <span>data is not actually obtained from the experiment represented in a line graph is defnitely that </span><span>a colored line with a broken line. It is a well known fact that to obtain the actual data from the experiment you there should be plotted points on the line. Hope it will help you! Regards.</span>
7 0
2 years ago
Read 2 more answers
A mass of 5.4 grams of aluminum (al) reacts with an excess of copper (ii) chloride (cucl2) in solution, as shown below. 3cucl2 2
horrorfan [7]

Answer:

Mass of copper produced is 19.07g

Explanation:

Let's bring out the chemical equation;

3CuCl2 + 2Al → 2AlCl3 + 3Cu

3           :   2       :   2       : 3

Upon confirming that the reaction is indeed balanced, we can proceed.

The questions asks to calculate mass of Cu formed when a mass of 5.4g of Al is being used.

From the equation, what is the relationship between Al and Cu?

2 mol of Al would react to form 3 mol of Cu

Expressing this in terms of mass, we have;

mass = no. of moles * molar mass

Mass of Aluminium = 2 * 26.98 = 53.98 g

Mass of Copper = 3 * 63.546 = 190.638g

This means;

53.98 g of Al would react to form 190.638g of Cu

So how much Cu would form from 5.4 g of Al?

This leads us to;

53.98 = 190.638

5.4 = x

Upon cross multiplication, we are left with;

x = (190.638 * 5.4) / 53.98

x = 19.07 g

Mass of copper produced is 19.07g

8 0
2 years ago
The decomposition of AB given here in this balanced equation 2AB (g)⟶ A2 (g) + B2 (g), has rate constants of 8.58 x 10-9 L/mol s
denis-greek [22]

Answer:

3.24 × 10^5 J/mol

Explanation:

The activation energy of this reaction can be calculated using the equation:

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

Where; Ea = the activation energy (J/mol)

R = the ideal gas constant = 8.3145 J/Kmol

T1 and T2 = absolute temperatures (K)

k1 and k2 = the reaction rate constants at respective temperature

First, we need to convert the temperatures in °C to K

T(K) = T(°C) + 273.15

T1 = 325°C + 273.15

T1 = 598.15K

T2 = 407°C + 273.15

T2 = 680.15K

Since, k1= 8.58 x 10-9 L/mol, k2= 2.16 x 10-5 L/mol, R= 8.3145 J/Kmol, we can now find Ea

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

ln(2.16 x 10-5/8.58 x 10-9) = Ea/8.3145 × (1/598.15 - 1/680.15)

ln(2517.4) = Ea/8.3145 × 2.01 × 10^-4

7.831 = Ea(2.417 × 10^-5)

Ea = 3.24 × 10^5 J/mol

8 0
2 years ago
Which conjugate pair is suited best to make this buffer? Which conjugate pair is suited best to make this buffer? Phosphoric aci
Juliette [100K]

Answer:

bkfhjjfrsxtr

Explanation:

4 0
2 years ago
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