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iren2701 [21]
3 years ago
8

A block is pulled 0.90 m to the right in 4.5 s. What is the block's average speed to the nearest tenth of m/s?

Chemistry
2 answers:
pshichka [43]3 years ago
8 0

Answer:

-

Explanation:

jeka57 [31]3 years ago
5 0

Answer:

The average speed of the blocks are 0.36 m/s.

Explanation:

Average speed is defined as the ratio of distance covered per unit time. So if it is said that blocks are pulled to 0.9 m in the right side. This means the blocks cover a distance of 0.9 m from the origin and that distance is covered in 2.5 s. Thus, the average speed can be calculated from the change in speed with respect to time. As at time t = 0 , the speed is also zero, and at time t = 2.5 s , the speed will be.

Since, in this case, the speed is equal to the average speed of blocks. So the average speed of the blocks will be 0.36 m/s.

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How to balance equations for reduction/oxidation reaction?
KATRIN_1 [288]

The first step in balancing any redox reaction is determining whether or not it is even an oxidation-reduction reaction, which requires that species exhibits changing oxidation statesduring the reaction. To maintain charge neutrality in the sample, the redox reaction will entail both a reduction component and an oxidation components and is often separated into independent two hypothetical <span>half-reactions </span>to aid in understanding the reaction. This requires identifying which element is oxidized and which element is reduced. For example, consider this reaction:

<span><span><span>Cu(s)+2A<span>g+</span>(aq)→C<span>u<span>2+</span></span>(aq)+2Ag(s)</span>(1)</span><span>(1)<span>Cu(s)+2A<span>g+</span>(aq)→C<span>u<span>2+</span></span>(aq)+2Ag(s)</span></span></span>

The first step in determining whether the reaction is a redox reaction is to splitting the equation into two hypothetical half-reactions. Let's start with the half-reaction involving the copper atoms:

<span><span><span>Cu(s)→C<span>u<span>2+</span></span>(aq)</span>(2a)</span><span>(2a)<span>Cu(s)→C<span>u<span>2+</span></span>(aq)</span></span></span>

The oxidation state of copper on the left side is 0 because it is an element on its own. The oxidation state of copper on the right hand side of the equation is +2. The copper in this half-reaction is oxidized as the oxidation states increases from 0 in Cu to +2 in Cu2+. Now consider the silver atoms

<span><span><span>2A<span>g+</span>(aq)→2Ag(s)</span>(2b)</span><span>(2b)<span>2A<span>g+</span>(aq)→2Ag(s)</span></span></span>

In this half-reaction, the oxidation state of silver on the left side is a +1. The oxidation state of silver on the right is 0 because it is an element on its own. Because the oxidation state of silver decreases from +1 to 0, this is the reduction half-reaction.

Consequently, this reaction is a redox reaction as both reduction and oxidation half-reactions occur (via the transfer of electrons, that are not explicitly shown in equations 2). Once confirmed, it often necessary to balance the reaction (the reaction in equation 1 is balanced already though), which can be accomplished in two ways because the reaction could take place in neutral, acidic or basic conditions.


6 0
3 years ago
Determine the number of atoms and the mass of zirconium, silicon and oxygen found in 0.3384 Mol of zircon ZrSiO4, a
sattari [20]

Explanation:

1 mol of ZrSiO₄ , have , 1 mol of Zr , 1 mol of Si and  4 mol of  Oxygen .

since ,

1 mol of a substance have 6.022 * 10 ²³ particles ,

From the question ,

0.3384 mol of zircon ZrSiO₄ ,

hence ,

from the above equation ,

1 mol of a substance have 6.022 * 10 ²³ particles ,

0.3384 mol = 0.3384 * 6.022 * 10 ²³ number of atoms

                  = 2.037 * 10 ²³ number of atoms

Hence ,

0.3384 mol of ZrSiO₄ will have , 0.3384 mol of Zr ,0.3384 mol of Si and  4*0.3384 mol of  Oxygen .

Hence ,

  • number of atoms of Zr = 0.3384 mol * 6.022 * 10 ²³ = 2.037 * 10 ²³ number of atoms
  • number of atoms of Si = 0.3384 mol * 6.022 * 10 ²³ = 2.037 * 10 ²³ number of atoms
  • number of atoms of O = 4 * 0.3384 mol * 6.022 * 10 ²³ = 8.148 * 10 ²³ number of atoms

The mass is calculated as -

since ,

Moles is denoted by given mass divided by the molecular mass ,  

Hence ,  

n = w / m

n = moles ,  

w = given mass ,  

m = molecular mass .

n = w / m

w = n * m

  • mass of Zr ,

Since we know , the molecular mass of Zr = 91 g/mol

moles of Zr = 0.3384 mol

using the above formula ,

w = n * m

w = 0.3384 mol  * 91 g/mol = 30.79 g

hence ,

The mass of Zr = 30.79 g

  • mass of Si ,

Since we know , the molecular mass of Si = 28 g/mol

moles of Si = 0.3384 mol

using the above formula ,

w = n * m

w = 0.3384 mol  * 28 g/mol = 9.475 g

hence ,

The mass of Si = 9.475 g

  • mass of O ,

Since we know , the molecular mass of O = 16 g/mol

moles of Si = 1.3536 mol

using the above formula ,

w = n * m

w = 1.3536 mol  * 16 g/mol = 21.65 g

hence ,  

The mass of O = 21.65 g

8 0
3 years ago
A 0.01 M aqueous NaOH solution has a hydroxide concentration of ————— M?
MA_775_DIABLO [31]

Answer:

Has a hydroxide concentration of 0.01M

Explanation:

When NaOH dissociates, it gives 1mol Na+ and 1mol OH- this multiplies the concentration value and we get 0.01×1= 0.01M

4 0
3 years ago
How many grams of oxygen can be produced from the decomposition of 100. g of kclo3?
aivan3 [116]
<span>100 g of KClO3 @ 122.55 g/mol = 0.816 moles of KClO3
   by the reaction
 2 KClO3 --> 2 KCl & 3 O2

   0.816 moles of KClO3 @ 3 moles O2 / 2 moles KClO3 = 1.224 moles of O2 can be made using molar mass
 1.224 moles of O2 @ 32.0 g/mol =
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6 0
3 years ago
Read 2 more answers
A cylinder with a movable piston contains 2.00 g of helium, He, at room temperature. More helium was added to the cylinder and t
mr Goodwill [35]

Answer : The mass of helium added to the cylinder was, 1.5 grams

Explanation :

Avogadro's law : It is defined as the volume of gas is directly proportional to the number of moles of gas at constant pressure and temperature.

V\propto n

or,

\frac{V_1}{n_1}=\frac{V_2}{n_2}

where,

V_1 = initial volume of gas = 2.00 L

V_2 = final volume of gas = 3.50 L

n_1 = initial moles of gas = \frac{\text{Mass of He}}{\text{Molar mass of He}}=\frac{2.00g}{4g/mol}=0.5mol

n_2 = final temperature of gas = ?

Now put all the given values in the above equation, we get:

\frac{2.00L}{0.5mol}=\frac{3.50L}{n_2}

n_2=0.875mol

Now we have to calculate the mass of helium were added to the cylinder.

\text{Mass of He}=\text{Moles of He}\times \text{Molar mass of He}

\text{Mass of He}=0.875mol\times 4g/mol=3.5g

Mass of helium added = 3.5 - 2.00 = 1.5 g

Thus, the mass of helium added to the cylinder was, 1.5 grams

8 0
3 years ago
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