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Alexxx [7]
3 years ago
8

What is the IMA of the lever pictured? 0.18 0.20 0.90 5

Chemistry
2 answers:
bekas [8.4K]3 years ago
8 0
1/5 would be your answer.
IceJOKER [234]3 years ago
6 0
<span>The ideal mechanical advantage represents the number of times the input force is multiplied under ideal conditions, that is with no friction. Actual mechanical advantage on the other hand stands for the number of times the input force is multiplied. 
Hence; IMA (ideal mechanical advantage)=Le/Lr
The Lr =0.3 +1.2 = 1.5 and Le= 0.3
       = 0.3/1.5 = 1/5;
therefore the correct answer is 0.2</span>
You might be interested in
CS2 (s) + 3 O2 (g) → CO2 (g) + 2 SO2 (g)
Natasha_Volkova [10]

Answer:

2.067 L ≅ 2.07 L.

Explanation:

  • The balanced equation for the mentioned reaction is:

<em>CS₂(g) + 3O₂(g) → CO₂(g) + 2SO₂(g),</em>

It is clear that 1.0 mole of CS₂ react with 3.0 mole of O₂ to produce 1.0 mole of CO₂ and 2.0 moles of SO₂.

  • At STP, 3.6 L of H₂ reacts with (?? L) of oxygen gas:

It is known that at STP: every 1.0 mol of any gas occupies 22.4 L.

<u><em>using cross multiplication:</em></u>

1.0 mol of O₂ represents → 22.4 L.

??? mol of O₂ represents → 3.1 L.

∴ 3.1 L of O₂ represents = (1.0 mol)(3.1 L)/(22.4 L) = 0.1384 mol.

  • To find the no. of moles of SO₂ produced from 3.1 liters (0.1384 mol) of hydrogen:

<u><em>Using cross multiplication:</em></u>

3.0 mol of O₂ produce → 2.0 mol of SO₂, from stichiometry.

0.1384 mol of O₂ produce → ??? mol of SO₂.

∴ The no. of moles of SO₂ = (2.0 mol)(0.1384 mol)/(3.0 mol) = 0.09227 mol.

  • Again, using cross multiplication:

1.0 mol of SO₂ represents → 22.4 L, at STP.

0.09227 mol of SO₂ represents → ??? L.

∴ The no. of liters of SO₂ will be produced = (0.09227 mol)(22.4 L)/(1.0 mol) = 2.067 L ≅ 2.07 L.

8 0
3 years ago
A
amm1812

Solution:

1) Separate out the half-reactions. The only issue is that there are three of them.

<span>Fe2+ ---> Fe3+ 
S2¯ ---> SO42¯ 
NO3¯ ---> NO</span>

How did I recognize there there were three equations? The basic answer is "by experience." The detailed answer is that I know the oxidation states of all the elements on EACH side of the original equation. By knowing this, I am able to determine that there were two oxidations (the Fe going +2 to +3 and the S going -2 to +6) with one reduction (the N going +5 to +2).

Notice that I also split the FeS apart rather than write one equation (with FeS on the left side). I did this for simplicity showing the three equations. I know to split the FeS apart because it has two "things" happening to it, in this case it is two oxidations.

Normally, FeS does not ionize, but I can get away with it here because I will recombine the Fe2+ with the S2¯ in the final answer. If I do everything right, I'll get a one-to-one ratio of Fe2+ to S2¯ in the final answer.

2) Balancing all half-reactions in the normal manner.

<span>Fe2+ ---> Fe3+ + e¯ 
4H2O + S2¯ ---> SO42¯ + 8H+ + 8e¯ 
3e¯ + 4H+ + NO3¯ ---> NO + 2H2O</span>

3) Equalize the electrons on each side of the half-reactions. Please note that the first two half-reactions (both oxidations) total up to nine electrons. Consequently, a factor of three is needed for the third equation, the only one shown below:

<span>3 [3e¯ + 4H+ + NO3¯ ---> NO + 2H2O]</span>

Adding up the three equations will be left as an exercise for the reader. With the FeS put back together, the sum of all the coefficients (including any that are one) in the correct answer is 15.

Problem #2: CrI3 + Cl2 ---> CrO42¯ + IO4¯ + Cl¯ [basic sol.]

Solution:

Go to this video for the solution

Problem #3: Sb2S3 + Na2CO3 + C ---> Sb + Na2S + CO

Solution:

1) Remove all the spectator ions:

<span>Sb26+ + CO32- + C ---> Sb + CO</span>

Notice that I did not write Sb3+. I did this to keep the correct ratio of Sb as reactant and product. It also turns out that it will have a benefit when I select factors to multiply through some of the half-reactions. I didn't realize that until after the solution was done.

2) Separate into half-reactions:

<span>Sb26+ ---> Sb 
CO32- ---> CO 
C ---> CO</span>

3) Balance as if in acidic solution:

<span>6e¯ + Sb26+ ---> 2Sb 
2e¯ + 4H+ + CO32- ---> CO + 2H2O 
H2O + C ---> CO + 2H+ + 2e¯Could you balance in basic? I suppose, but why?</span>

4) Use a factor of three on the second half-reaction and a factor of six on the third.

<span>6e¯ + Sb26+ ---> 2Sb 
3 [2e¯ + 4H+ + CO32- ---> CO + 2H2O] 
6 [H2O + C ---> CO + 2H+ + 2e¯]The key is to think of 12 and its factors (1, 2, 3, 4, 6). You need to make the electrons equal on both sides (and there are 12 on each side when the half-reactions are added together). You get 12 H+ on each side (3 x 4 in the second and 6 x 2 in the third). You get six waters with 3 x 2 in the second and 6 x 1 in the third.Everything that needs to cancel gets canceled!</span>

5) The answer (with spectator ions added back in):

<span>Sb2S3 + 3Na2CO3 + 6C ---> 2Sb + 3Na2S + 9CO</span>

6) Here's a slightly different take on the solution just presented.

<span>a) Write the net ionic equation:<span>Sb26+ + CO32- + C ---> Sb + CO</span>b) Notice that charges must be balanced and that we have zero charge on the right. So, do this:<span>Sb26+ + 3CO32- + C ---> Sb + CO</span>c) Now, balance for atoms:<span>Sb26+ + 3CO32- + 6C ---> 2Sb + 9CO</span>d) Add back the sodium ions and sulfide ions to recover the molecular equation.<span>Sb2S3 + 3Na2CO3 + 6C ---> 2Sb + 3Na2S + 9CO</span></span>

7) Here's a discussion of a wrong answer to the above problem.

However, after reading the above wrong answer example, look at problem #10 below for an instance of having to add in a substance not included in the original reaction.

Problem #4: CrI3 + H2O2 ---> CrO42¯ + IO4¯ [basic sol.]

Solution:

1) write the half-reactions:

<span>Cr3+ ---> CrO42¯ 
I33¯ ---> IO4¯ 
H2O2 ---> H2O</span>

I wrote the iodide as I33¯ to make it easier to recombine it with the chromium ion at the end of the problem.

2) Balance as if in acidic solution:

<span>4H2O + Cr3+ ---> CrO42¯ + 8H+ + 3e¯ 
12H2O + I33¯ ---> 3IO4¯ + 24H+ + 24e¯ 
2e¯ + 2H+ + H2O2 ---> 2H2O</span>

I used water as the product for the hydrogen peroxide half-reaction because that gave me a half-reaction in acid solution. It will all go back to basic at the end of the problem.

3) Recover CrI3 by combining the first two half-reactions from just above:

<span>16H2O + CrI3 ---> 3IO4¯ + CrO42¯ + 32H+ + 27e¯</span>

4) Equalize the electrons:

<span>2 [16H2O + CrI3 ---> 3IO4¯ + CrO42¯ + 32H+ + 27e¯] 
27 [2e¯ + 2H+ + H2O2 ---> 2H2O]leads to:32H2O + 2CrI3 ---> 6IO4¯ + 2CrO42¯ + 64H+ + 54e¯ 
54e¯ + 54H+ + 27H2O2 ---> 54H2O</span>

5) Add the half-reactions together. Strike out (1) electrons, (2) hydrogen ion and (3) water. The result:

<span>2CrI3 + 27H2O2 ---> 2CrO42¯ + 6IO4¯ + 10H+ + 22H2O</span>

6) Add 10 hydroxides to each side. This makes 10 more waters on the right, so combine with the water alreadyon the right-hand side to make 32:

<span>2CrI3 + 27H2O2 + 10OH¯ ---> 2CrO42¯ + 6IO4¯ + 32H2O</span>



3 0
3 years ago
HgS + O2 → HgO + SO2
Igoryamba

Answer:

2HgS + 3O2 → 2HgO + 2SO2

The coefficients are: 2, 3, 2, 2

Explanation:

HgS + O2 → HgO + SO2

The equation can be balance as follow:

Put 3 in front of O2 as shown below:

HgS + 3O2 → HgO + SO2

Now we can see that there are 6 atoms of O on the left side of the equation and a total of 3 atoms on the right side. It can be balance by putting 2 in front of HgO and SO2 as shown below:

HgS + 3O2 → 2HgO + 2SO2

Now we have 2 atoms of both Hg and S on the right side and 1atom each on the left. It can be balance by putting 2 in front of HgS as shown below:

2HgS + 3O2 → 2HgO + 2SO2

Now the equation is balanced.

The coefficients are: 2, 3, 2, 2

The law of conservation of mass(matter) states that matter(mass) can neither be created nor destroyed during a chemical reaction but changes from one form to another. An unbalanced equation suggests that matter has been created or destroyed. While a balanced equation proofs that matter can never be created but changes to different form. This is the more reason we have count the atoms of an element on both side of the equation to see if they are balanced irrespective of the new form they assume in the product

5 0
3 years ago
When the pH of a solution changes from a pH of 5 to a pH of 3, the hydronium ion concentration is
gizmo_the_mogwai [7]

Answer : The correct option is, (D) 100 times the original content.

Explanation :

As we are given the pH of the solution change. Now we have to calculate the ratio of the hydronium ion concentration at pH = 5 and pH = 3

As we know that,

pH=-\log [H_3O^+]

The hydronium ion concentration at pH = 5.

5=-\log [H_3O^+]

[H_3O^+]=1\times 10^{-5}M      ..............(1)

The hydronium ion concentration at pH = 3.

3=-\log [H_3O^+]

[H_3O^+]=1\times 10^{-3}M      ................(2)

By dividing the equation 1 and 2 we get the ratio of the hydronium ion concentration.

\frac{[H_3O^+]_{original}}{[H_3O^+]_{final}}=\frac{1\times 10^{-5}}{1\times 10^{-3}}=\frac{1}{100}

100\times [H_3O^+]_{original}=[H_3O^+]_{final}

From this we conclude that when the pH of a solution changes from a pH of 5 to a pH of 3, the hydronium ion concentration is 100 times the original content.

Hence, the correct option is, (D) 100 times the original content.

7 0
3 years ago
Read 2 more answers
Relative dating methods help scientists to _____.
klasskru [66]
<span>identify the order in which rock units formed</span>
8 0
3 years ago
Read 2 more answers
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