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Alona [7]
1 year ago
5

What are the two main classes of simple machines?

Chemistry
2 answers:
Vlad [161]1 year ago
7 0

Answer:

Palanca.

Torno.

Polea.

Plano inclinado.

Cuña.

Tornillo.

Explanation:

Len [333]1 year ago
6 0
The lever family and the inclined plane family.
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Will give brainliest if correct. Which of these bonds should be classified as a polar covalent bond?
quester [9]

Answer:

The answer is O-H.

Explanation:

This is because when you subtract their EN values you get 1.4, and that is in the range of the polar covalent bond values.

4 0
2 years ago
How many grams of kbr must be dissolved to make 25.0 ml of a solution that has a molarity of 0.85 m?
lyudmila [28]
Molarity is defined as the number of moles of solute in 1 L of solution 
molarity of solution to be prepared is 0.85 M
this means that there should be 0.85 mol of KBr in 1 L of solution 
if 1 L contains - 0.85 mol
then 25.0 mL should contain - 0.85 mol / 1000 mL x 25.0 mL = 0.0213 mol 
mass of KBr - 0.0213 mol x 119 g/mol = 2.53 g
mass of KBr that should be dissolved in 25.0 mL is 2.53 g

4 0
3 years ago
1,4-Pentadiene has a AHhydro = -254 kJ/mol while trans-1,3-pentadiene has a AHhydra = -226 kJ/mol. Explain this difference in he
julsineya [31]

Answer:

trans-1,3-pentadiene is more stable than 1,4-pentadiene due to presence of a conjugated double bond.

Explanation:

Here, \Delta H_{hydro}=H(hydrogenated pdt.)-H(diene)

H(hydrogenated pdt.) is same for both 1,4-pentadiene and 1,3-pentadiene as they both produce pentane after hydrogenation

H(diene) depends on stability of diene.

More stable a diene, lesser will be it's H(diene) value (more neagtive).

trans-1,3-pentadiene is more stable than 1,4-pentadiene due to presence of a conjugated double bond.

Hence, \Delta H_{hydro} is higher (less negative) for trans-1,3-pentadiene

5 0
3 years ago
The density of aluminum is 2.7 g/mL if I add a piece of aluminum to a graduated cylinder of water in the water level increases b
viva [34]
The answer would be 5.5g because you have to subtract 8.2 minus 2.7 and you get 5.5g so basically the answer is C.
8 0
3 years ago
The rate constant for the second-order reaction: 2NOBr(g) → 2NO(g) + Br2(g) is 0.80/(M · s) at 10°C. Starting with a concentrati
a_sh-v [17]

Answer : The concentration of NOBr after 95 s is, 0.013 M

Explanation :

The integrated rate law equation for second order reaction follows:

k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)

where,

k = rate constant = 0.80M^{-1}s^{-1}

t = time taken  = 95 s

[A] = concentration of substance after time 't' = ?

[A]_o = Initial concentration = 0.86 M

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

0.80=\frac{1}{95}\left (\frac{1}{[A]}-\frac{1}{(0.86)}\right)

[A] = 0.013 M

Hence, the concentration of NOBr after 95 s is, 0.013 M

4 0
2 years ago
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