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3 years ago

How many atoms are in 12 moles of iron?

1 answer:

3 0

Explanation: The atomic mass of iron (Fe) is 55.85 amu. The atomic mass of radium (Ra) is 226 amu. One mole of atoms of any element contains 6.022 X 1023 atoms, regardless of the type of element. The mass of one mole of an element depends on what that element is, and is equal to the atom mass of that element in grams.

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What color change is exhibited by phenolphthalein during a titration of aqueous acetic acid with aqueous sodium hydroxide ? a. c

Agata [3.3K]

Phenolphthalein during a titration

4 0

3 years ago

What is the concentration of H+ ions at a pH = 11?

natta225 [31]

Answer:

$\huge 1 × {10}^{-11} \: \: M$

Explanation:

The pH of a solution can be found by using the formula

$pH = - log [ {H}^{+} ]$

Since we are finding the H+ ions we find the antilog of the pH

So we have

$11 = - log({H}^{+}) \\ {H}^{+} = {10}^{ - 11}$

We have the final answer as

$1 × {10}^{-11} \: \: M$

Hope this helps you

6 0

3 years ago

Why in a rollercoaster when the cart is going up it decreases the speed of the cart but when it's going down it increases?

alukav5142 [94]

Because gravity is trying to pull it down not up and when it's going down it's going with gravity not against it.

8 0

3 years ago

PLEASE HELP ASAP :)

Vinvika [58]

is night i will answer you in the morning

8 0

3 years ago

Suppose that 3.33 g of acetone at 25.0 °C condenses on the surface of a 44.0-g block of aluminum that is initially at 25 °C. If

jeyben [28]

Answer:

68.6 °C

Explanation:

From conservation of energy, the heat lost by acetone, Q = heat gained by aluminum, Q'

Q = Q'

Q = mL where Q = latent heat of vaporization of acetone, m = mass of acetone = 3.33 g and L = specific latent heat of vaporization of acetone = 518 J/g

Q' = m'c(θ₂ - θ₁) where m' = mass of aluminum = 44.0 g, c = specific heat capacity of aluminum = 0.9 J/g°C, θ₁ = initial temperature of aluminum = 25°C and θ₂ = final temperature of aluminum = unknown

So, mL = m'c(θ₂ - θ₁)

θ₂ - θ₁ = mL/m'c

θ₂ = mL/m'c + θ₁

substituting the values of the variables into the equation, we have

θ₂ = 3.33 g × 518 J/g/(44.0 g × 0.9 J/g°C) + 25 °C

θ₂ = 1724.94 J/(39.6 J/°C) + 25 °C

θ₂ = 43.56 °C + 25 °C

θ₂ = 68.56 °C

θ₂ ≅ 68.6 °C

So, the final temperature (in °C) of the metal block is 68.6 °C.

5 0

3 years ago