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cricket20 [7]
3 years ago
13

Express this amount in dozens. (Use decimal form.) 75 pies

Chemistry
1 answer:
MAXImum [283]3 years ago
3 0
You just divide 75 by 12
6.25 dozens
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Is water wet? I honestly really need to know-
Eva8 [605]

Answer:

yesh water is whet.

Explanation:

4 0
3 years ago
Read 2 more answers
Which of these is not true of Bohr's model of the atom ?
adelina 88 [10]
B and D are not true of Bohr's model of the atom. Bohr's model assumed that electron of a hydrogen atom travels in discrete elliptical orbits (not circular, as you have probably been taught) whose size depended on the allowed energy levels of the electron.
7 0
3 years ago
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If 35.5 mL of 0.23 M HCl is required to completely neutralize 20.0 mL of NH3, what is the concentration of the NH3 solution? Sho
V125BC [204]
NH3 is neutralised by the equation:

HCL + NH3 -> NH4CL

In this equation there is a one to one relationship in terms of the number of moles of each reactant. I.e. To neutralise 1 mole of NH3 we require 1 mole of HCL.

To calculate the concentration of NH3 required, we must first calculate the number of moles of HCL used.

volume HCL = 35.5mL = 0.0355 litres 
concentration HCL = 0.23M = 0.23 mole/litre 
Note that the term "M" for concentration simply means moles/litre
number moles = concentration x volume
number moles HCL = 0.0355 x 0.23 = 0.008165 moles HCL
based on the equation, we know the number of moles of NH3 must be the same

So, 
moles NH3, n = 0.008165
volume NH3, v = 20.0mL = 0.020 litres
n = c x v
c = n / v
c = 0.008165 / 0.020
=0.41
i.e. the concentration of NH3 would be 0.41 moles/litre or 0.41M

This intuitively makes sense because there is less volume of NH3 required to be neturalised, in a one-to-one mole relationship. So the concentration of NH3 would need to be higher than that of HCL.

6 0
3 years ago
How much energy (in J) is lost when a sample of iron with a mass of 26.4 g cools from 74.0 ∘C to 26.0 ∘C?
Greeley [361]

Answer:

Q=-526.6J

Explanation:

Hello,

In this case, for the computation of the energy loss when the cooling process is carried out, we use the shown below equation:

Q=mCp\Delta T

Whereas we need the mass, specific heat and change in temperature of iron within the process. Thus, the only value we need is the specific heat that is 0.444 J/(g°C), therefore, we compute the heat loss:

Q=26.4g*0.444\frac{J}{g\°C}*(26.0\°C-74.0\°C)\\ \\Q=-526.6J

Negative sign points out the loss due to the cooling.

Regards.

8 0
4 years ago
Write a paragraph explaining the effect of surface area on the rate of reaction.
nexus9112 [7]
(For a bit of context I will use the reaction between HCl and Mg as an example)

The larger the surface area of the magnesium metal, the more particles are exposed to collide with the aqueous HCl particles to cause the reaction to occur. This increases the frequency per second of collisions, speeding up the rate of reaction.

The effect of a catalyst is to reduce the minimum collision energy which allows the reaction to happen. This does not increase the number of collisions per second, but increases the percentage of successful collisions, which consequently causes the rate of reaction to increase .

I have drawn diagrams showing the effect of surface area, but there isn't really a meaningful diagram that I know of to show the impact of a catalyst (at least not at GCSE level).


6 0
3 years ago
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