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2 years ago

Is dihydrogen monoxide a real substance? Do you think it should be banned?

1 answer:

5 0

Yes it’s a real substance and yes it should be banned because it kills people without warning.It has no taste,smell,and you can’t see it.

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What is the nucleus of an atom split

larisa [96]

In nuclear physics and nuclear chemistry, nuclear fission is either a nuclear reaction or a radioactive decay process in which the nucleus of an atom splits<span> into smaller parts (lighter </span>nuclei<span>). Hope this helps</span>

4 0

3 years ago

Read 2 more answers

Using the following equation for the combustion of octane calculate the heat associated with the formation of 100.0 g of carbon

natali 33 [55]

Answer:

The right solution is "-602.69 KJ heat".

Explanation:

According to the question,

The 100.0 g of carbon dioxide:

= $\frac{100.0 \ g}{114.33\ g/mole}$

= $0.8747 \ moles$

We know that 16 moles of $CO_2$ formation associates with -11018 kJ of heat, then

0.8747 moles $CO_2$ formation associates with,

= $-\frac{0.8747}{16}\times 11018 \ KJ \ of \ heat$

= $-0.0547\times 11018$

= $-602.69 \ KJ \ heat$

8 0

2 years ago

A chemist mixes 75.0 g of an unknown substance at 96.5°C with 1,150 g of water at 25.0°C. If the final temperature of the system

AleksAgata [21]

To do this problem it is necessary to take into account that the heat given by the unknown substance is equal to the heat absorbed by the water, but considering the correct sign:

$-m\cdot c_e\cdot \Delta T = m_w\cdot c_e_w\cdot \Delta T_w$

Clearing the specific heat of the unknown substance:

$c_e = \frac{m_w\cdot c_e_w\cdot \Delta T_w}{m\cdot \Delta T} = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot (37.1 - 25.0)^\circ C}{75\ g\cdot (37.1 - 96.5)^\circ C}$

$c_e = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot 12.1^\circ C}{75\ g\cdot (-59.4)^\circ C} = \bf 13.07\frac{J}{g\cdot ^\circ C}$

6 0

2 years ago

Transferring or sharing electrons between atoms forms

3241004551 [841]

It forms something called a bond.

4 0

2 years ago

If 2.60 g of NaBr are dissolved in enough water to make 160. mL of solution, what is the molar concentration of

navik [9.2K]

This problem has two parts; the first one asking for the concentration of NaBr given both its mass and volume and the second one asking for its volume given both mass and concentration. The answers turn out to be 0.158 M and 211 mL.

<h3>Molarity</h3>

In chemistry, the use of units of concentration depends on both the substances to analyze and their amounts. In such a way, for molarity, one needs the following relationship between the moles of solute and volume of solution:

$M=\frac{n}{V}$

Thus, for the first part of the problem we first calculate the moles in 2.60 g of NaBr via its molar mass:

$2.60g*\frac{1mol}{102.89g} =0.0253mol$

Next, we convert the 160. mL to L by dividing by 1000 in order to obtain 0.160 L to subsequently calculate the molarity:

$M=\frac{0.0253mol}{0.160L}=0.158M$

Next, since the moles remain the same and for the second part we are asked for the volume given the concentration, one can solve for the volume so as to obtain:

$V=\frac{n}{M} =\frac{0.158M}{0.120mol/L}\\ \\V=0.211L$

That in milliliters turns out to be:

$V=0.211L*\frac{1000mL}{1L}=211mL$

Learn more about molarity: brainly.com/question/10053901

6 0

1 year ago