Convert 5.626 1 to cl.

What is the chemical name of the compound Na2CO3? Use the list of polyatomic ions and the periodic table to help you answer.

m_a_m_a [10]

Answer:

I believe the answer is B. Sodium carbonate

5 0

3 years ago

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I will give brainliest. If you burn the carbon in limited air, the reaction is

Fynjy0 [20]

This reaction is different in that the carbon undergoes an incomplete combustion as opposed to complete combustion where carbon is fully oxidized. A caveat: incomplete combustion products in general can be difficult to predict without sufficient information, as it's not uncommon to obtain a mixture of different products.

Here, we are told that solid carbon is burned in limited air to produce a gas. I am presuming that, in the equation that's given, the "0" represents a blank where you must fill in a chemical symbol. In this case, our equation would be: 2C(s) + O₂(g) → 2CO(g).

There is not enough information here to provide the numerical answers to the two questions. From the words in the question (e.g., "is different" and "this time"), it would seem that this question is an excerpt from a larger or preceding question where specific numbers had been provided or computed.

However, it's possible to make some general observations on how one may go about answering these questions if one had more information.

Since we're to assume that oxygen is the limiting reagent, if one is given the amount of solid carbon (either in mass, moles, or number of atoms), it's possible to determine the moles of CO(g) that's produced since C and CO have an equal stoichiometric ratio. So, for example, if one burns 2 moles of C(s), then 2 moles of CO(g) would be produced.

But, there is still not enough information to compute the volume of CO gas if this is the line of questioning. We don't know, for instance, the temperature or pressure of the reaction conditions. In fact, the only way it would be possible to answer this would be if you were given beforehand a conversion factor that relates the volume of CO(g) to its quantity (e.g., to assume that one mole of gas occupies x liters).

As for the second question, this would depend on what you know about the quantity of the C(s) reacted and/or the quantity (or volume, from question a) of CO(g) produced. If you can get the number of moles of C(s) reacted or CO(g) produced, the number of moles of O₂(g) used up: It would be half the number of moles of C(s) reacted or half the number of moles of CO(g) produced). Again, it's impossible to determine the volume of O₂(g) using just the information provided here, so I suspect that you must have further information relating gas quantity to volume. As we did with CO(g), the volume of O₂(g) used up can be found using whatever conversion factor you have.

If you have any further information or questions, please feel free to follow up.

6 0

2 years ago

Chapter 5. You must show all your work. Solve the following problems. (a) (8 points) When a cold drink is taken from a refrigera

GuDViN [60]

Answer:

see explanation below

Explanation:

To do this, we need to use the Newton's law of cooling which is:

dT/dt = (k - Ts)

Where:

Ts: temperature of surroundings (In this case, 20 °C)

k: constant

t: time

Now, after we do the integrals of this, the general expression would be:

T(t) = Ce^kt + Ts (1)

Now, with the first data, we need to calculate the value of C. This value will be the same after time has passed. You can see this as the concentration of the drink. As it's not experimenting any reaction, it's concentration remain the same, and only the temperature will change.

Now, to get the value of C, we can begin with the fact that at time = 0, temperature was 5°C so:

T(0) = Ce^k*0 + Ts

Replacing:

5 = Ce^1 + 20

5 - 20 = C*1

C = -15

Now that we have this, we can solve the first part of the problem

(i):

First, we need to get the value of k, we know the final temperature at t = 25, so we can solve for k, which is constant too, and then, calculate the temperature for t = 50 min

solving for k, with T = 10 °C, C = -15, Ts = 20 °C and t = 25 min:

10 = -15e^25k + 20

10 - 20 = -15e^25k

-10/-15 = e^25k

ln(-10/-15) = 25k

k = -0.405465/25

k = -0.0162

Now that we have k, let's calculate T after t = 50

T = -15e^(-0.0162)*50 + 20

T = -6.67 + 20

T(50) = 13.33 °C

(ii)

For this part, we only need to solve for t:

18 = -15e^(-0.0162)t + 20

18 - 20 / -15 = e^-0.0162t

0.1333 = e^-0.0162t

ln(0.1333) = -0.0162t

t = -2.0145/-0.0162

t = 124.37 min

8 0

3 years ago

3% Hydrogen Peroxide has an oral LD50 of 900 mg/kg. Acetic Acid has an oral LD50 of 3310 mg/kg. Which one is more hazardous to c

Maurinko [17]

LD50 is defined as the lethal dose 50% which describes the amount of material required to kill 50% of the testing population. It is given in units of mg of chemical per kg of bodyweight of the recipient.

Comparing hydrogen peroxide and acetic acid, we see that peroxide has a lower LD50 of 900 mg/kg, with acetic acid having LD50 = 3310 mg/kg. When comparing LD50 values, the smaller value will be the more toxic compound. What this means is that in this case, a smaller amount of peroxide is required to kill 50% of the testing population compared to acetic acid.

Therefore, 3% hydrogen peroxide is more hazardous to consume.

7 0

3 years ago

Are solar and lunar eclipses processed by the movements of the Earth and moon?

Inga [223]

Explanation:

As the Earth rotates on its axis and revolves around the Sun, several different effects are produced. When the new moon comes between the Earth and the Sun along the ecliptic, a solar eclipse is produced. When the Earth comes between the full moon and the Sun along the ecliptic, a lunar eclipse occurs.

7 0

3 years ago

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