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Effectus [21]
3 years ago
6

Chemistry gcse

Chemistry
1 answer:
mart [117]3 years ago
8 0
Yes, anything with carbonate, hydrogen carbonate (bicarbonate) at the end is a carbonate.

Examples:NaHCO3 (Sodium hydrogen carbonate or Sodium bicarbonate)Na2CO3 (Sodium carbonate)
You might be interested in
PLEASE HELP!!!! WILL MARK BRAINLIEST!!!! A rigid steel container with a volume of 30 L is filled with oxygen to a pressure of 9.
GuDViN [60]

Answer:

12 atm

Explanation:

First, let us convert Celcius into Kelvin: 28.0 °C = 301.15 K and 129.0 °C = 402.15 K

For this question we must employ the Combined Gas Law: \frac{P_1V_1}{P_2V_2}=\frac{n_1RT_1}{n_2RT_2}, where P_1 is the initial pressure and P_2 is the new pressure.

We know that intitially, P=9 atm, V=30 L, and T=301.15K. From our problem, only temperature and pressure changes, while the number of moles, volume and the gas constant, R, stay the same, so they are irrelevant.

Thus, the filled out Combined Gas Law would be:

\frac{9 atm}{P_2}=\frac{301.15K}{402.15K}, where the volume, moles of gas, and R are cancelled out.

We can manipulate this equation to derive the new pressure. We find that

9atm≈0.74885P_2.  

This means that

P_2≈9/0.74885≈12 atm

4 0
2 years ago
On the axes provided, label pressure on the horizontal axis from O mb to 760 mb and volume on the vertical axis from O to 1 mL.
Delicious77 [7]

Answer:

  • Please, find the graph with the labels and points located on the axes in the picture attached.

Explanation:

This is how you meet all the instructions and some important comments to understand how this kind of graphs word:

<u>1) Label pressure on the horizontal axis from O mb to 760 mb and volume on the vertical axis from O to 1 mL. </u>

The horizontal axis is used to record the independent variable and the vertical axis is used to record the dependent variable. The axes most be properly labeled with the name of the variable and the units.

In this case the origin is the point (0,0) which means that the axes cross each other, perpendicularly, at a pressure of 0 mb and a volume of 0.0 mililiters.

<u>2) Assign values to axes divisions in such a way that you occupy almost all the space on both axes. </u>

A good graph searches to occupy the whole space on both cases; to do that, find the maximum value for each variable, pressure and volume, and choose the values of the marks.

The range of the pressure (horizontal axis) is [90, 760 mb], so you should choose big divisions (marks) of 100 mb, and assign 800 mb to the right most mark on the horizontal axis. Then, you can divide each interval of 100 mb into 10 spaces, with small divisions of 10 mb (my graph uses 4 spcaes, with small divisions of 25 mb, but I recommend you use small divisions of 10 mb).

The range of the volume (vertical axis) is [0.1, 0.8], so you should choose only divisions with value of 0.1 ml.

<u>3) Now locate and label the points: </u>

  • (90, 0.9) ⇒ 90 mb, 0.9 ml
  • (100, 0.8) ⇒ 100 mb, 0.8 ml
  • (400, 0.2) ⇒ 400 mb, 0.2 ml
  • (600, 0.15) ⇒ 600 mb, 0.15 ml
  • (760, 0.1) ⇒ 760 mb, 0.1 ml

The points of the kind (x, y) are called ordered pairs, which means that the order matters, because it has a meaning: the first number represents the independent variable and the second number represents the dependent variable.

So, in the point (90, 0.9), 90 is a pressure of 90 mb and 0.9 is a volume of 0.9 ml.

To locate (600, 0.15), since the horizontal marks have value of 0.1, you must locate the second coordinate of your point between the marks 0.1 and 0.2 ml.

With that you can now locate each point on your graph.

5 0
3 years ago
Read 2 more answers
The enzyme urease catalyzes the breakdown of urea in the body. Urease breaks urea down to 2NH3+CO2. This is an example of a hydr
Alika [10]

Answer:

Look at the picture.

Explanation:

On stage one binding of a substrate occurs (and also the geometry of active site may change) and water comes to the site. On stage two the hydrolisis takes place and on stage 3 products deabsorb from the enzyme.

8 0
3 years ago
Calculate the wavelength of a photon of light having a frequency of 3.50 x 1015 Hz.
Aliun [14]

Answer:

Wavelength, \lambda=8.57\times 10^{-8}\ m

Explanation:

Given that,

Frequency, f=3.5\times 10^{15}\ Hz

We need to find the wavelength of a photon of light. The relation between frequency and wavelength is as follows :

c=f\lambda\\\\\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{3.5\times 10^{15}}\\\\\lambda=8.57\times 10^{-8}\ m

So, the wavelength of the light is 8.57\times 10^{-8}\ m.

4 0
2 years ago
What type of mirror could have resulted in the image of the giraffe above?
natka813 [3]

Answer:

A convex mirror

Explanation:

Good luck on the test m8!

8 0
2 years ago
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