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Len [333]
2 years ago
11

Which gas is produced when we add a pinch of baking soda in vinegar​

Chemistry
2 answers:
Luda [366]2 years ago
8 0
When you add a pinch of baking soda into vinegar, carbon dioxide gas is produced
iren [92.7K]2 years ago
7 0

Hey there!

When we add a pinch of baking soda to vinegar we can see the formation/production of <u>carbon </u><u>dioxide</u><u> </u><u>gas </u><u>(</u><u>CO2)</u><u>.</u>

Hope it helps ya!

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Balance the following reaction. A coefficient of "1" is understood. Choose option "blank" for the correct answer if the coeffici
goblinko [34]
 The balanced equation for

 Ca(OH)2  +  H3PO4→  Ca3(PO4)2  +  H2O    is

 3 Ca(OH)2  +2 H3PO4→  Ca3(Po4)2  +  6  H2O

3 moles of  Ca(OH)2  reacted with 2  moles of H3PO4  to  form  1 mole  of  Ca3(PO4)2   and  6  moles  of H2O
6 0
3 years ago
Read 2 more answers
What are three applications of the<br>coal ​
labwork [276]

Three uses of coal:

1. Electricity generation

2. Steel production

3. Cement manufacturing and as a liquid fuel

Hope it helps!

3 0
2 years ago
What would you multiply "moles of oxygen" by to get the units "grams of oxygen"?
photoshop1234 [79]

Answer:

AM

Explanation:

to go from moles to grams you multiply by the Atomic Mass or Molar Mass (Atomic Mass for an element and Molar Mass for a compound).

The formula is:

Mass = moles * MM

6 0
3 years ago
g Consider an ideal atomic gas in a cylinder. The upper part of the cylinder is a moveable piston of negligible weight. The heig
kumpel [21]

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

A cylindrical weight with a mass (m) of 3 kg is dropped, that is, its initial velocity (u) is 0 m/s and travels 10 m (s). Assuming the acceleration (a) is that of gravity (9.8 m/s²). We can calculate the velocity (v) of the weight in the instant prior to the collision with the piston using the following kinematic equation.

v^{2} = u^{2} + 2as = 2 (9.8 m/s^{2} ) (10m) \\\\v = 14 m/s

The object with a mass of 3 kg collides with the piston at 14 m/s, The kinetic energy (K) of the object at that moment is:

K = \frac{1}{2} m v^{2} = \frac{1}{2} (3kg) (14m/s)^{2} = 294 J

The kinetic energy of the weight is completely converted into heat transferred into the gas cylinder. Thus, Q = 294 J.

Given all the process is at 250 K (T), we can calculate the change of entropy of the gas using the following expression.

\Delta S_{gas} = \frac{Q}{T} = \frac{294 J}{250K} = 1.18 J/K

The change in the entropy of the environment, has the same value but opposite sign than the change in the entropy of the gas. Thus, \Delta S_{env} = -1.18 J/K

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

Learn more: brainly.com/question/22655760

6 0
2 years ago
Hydrogen's sole electron occupies the 1s orbital but can be excited to the 4p orbital. List all the orbitals this electron might
vesna_86 [32]
4p, 3d, 3p, 3s, 2p, 2s and 1s orbitals may be occupied during de-excitation.<span />
8 0
3 years ago
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