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iVinArrow [24]
2 years ago
6

In which 2 states of matter can a substance take the shape of its container?

Chemistry
2 answers:
Natasha2012 [34]2 years ago
5 0

Answer:

solid and liquid

Explanation:

Vinvika [58]2 years ago
3 0
The answer would be Solid and Liquid
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Combined gas law problem: a balloon is filled with 500.0 mL of helium at a temperature of 27 degrees Celsius and 755 mmHg. As th
Papessa [141]
The statement of the combined gas law for a fixed amount of gas is,
PV/T = constant
Here, the units of pressure and volume must be consistent and the temperature must be the absolute temperature (Kelvin or Rankine).
0.65 atm is equivalent to 494 mmHg
Using the equation:
(755 x 500) / (27 + 273) = (494 x V) / (-33 + 273)
V = 3396 ml = 3.4 liters
8 0
3 years ago
Which machine is the least simple? a bike a screw an inclined plane a lever
Ivenika [448]

Answer: A bike

A bike is composed of various simple machines. It has a wheel and axle as one component, and it also has screws to hold the various parts together, along with levers and pulleys that are connected to the pedals. So all of these simple machine concepts work together to help transport the rider from point A to point B.

The other answer choices of screw, inclined plane, and lever, are fairly simple machines that don't have many things going on at once compared to a bike.

4 0
3 years ago
Wat helps measure 26 ML of a liquid
kotegsom [21]

Answer:

6 liquids

Explanation:

because yea

8 0
3 years ago
Read 2 more answers
A solution of 0.0027 M K2CrO4 was diluted from 3.00 mL to 100. mL. What is the molarity of the new solution?
Stells [14]

Answer:

[K₂CrO₄] → 8.1×10⁻⁵ M

Explanation:

First of all, you may know that if you dilute, molarity must decrease.

In the first solution we need to calculate the mmoles:

M = mmol/mL

mL . M = mmol

0.0027 mmol/mL . 3mL = 0.0081 mmoles

These mmoles  of potassium chromate are in 3 mL but, it stays in 100 mL too.

New molarity is:

0.0081 mmoles / 100mL = 8.1×10⁻⁵ M

4 0
2 years ago
Nitrogen and hydrogen combine to form ammonia in the Haber process. Calculate (in kJ) the standard enthalpy change ∆H° for the r
Kazeer [188]

Answer:

∆H° rxn = - 93 kJ

Explanation:

Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds  broken minus bonds formed (H according to Hess Law.

We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.

              N₂ (g)   +            3H₂ (g)   ⇒                          2NH₃ (g)

1 N≡N = 1(945 kJ/mol)     3 H-H = 3 (432 kJ/mol)       6 N-H = 6 ( 389 kJ/mol)

∆H° rxn = ∑  H bonds broken  - ∑ H bonds formed

∆H° rxn = [ 1(945 kJ)   + 3 (432 kJ) ] - [ 6 (389 k J]

∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ

be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond,  N≡N,  we have to use this one .

8 0
3 years ago
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