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Vlad1618 [11]
3 years ago
8

Self registering thermometers:

Chemistry
2 answers:
Katyanochek1 [597]3 years ago
6 0

Answer:

ur answer would be (A)

Explanation:

son4ous [18]3 years ago
4 0

Answer:

record only the maximum and minimum temperatures they are exposed to

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The volume of a given mass of agas is 360cm cubic at 50 degrees and 700 milimetres Hg .find it's volume at s.t.p​
saveliy_v [14]

Answer:

V₂  =279.9 cm³

Explanation:

Given data:

Initial volume = 360 cm³

Initial temperature = 50°C

Initial pressure = 700 mmHg

Final volume = ?

Final temperature = 273 k

Final pressure = 1 atm

Solution:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Solution:

<em>We will convert the mmHg to atm.</em>

700/760 = 0.92 atm

<em>and °C to kelvin.</em>

50+273 = 323 K

P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁ T₂/ T₁ P₂

V₂ = 0.92 atm × 360 cm³ × 273 K / 323 K ×1 atm

V₂  =  290417.6 atm .cm³.  K  / 323 k. atm

V₂  =279.9 cm³

8 0
3 years ago
Why does brining a turkey make it juicier?
olya-2409 [2.1K]

Answer:

hope this answer will help you.

6 0
3 years ago
Which is a diatomic molecule?<br> A. Ar<br> B. CO<br> C. CO2<br> D. NaCl
aleksley [76]
NaCI is the answer, in what my teacher taught me. Hope I helped :)
5 0
3 years ago
Does the law of inertia apply to both moving and non-moving objects
zhuklara [117]
Yes becuase it says it maintains its velocity if it is moving or at rest
8 0
3 years ago
A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?
Nastasia [14]

Answer:

At -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

Explanation:

Let's assume the gas behaves ideally.

As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

                                          \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

where P_{1} and P_{2} are initial and final pressure respectively.

           V_{1}  and V_{2} are initial and final volume respectively.

           T_{1} and T_{2} are initial and final temperature in kelvin scale respectively.

Here P_{1}=150.0kPa , V_{1}=1.75L , T_{1}=(273-23)K=250K, P_{2}=210.0kPa and V_{2}=1.30L

Hence    T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}

            \Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}

            \Rightarrow T_{2}=260K

            \Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}

So at -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

5 0
3 years ago
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