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kherson [118]
2 years ago
5

What is the temperature of the gas inside a 750 ml balloon filled with 0.30 g hydrogen gas? The pressure of the balloon is 1.2 a

tm.
Chemistry
1 answer:
zhenek [66]2 years ago
4 0
Use the ideal gas law: PV = nRT, and solve for T,

T = PV/nR.

We should convert any of our parameters to units that would be easy to use with a known R value. One option would be to have the pressure in atm, volume in L, and convert the mass of the hydrogen gas to moles:

P = 1.2 atm
V = 750 mL = 0.750 L
n = (0.30 g H2)/(2.0159 g/mol) = 0.1488 mol H2
R = 0.0821 L•atm/mol•K.

Solving for T,

T = (1.2 atm)(0.750 L)/(0.1488 mol H2)(0.0821 L•atm/mol•K) = 73.671 K.

If you opt to leave your temperature in Kelvin, I would go with 74 K, as that has two significant figures like the rest of the values.

If you opt to go with Celsius, then the answer would be a bit awkward if you had to follow two sig figs: 73.671 - 273.15 = -199.49 ≈ -2.0 × 10² °C.

Pick your poison, I suppose.
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A square chunk of plastic has a length of 5 cm, width of 5 cm and height of 5 cm. It has a mass of 200 g. What is it’s density
vichka [17]

Answer:

1.6g/mL

Explanation:

Density equation is D=m/v

Density = g/mL

m=mass of sample in grams

v = volume of sample in mL

The volume of a square can be calculated by V=l*w*h.

In this case it is 5cm*5cm*5cm = 125cm^3

Since we know that 1cm^3 ~ 1mL we can convert the volume to mL as so:

125cm^3 (1mL/(1cm^3)) = 125mL

Then simply plug into the density equation:

D=200g/125mL = 1.6g/mL

3 0
2 years ago
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WITCHER [35]

I hope it helps you ❤️❤️❤️❤️

4 0
3 years ago
Read 2 more answers
4. DBearded waste of Co-60 must be stored until it is no longer radioactive. Cobalt-60
Bingel [31]

464 g radioisotope was present when the sample was put in storage

<h3>Further explanation</h3>

Given

Sample waste of Co-60 = 14.5 g

26.5 years in storage

Required

Initial sample

Solution

General formulas used in decay:  

\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}

t = duration of decay  

t 1/2 = half-life  

N₀ = the number of initial radioactive atoms  

Nt = the number of radioactive atoms left after decaying during T time  

Half-life of Co-60 = 5.3 years

Input the value :

\tt 14.5=No.\dfrac{1}{2}^{26.5/5.3}\\\\14.5=No.\dfrac{1}{2}^5\\\\No=\boxed{\bold{464~g}}

8 0
2 years ago
Assuming that the experiments performed in the absence of inhibitors were conducted by adding 5 μl of a 2 mg/ml enzyme stock sol
prohojiy [21]

Hey there!:

From the given data ;

Reaction  volume = 1 mL , enzyme content = 10 ug ( 5 ug in 2 mg/mL )

Enzyme mol Wt = 45,000 , therefore [E]t is 10 ug/mL , this need to be express as "M" So:

[E]t in molar  = g/L * mol/g

[E]t  = 0.01 g/L * 1 / 45,000

[E]t = 2.22*10⁻⁷

Vmax = 0.758 umole/min/ per mL

= 758 mmole/L/min

=758000 mole/L/min => 758000 M

Therefore :

Kcat = Vmax/ [E]t

Kcat = 758000 / 2.2*10⁻⁷ M

Kcat = 3.41441 *10¹² / min

Kcat = 3.41441*10¹² / 60 per sec

Kcat = 5.7*10¹⁰ s⁻¹

Hence   kcat of   xyzase is  5.7*10¹⁰ s⁻¹


Hope that helps!



4 0
3 years ago
How can endothermic and exothermic reactions help others (society)
AnnZ [28]
Here try this presentation, it should help :) 

https://prezi.com/6ofdo3fadl9a/exothermic-and-endothermic-reactions-in-everyday-life/
4 0
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