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Natali [406]
3 years ago
15

A weather balloon is released into the sky to take data. At ground level the temperature was 32 degrees celcius and the balloon

was filled to a volume of 267.4 L. The balloon rose into the sky in which the temperature dropped to 13 degrees celcius. What is the new volume of the balloon?
Chemistry
1 answer:
rjkz [21]3 years ago
8 0

Answer:I’m not sure

Explanation:free trial. oKay

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pickupchik [31]

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5 0
2 years ago
Identify the solute and solvent in a dilute aqueous solution of potassium chloride
erastova [34]
Solute=potassium chloride, solvent=water
6 0
2 years ago
PLEASE HELP I WILL GIVE BRAINLYEST!!!!
adoni [48]
I think that it is true to the bone
8 0
3 years ago
What is the ph of a 0.001 62 m naoh solution?
marin [14]
The answer 
first of all, we should know that NaOH is a strong base. For such a product, the conentration of the OH -  is equivalent to the concentration of the NaOH itself.
that means:
[ OH -] = [ NaOH] =<span>0.001 62 
and for a strong basis, pH can be calculated as pH = 14 + log </span>[ OH -] 
first we compute log [ OH -] :
log [ OH -] = log (0.001 62)= -2.79

finally pH = 14 -2.79 = 11.20
4 0
3 years ago
Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm8 at 27 ∘C. (R =
maxonik [38]

Answer:

Total pressure =  56.77 bar

Explanation:

Given data:

Mass of dioxygen = 8 g

Mass of dihydrogen = 4 g

Volume of vessel = 1 dm³

Temperature = 27°C (27+273 = 300 K)

R = 0.083 bar.dm³ / mol.K

Total pressure = ?

Solution:

Number of moles of dioxygen:

Number of moles = mass/molar mass

Number of moles = 8 g/ 32 g/mol

Number of moles = 0.25 mol

Pressure of dioxygen:

PV = nRT

P = nRT/V

P = 0.25 mol × 0.083 bar.dm³ / mol.K × 300 K / 1 dm³

P = 6.97 bar.dm³  /1 dm³

P = 6.97 bar

Number of moles of dihydrogen:

Number of moles = mass/molar mass

Number of moles = 4 g/ 2 g/mol

Number of moles = 2 mol

Pressure of dihydrogen:

PV = nRT

P = nRT/V

P = 2 mol × 0.083 bar.dm³ / mol.K × 300 K / 1 dm³

P = 49.8 bar.dm³  /1 dm³

P = 49.8 bar

Total pressure of mixture in a vessel:

Total pressure = P (O₂) + P(H₂)

Total pressure =  6.97 bar + 49.8 bar

Total pressure =  56.77 bar

7 0
3 years ago
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