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expeople1 [14]
2 years ago
6

what is the difference between aqueous hydrogen fluoride and hydrofluoric acid? do they have different formulas or are they just

the same but different names?
Chemistry
1 answer:
larisa86 [58]2 years ago
6 0

Answer:

Hydrofluoric acid is aqueous hydrogen fluoride (dissolved in water). Hydrogen fluoride will form hydrofluoric acid when it comes in contact with water.

Explanation:

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Which phase change results with increased kinetic energy?
weqwewe [10]

An increase in kinetic energy corresponds to an increase in temperature. Out of boiling, condensation, freezing, and precipitation, boiling is the only one that indicates an increase in temperature.

3 0
3 years ago
Order the elements from smallest to largest atomic number
postnew [5]

the largest is francium, Z=87 , or caesium, Z=55 ; and the smallest is definitely helium, Z=2 .

Explanation:

make me Brainliest

8 0
3 years ago
A eudiometer contains a 65.0 ml sample of a gas collected
SCORPION-xisa [38]

Answer:

53.1 mL

Explanation:

Let's assume an ideal gas, and at the Standard Temperature and Pressure are equal to 273 K and 101.325 kPa.

For the ideal gas law:

P1*V1/T1 = P2*V2/T2

Where P is the pressure, V is the volume, T is temperature, 1 is the initial state and 2 the final state.

At the eudiometer, there is a mixture between the gas and the water vapor, thus, the total pressure is the sum of the partial pressure of the components. The pressure of the gas is:

P1 = 92.5 - 2.8 = 89.7 kPa

T1 = 23°C + 273 = 296 K

89.7*65/296 = 101.325*V2/273

101.325V2 = 5377.45

V2 = 53.1 mL

6 0
3 years ago
The gram formula mass of NH4Cl is(1) 22.4 g/mole (3) 53.5 g/mole(2) 28.0 g/mole (4) 95.5 g/mole
g100num [7]
The answer is (3) 53.5 g/mol. The gram formula mass means that the mass of one mol compound. 1N=14, 4H=4, 1Cl=35.5. So the gram formula mass of NH4Cl=14+4+35.5=53.5 g/mol.
7 0
3 years ago
Read 2 more answers
a solution with a transmittance of 0.44 is analyzed in a spectrophotometer with 6% stray light. calculate the absorbance reporte
IgorLugansk [536]

The absorbance reported by the defective instrument was 0.3933.

Absorbance A = - log₁₀ T

Tm = transmittance measured by spectrophotometer

Tm = 0.44

Absorbance reported in this equipment = -log₁₀ (0.44) = 0.35654

True absorbance can be calculated by true transmittance, Tm = T+S(α-T)

S = fraction of stray light = 6%= 6/100 = 0.06

α= 1, ideal case

T = true transmittance of the sample

Tm = T+S(α-T)

now, T= Tm-S/ 1-S = 0.44-0.06/ 1-0.06 = 0.404233

therefore, actual reading measured is A = -log₁₀ T = -log₁₀ (0.404233)

i.e; 0.3933

To know more about transmittance click here:

brainly.com/question/17088180

#SPJ4

3 0
1 year ago
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