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olga_2 [115]
2 years ago
14

How much 2 M HBr is needed to neutralize 380 mL of 0.1 M NH4OH?

Chemistry
1 answer:
Nata [24]2 years ago
8 0

Answer:

19ml

Explanation:

trust me

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Micky Mo is a 64-year-old male admitted to the emergency room for asthma. His laboratory results are as follows: pH 7.31, pCO2 h
Salsk061 [2.6K]

Answer:

Micky Mo is suffering from respiratory acidosis.

Explanation:

The pCO2 level in micky"s body is higher than normal it means the excess amount of  CO2 will reacts with water to generate carbonic acid(H2CO3).

  On the other hand according to the question total HCO3- also higher than normal.As a result the excess HCO3- will react with proton to form carbonic acid which is in turn dissociate to generate CO2 and H2O to maintain normal acid base homeostasis.

 From that point of view it can be said Micky Mo is suffering from respiratory acidosis.

5 0
2 years ago
The isotope 64Cu has t1/2 of 12.7 hours. If the initial concentration of this isotope in an aqueous solution is 845 ppm, what wi
Kisachek [45]

Answer:

A = 679.2955 ppm

Explanation:

In this case, we already know that 64Cu has a half life of 12.7 hours. The expression to use to calculate the remaining solution is:

A = A₀ e^-kt

This is the expression to use. We have time, A₀, but we do not have k. This value is calculated with the following expression:

k = ln2 / t₁/₂

Replacing the given data we have:

k = ln2 / 12.7

k = 0.0546

Now, let's get the concentration of Cu:

A = 845 e^(-0.0546*4)

A = 845 e^(-0.2183)

A = 845 * 0.8039

A = 679.2955 ppm

This would be the concentration after 4 hours

7 0
2 years ago
Calculate the mass of water vapour present in a room of volume 400 m3 that contains air at 27 °C on a day when the relative humi
Alex73 [517]

Answer:

Mass of water = 6251. 706g or 6.25Kg

Explanation:

Relative humidity = (actual vapor pressure/saturation vapor pressure) * 100%

Actual vapor pressure, Pw = relative humidity * saturation vapor pressure

Pw = 60% * (35.6 *0.001)atm = 0.0216atm

Note: 1mbar = 0.001atm

Using the ideal gas equation: PV=nRT; where P = Pw= 0.02136atm, V= (400

* 1000)dm^3, R= 0.082 atmdm^3/kmol, T= (27+273)K, n = number of moles

Note: 1m^3 = 1000dm^3,R is the molar gas constant.

Making n subject of the formula, n = PV/RT

n= (0.02136 * 400000)/(0.082 * 300) = 347.317 moles

Mass (g) = number of moles (n) * molar mass

molar mass of water=18g

Mass of water = 347.317 * 18 = 6251. 706g or 6.25Kg

6 0
2 years ago
Identify the term that applies to each definition.
Illusion [34]

Answer:

A. Reference blank

B. Cuvettes

C. Transmittance

D. Absorbance

E. Wavelength

Explanation:

A reference blank is a sample prepared using the solvent and any other chemicals in the sample solutions, but not the absorbing substance.

A square-shaped container, typically made of quartz, designed to hold samples in a spectrophotometer is known as Cuvettes.

A measurement of the amount of light that passes through a sample or percentage of light transmitted by the sample, with the respective intensities of the incident and transmitted beams is called Transmittance.

The measurement of the amount of light taken in by a sample is known as Absorbance

The wavelength is also the distance travelled by the wave during a period of oscillation. In spectrophotometry, the unit is inversely proportional to energy and commonly measured in nanometers

6 0
2 years ago
a drop of water weighing 0.48 g condenses on the surface of a 55-g block of aluminum that is initially at 25C. if the heat durin
Gwar [14]

Mass of water vapor is 0.48 gms.

Weight of the aluminium block is 55 gms.

Heat of vaporization of water at 25 degree Celsius is 44.0 kJ/mol.

The amount of heat given by the condensation is:

q = Heat of vaporization × Mass of vapor / Molar mass of steam

= 44.0 kJ/mol ×0.48 g / 18 g/mol

= 1.173 kJ

Now the final temperature of the metal block is calculated by the formula:

q = m × c × ΔT

q = m × c × (T₂ - T₁)

Here, q is the amount of heat, m is the mass of the metal, c is the specif heat of the metal, T₁ is initial temperature, and T₂ is the final temperature.

Now, substituting the values we get,

1.173 kJ = 55 g × 0.903 J/g. ° C * (T₂ - 25°C)

1.173 × 10³ J = 55 g × 0.903 J/g. degree C × (T₂ - 25°C)

1.173 × 10³ = 49.665 ° C * (T₂ - 25° C)

T₂ = 49° C.

Thus, the final temperature of the metal block is 49° C.

8 0
2 years ago
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