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Vadim26 [7]
3 years ago
13

2 . what is the formula weight of (nh4)2so4? 118 amu 116 amu 100 amu 132 amu

Chemistry
1 answer:
Nataly [62]3 years ago
6 0
Hey there!

(NH₄)₂SO₄ = 14 * 2 + 1 * 8 + 32+ 16 * 4 => 132 amu


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Substance A has the following properties.
givi [52]

A curve of temperature vs. time for the entire heating process.

The sample is heated up to 100.°C, therefore, the heat and time required to heat the sample to its boiling point, the heat and time required to boil the sample, and the heat and time required to heat the sample from its boiling point to 100.°C are needs to be calculated.

i ) Calculating the heat and time required to heat the sample to its boiling point:

Boiling point = 85°C

C(liquid) = 2.5 J/g °C

The heat required up to melting the sample is calculated in the previous parts. Therefore, the heat required to heat the sample from -20°C to 85°C can be calculated as,

Therefore, T f = 85°C  and T i = - 20°C

Plug in the values in the specific heat formula to calculate the heat energy required to heat the sample to its melting point,

q3 = 25 g ×  2.5 J/g °C × [85 - (-20)]°C

     = 25 J/°C ×[85+20]°C

     = 6562.5 J

The total heat energy required for heating the sample from initial temperature to boiling point is:-

q1 + q2 + q3 = 500 J + 4500 J + 6562.5 J

                    = 11562.5 J

The Rate of heating = 450 J/min

450. J = 1 min

   11562.5 J = ? min

11562.5 J × 1min/450 J = 25.69 min

ii) Calculating the heat and time required to boil the sample:

∆H Vap = 500 J/g

The boiling is the phase change from liquid to gas at 85°C, therefore, the heat required to boil the sample can be determined

q4= m × ∆Hvap

    = 25 g × 500 J/g

   = 12500 J

Thus, total heat required to this phase change is q1 + q2 + q3 + q4  = 500 J + 4500 J +6562.5  J + 12500 J = 24062.5 J

The Rate of heating = 450 J / min

450 J = 1 min

24062.5 J = ? min

24062.5J ×  1min / 450 J = 53.47 min

iii) Calculating the heat and time required to heat the sample from its boiling point to 100°C

C gas = 0.5 J / g °C

The heat required to boil the sample is calculated in the previous parts. Therefore, the heat required to heat the sample from 85°C to 100°C can be calculated as,

Therefore, T f = 100.°C  and T i = 85°C

q5 = 25 g ×  0.5 J / g °C × [100 - 85] °C

    = 25 J / °C ×15 °C

    = 187.5  J

The total heat energy required for heating the sample from initial temperature to 100°C is

q1 + q2 + q3 + q4 + q5 = 500 J + 4500 J + 2625J + 12500 J + 187.5 J

                                      =24250 J

The Rate of heating = 450 J / min

  450. J = 1 min

 24250 J=? min

Thus, heating the sample to 100.°C takes a total of 53.89 min.

iv) Draw a curve of temperature vs. time for the entire heating process:-

Temperature °C     Temperature K     Heat energy (J)     Time (min)

 -40 °C                       233                             0                     0

-20 °C                          253                          500                  1.11    

Melting -20 °C             253                        5000                   11.11

85 °C                         358                         11562.5              25.69

Boiling 85 °C             358                           24062.5          53.475              

100  °C                       373                             24250          53.89

Hence, the graph for the result is in the image.

Learn more about temperature here:-brainly.com/question/24746268

#SPJ4

4 0
1 year ago
How much heat, in joules, must be added to a 75.0-g iron block with a specific heat of 0.449 J/g °C to increase its temperature
svp [43]

Answer:

50,849.25 Joules

Explanation:

The amount of heat, Q, required to raise the temperature of a body with mass, m, and specific heat capacity, c is given by:

Q = mcΔT, where ΔT represents the change in temperature.

In the case of the iron block:

m = 75 g

c = 0.449 J/g °C

ΔT = 1535 - 25 = 1510 °C

Therefore,

Q = 75 g x 0.449 J/g °C x 1510 °C

     = 50,849.25 Joules

<em>Hence, </em><em>50,849.25 Joules </em><em> of heat must be added to  a 75.0-g iron block with a specific heat of 0.449 J/g °C to increase its temperature from 25 °C to its melting temperature of 1535 °C</em>

5 0
2 years ago
In a 0.20 M solution, a weak acid is 3.0% dissociated.(a) Calculate the [H₃O⁺] , pH , [OH⁻], and pOH of the solution.
marin [14]

H₃O⁺ = 6.0×10^{-3}

OH = 1.7×10^{-12}

pH = 2.22

pOH = 11.78

<h3>What is pH?</h3>

The term pH, which originally stood for "potential of hydrogen" (or "power of hydrogen"), is used in chemistry to describe how acidic or basic an aqueous solution is. Lower pH values are summarized for acidic solutions (solutions with higher H+ ion concentrations) than for basic or alkaline solutions.

The pH scale is inversely indicates to the concentration of hydrogen ions in the solution and is logarithmic.

⇒pH = -log(a_{H+})

Acidic solutions are those with a pH below 7, and basic solutions are those with a pH above 7, at a temperature of 25 °C (77 °F). At this temperature, solutions with a pH of 7 are neutral (e.g. pure water). The pH neutrality relies on temperature, falling below 7 if the temperature rises above 25 °C.

Learn more about pH

brainly.com/question/12609985

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8 0
2 years ago
What is the vapor pressure of CS2CS2 in mmHgmmHg at 26.5 ∘C∘C? Carbon disulfide, CS2CS2, has PvapPvap = 100 mmHgmmHg at −−5.1 ∘C
Digiron [165]

Answer: 26.5 mm Hg

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1= initial pressure at 26.5^oC = ?

P_2 = final pressure at -5.1^oC = 100 mm Hg

= enthalpy of vaporisation = 28.0 kJ/mol =28000 J/mol

R = gas constant = 8.314 J/mole.K

T_1= initial temperature = 26.5^oC=273+26.5=299.5K

T_2 = final temperature =-5.1^oC=273+(-5.1)=267.9K

Now put all the given values in this formula, we get

\log (\frac{P_1}{100})=\frac{28000}{2.303\times 8.314J/mole.K}[\frac{1}{299.5}-\frac{1}{267.9}]

\log  (\frac{P_1}{100})=-0.576

\frac{P_1}{100}=0.265

P_1=26.5mmHg

Thus the vapor pressure of CS_2CS_2 in mmHg at 26.5 ∘C is 26.5

7 0
3 years ago
What happens to a gas that is enclosed in a rigid container when the temperature of the gas is increased?
yanalaym [24]
A) - heat makes particles move faster, which usually makes the substance as a whole expand, but if the gas cannot expand the pressure will increase instead
7 0
3 years ago
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