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boyakko [2]
2 years ago
8

You are working with a concentrated solution of ammonium hydroxide. Which

Chemistry
1 answer:
yulyashka [42]2 years ago
7 0

Rubber gloves and other chemical resistant protective clothing are necessary when handling concentrated ammonium hydroxide.

<h3 /><h3>Why it is necessary to handle concentrated solution of ammonium hydroxide with care ?</h3>

  • Ammonia is an irritant and corrosive to the skin, eyes, respiratory tract and mucous membranes.

  • May cause severe chemical burns to the eyes, lungs and skin.

  • Skin and respiratory related diseases could be aggravated by exposure.

The extent of injury produced by exposure to ammonia depends on the duration of the exposure, the concentration of the liquid or vapor and the depth of inhalation.

Hence, Rubber gloves and other chemical resistant protective clothing are necessary when handling concentrated ammonium hydroxide.

<h3 />

Learn more about Ammonium Hydroxide here ;

brainly.com/question/3383763

#SPJ1

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If 5.100 g of c6h6 is burned and the heat produced from the burning is added to 5691 g of water at 21 °c, what is the final temp
emmasim [6.3K]
  <span>C6H12 = 6x12 + 6x1 = 78. 
The equation indicates that 2x78 = 156g benzene will produce 6542kJ. 
Using proportions you can then calculate that 
x/6542kJ = 7.9g / 156g 
x = 331.3kJ = 331300J. 

heat = mass x ΔT x 4.18J/g° 
ΔT = 331300J / (5691g x 4.18J/g°) = 13.9° 

final temp = 21 + 14° = 35°C</span>
4 0
4 years ago
Wet steam at 1100 kPa expands at constant enthalpy (as in a throttling process) to 101.33 kPa, where its temperature is 105°C. W
nikklg [1K]

Answer:

There are 3 steps of this problem.

Explanation:

Step 1.

Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.

Step 2.

Enthalpy of saturated liquid Haq = 781.124 J/g

Enthalpy of saturated vapour Hvap = 2779.7 J/g

Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g

Step 3.

In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy

So, H1=H2

     H2= (1-x)Haq+XHvap.........1

    Putting the values in 1

    2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}

                        = 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)

1904.976 (J/g) = x1998.576 (J/g)

                     x = 1904.976 (J/g)/1998.576 (J/g)

                     x = 0.953

So, the quality of the wet steam is 0.953

                   

7 0
3 years ago
Read 2 more answers
Gravity is due to:
Cerrena [4.2K]

Answer:

I'm probably wrong but I wanna say C.

Explanation:

5 0
4 years ago
4. Two double stranded fragments of DNA are exactly the same length. At 89 C, fragment A has completely denatured, which means t
nignag [31]

Answer:

The correct answer is "Fragment B likely has a higher Guanosine/Citosine content".

Explanation:

Guanosine/Citosine content, or GC content, refers to how many molecules of guanosine and citosine have a DNA fragment, respect to the content of adenine and thymine. The higher the GC content, the higher the temperature needed to denature the fragment of DNA. This happens because guanosine and citosine establish three hydrogen bonds, while adenine and thymine establish two hydrogen bonds when they bind together. Therefore, if fragment A and B are the same length, but at 89 C only fragment A is completely denatured, fragment B likely has a higher GC content.

4 0
3 years ago
CAN SOMEONE HELP ME PLZ AND THANKS WILL MARK U AS BRAINLIEST
jekas [21]

Explanation:

2. 2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O

First, we need to find the number of moles of CO_2 at 300K and 1.5 atm using the ideal gas law:

n= \dfrac{PV}{RT}= \dfrac{(1.5\:\text {atm})(33\:L)}{(0.082\:\text{L-atm/mol-K})(300K)}

=2.0\:\text{mol}\:CO_2

Now use the molar ratios to find the number of moles of ethane to produce this much CO_2.

2.0\:\text{mol}\:CO_2 \times \left(\dfrac{2\:\text{mol}\:C_2H_6}{4\:\text{mol}\:CO_2}\right)

=1.0\:\text{mol}\:C_2H_6

Finally, convert this amount to grams using its molar mass:

1.0\:\text {mol}\:C_2H_6 \times \left(\dfrac{30.07\:\text g\:C_2H_6}{1\:\text{mol}\:C_2H_6} \right)

=30.1\:g\:C_2H_6

3. 3Zn + 2H_3PO_4 \rightarrow 3H_2 + Zn_3(PO_4)_2

Convert 75 g Zn into moles:

75\:\text g\:Zn \times \left(\dfrac{65.38\:\text g\:Zn}{1\:\text{mol}\:Zn}\right)=1.1\:\text{mol}\:Zn

Then use the molar ratios to find the amount of H2 produced.

1.1\:\text{mol}\:Zn \times \left(\dfrac{3\:\text{mol}\:H_2}{3\:\text{mol}\:Zn}\right)=1.1\:\text{mol}\:H_2

Now use the ideal gas law PV=nRT to find the volume of H2 produced at 23°C and 4 atm:

V= \dfrac{nRT}{P}= \dfrac{(1.1\:\text{mol}\:H_2)(0.082\:\text{L-atm/mol-K})(296K)}{4\:\text{atm}}

=8.9\:\text L\:H_2

8 0
3 years ago
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