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ira [324]
1 year ago
6

Does the pH increase or decrease, and does it do so to a large or small extent, with each of the following additions?(c) 5 drops

of 0.1 M NaOH to 100 mL of 0.5 M HCl
Chemistry
1 answer:
Umnica [9.8K]1 year ago
6 0

The pH decreases to a large or small extent with each of the given additions.

<h3>What is common name of NaOH?</h3>

The common name of NaOH is sodium hydroxide. Lye and caustic soda are other names for sodium hydroxide, an inorganic substance having the formula NaOH. It is a white, solid ionic substance made up of the cations sodium (Na+) and the anions hydroxide (OH). Sodium hydroxide is a chemical that manufacturers utilize to make things like soap, rayon, paper, explosives, colors, and petroleum products. Processing cotton fabrics, metal cleaning and processing, oxide coating, electroplating, and electrolytic extraction are further uses for sodium hydroxide. A caustic metallic base is sodium hydroxide (NaOH), sometimes referred to as lye or caustic soda. Caustic soda, an alkali, is commonly employed in a variety of sectors, primarily as a potent chemical base in the production of pulp and paper, textiles, drinking water, and detergents.

To learn more about sodium hydroxide, visit:

brainly.com/question/24010534

#SPJ4

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Which of the elements must be available in water for the metamorphosis of tadpoles?
Grace [21]
The element iodine (I) is important for the fast and hastened metamorphosis of frog-tadpoles. As amphibians, the tadpoles can live in water and land but when they are born they are iodine-deficient. Tadpoles that do not receive ample amount of iodine become tadpoles until the end of their days. 

8 0
3 years ago
This is agree or disagree I need help ASAP!!
PIT_PIT [208]
All of them are agree :)
8 0
3 years ago
Read 2 more answers
PLEASE SOMEONE HELP ME WITH THIS!
Bad White [126]

Answer:

1) 1.202 L , 2) 1.291 dg , 3) 204.877  and 4) 1.04x10^{3\\}

Explanation:

You need to review about conversion factors and how to use them in the correct order. You can cancel the units and get the ones that you need if you use the appropriate conversion factors, remember is a number that you can use to multiply or divide.

For your exercise:

1) The conversion factor is: 1 L = 1000 mL

You will need to divide by 1000 mL to obtain liters L

1202.57120 mL x  \frac{1 L}{1000 mL} = 1.202 L

2) The conversion factor is: 1 g = 10 dg

0.1290743 g x \frac{10 dg}{ 1 g} = 1.291 dg

For the next exercises, you need to follow some rules:

1.  All numbers  that are different from Zero (non-zero digits) are significant figures.

2.The Zeros between non-zeros digits (Imbedded zeros) always are significant, 2007.

3. If you want to be specific and want some zeros to be significant you need to add a decimal point. For example 500. or 500.0

4. Leading zeros (to the left) are not significant.

5. Trailing zeros (zeros to the right) in a whole number without decimal point are not significant.

3) 843.062  - 638.1848  = 204.8772

Now if we round to 6 significant figures we get 204.877

4)123.0 x 8.43 = 1036.89

Now we round to 3 significant figures because 8.43 has the least significant figures.  

1.04x10^{3}

4 0
3 years ago
A carbon molecule that has a different arrangement of atoms is known as a/an
maw [93]
The correct response I believe is A. Isomer. If a carbon molecule possess more than one arrangement of how its atoms can be arranged, those other arrangements are known as isomers.
3 0
3 years ago
Never mind jhvjycdtrsesetdfyguhbjnk
OlgaM077 [116]

Complete Question

methanol can be synthesized in the gas phase by the reaction of gas phase carbon monoxide with gas phase hydrogen, a 10.0 L reaction flask contains carbon monoxide gas at 0.461 bar and 22.0 degrees Celsius. 200 mL of hydrogen gas at 7.10 bar and 271 K is introduced. Assuming the reaction goes to completion (100% yield)

what are the partial pressures of each gas at the end of the reaction, once the temperature has returned to 22.0 degrees C express final answer in units of bar

Answer:

The partial  pressure of  methanol is  P_{CH_3OH_{(g)}} =0.077 \  bar

The partial  pressure of carbon monoxide is  P_{CO} = 0.382 \ bar

The partial  pressure at  hydrogen is  P_H =  O \  bar

Explanation:

From the question we are told that

  The volume of the  flask is  V_f = 10.0 \  L

   The initial pressure of carbon monoxide gas is  P_{CO} = 0.461 \ bar

   The initial  temperature of carbon monoxide gas is T_{CO} = 22.0^oC

   The volume of the hydrogen gas is  V_h  =  200 mL = 200 *10^{-3} \  L

    The initial  pressure of the hydrogen is P_H  =  7.10 \  bar

    The initial temperature of the hydrogen  is  T_H = 271 \  K

The reaction of  carbon monoxide and  hydrogen is  represented as

         CO_{(g)} + 2H_2_{(g)} \rightarrow CH_3OH_{(g)}

Generally from the ideal gas equation the initial number of moles of carbon monoxide is  

        n_1  =  \frac{P_{CO} *  V_f }{RT_{CO}}

Here R is the gas constant with value  R  = 0.0821 \ L \cdot atm \cdot mol^{-1} \cdot K

=>     n_1  =  \frac{0.461  *  10 }{0.0821 * (22 + 273)}

=>     n_1  = 0.19

Generally from the ideal gas equation the initial number of moles of Hydrogen  is  

       n_2  =  \frac{P_{H} *  V_H }{RT_{H}}

      n_2  =  \frac{ 7.10 *  0.2 }{0.0821 * 271 }

=> n_2  =  0.064

Generally from the chemical equation of the reaction we see that

        2 moles of hydrogen gas reacts with 1 mole of CO

=>      0.064 moles of  hydrogen gas will react with  x  mole of  CO

So

          x = \frac{0.064}{2}

=>       x = 0.032 \ moles \ of  \  CO

Generally from the chemical equation of the reaction we see that

        2 moles of hydrogen gas reacts with 1 mole of CH_3OH_{(g)}

=>      0.064 moles of  hydrogen gas will react with  z  mole of  CH_3OH_{(g)}

So

          z = \frac{0.064}{2}

=>       z = 0.032 \ moles \ of  \ CH_3OH_{(g)}

From this calculation we see that the limiting reactant is hydrogen

Hence the remaining CO after the reaction is  

          n_k = n_1 - x

=>       n_k = 0.19  - 0.032

=>       n_k = 0.156

So at the end of the reaction , the partial pressure for  CO is mathematically represented as

      P_{CO} = \frac{n_k  *  R *  T_{CO}}{V}

=>    P_{CO} = \frac{0.158   *  0.0821 *  295}{10}

=>    P_{CO} = 0.382 \ bar

Generally the partial pressure of  hydrogen is  0 bar because hydrogen was completely consumed given that it was the limiting reactant

Generally the partial  pressure of the methanol is mathematically represented as

         P_{CH_3OH_{(g)}} = \frac{z  *  R *  T_{CO}}{V_f}

Here  T_{CO} is used because it is given the question that the   temperature  returned to 22.0 degrees C

So

      P_{CH_3OH_{(g)}} = \frac{0.03 * 0.0821 *  295}{10}

     P_{CH_3OH_{(g)}} =0.077 \  bar

6 0
3 years ago
Read 2 more answers
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