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vova2212 [387]
3 years ago
6

How can a prairie dog cause physical weathering

Chemistry
2 answers:
olga nikolaevna [1]3 years ago
6 0
A prairie dog can cause physical weathering by digging burrows.

Radda [10]3 years ago
5 0
• Animals such as prairie dogs, can cause physical weathering by digging burrows.
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"To determine the amount of heroin in the mixture, you dissolve 1.00 g of the white powdery mixture in water in a 100.0-mL volum
UkoKoshka [18]

Explanation:

Formula to calculate osmotic pressure is as follows.

 Osmotic pressure = concentration × gas constant × temperature( in K)

Temperature = 25^{o} C

                      = (25 + 273) K

                      = 298.15 K  

Osmotic pressure = 531 mm Hg or 0.698 atm     (as 1 mm Hg = 0.00131)

Putting the given values into the above formula as follows.

       0.698 = C \times 0.082 \times 298.15 K


               C = 0.0285

This also means that,

  \frac{\text{moles}}{\text{volume (in L)}} = 0.0285

So,     moles = 0.0285 × volume (in L)

                      = 0.0285 × 0.100

                     = 2.85 \times 10^{-3
}

Now, let us assume that mass of C_{12}H_{23}O_{5}N = x grams

And, mass of C_{12}H{22}O_{11} = (1.00 - x)

So, moles of C_{12}H_{23}O_{5}N = \frac{mass}{\text{molar mass}}

                              = \frac{x}{369}

Now, moles of C_{12}H_{22}O_{11} = \frac{(1.00 - x)}{342}

                   = \frac{x}{369} + \frac{(1.00 - x)}{342}

                  = 2.85 \times 10^{-3}

             = x = 0.346

Therefore, we can conclude that amount of C_{12}H_{23}O_{5}N present is 0.346 g  and amount of C_{12}H_{22}O_{11} present is (1 - 0.346) g = 0.654 g.

4 0
3 years ago
Pls help me I don’t know what to dooooo
Ivan
Molarity = moles of solute/volume of solution in liters.

The solute here is NaCl, of which we have 46.5 g. To calculate the molarity of an NaCl solution, we need to know the number of moles of NaCl. To convert from grams to moles, we divide the mass by the molar mass of NaCl. The molar mass of NaCl is the sum of the atomic masses of Na and Cl: 23 amu + 35 amu = 58 amu. For our purposes, we can regard amu as equivalent to grams/mole.

(46.5 g)/(58 g/mol) = 0.8017 moles NaCl.

Now that we know both the number of moles of our NaCl solute and the volume of the solution, we can calculate the molarity:

(0.8017 moles NaCl)/(2.2 L) = 0.364 M.
5 0
3 years ago
An individual is hospitalized and the initial blood work indicates high levels of hco3- in the blood and a ph of 7. 47. This wou
expeople1 [14]

An individual is hospitalized and the initial blood work indicates high levels of HCO_{3} ^{-} in the blood and a pH of 7. 47. This would indicate the individual probably has compensated respiratory acidosis.

A chronic illness usually leads to compensated respiratory acidosis because the kidneys have time to adjust to the delayed onset. Even if the PCO_{2} is elevated in a compensated respiratory acidosis, the pH is within the usual range.

The kidneys counteract a respiratory acidosis by increasing the amount of HCO_{3} that tubular cells reabsorb from the tubular fluid, the amount of H^{+} that collecting duct cells secrete while also producing HCO_{3} , and the amount of NH_{3} buffer that is formed through ammoniagenesis.

Respiratory acidosis is frequently brought on by hypoventilation as a result of: breathing depression , paralysis of the respiratory muscles, diseases of the chest wall , abnormalities of the lung parenchyma and abdominal squeezing.

Learn more about Respiratory acidosis here;

brainly.com/question/9694207

#SPJ4

5 0
1 year ago
17) A 0.20 M solution of HCl with a volume of 15.0 mL is exactly neutralized by the 0.10 M solution of NaOH with what volume? HI
galina1969 [7]

Answer: A 0.20 M solution of HCl with a volume of 15.0 mL is exactly neutralized by the 0.10 M solution of NaOH with 3 mL volume.

Explanation:

Given: M_{1} = 0.20 M,      V_{1} = 15.0 mL

M_{2} = 0.10 M,            V_{2} = ?

Formula used is as follows.

M_{1}V_{1} = M_{2}V_{2}

Substitute the values into above formula s follows.

M_{1}V_{1} = M_{2}V_{2}\\0.20 M ]times 15.0 mL = 0.10 M ]times V_{2}\\V_{2} = 30 mL

Thus, we can conclude that a 0.20 M solution of HCl with a volume of 15.0 mL is exactly neutralized by the 0.10 M solution of NaOH with 3 mL volume.

4 0
3 years ago
Calculate the pH of a buffer prepared by mixing 20.0 mL of 0.10 M acetic acid and 55.0 mL of 0.10 M sodium acetate
MatroZZZ [7]

Answer:

Calculate the pH of a buffer prepared by mixing 30.0 mL of 0.10 M acetic acid and 40.0 mL of 0.10 M sodium acetate.

3 0
3 years ago
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