What happens in a subduction zone?

Scientific theories are usually

jasenka [17]

Answer:

B

Explanation:

Cuz they are supported once discovered and then widely spread then accepted

4 0

3 years ago

Read 2 more answers

In Experiment 9, a 1.05 g sample of a mixture of NaCl (MW 58.45) and NaNO2 (MW 69.01) is reacted with excess sulfamic acid. The

Kaylis [27]

Answer:

There is 76.6 mL of nitrogen collected

Explanation:

Step 1: Data given

Mass of the sample = 1. 05 grams

The sample is 40.00% by mass NaNO2

MW of NaCl = 58.45 g/mol

MW of NaNO2 = 69.01 g/mol

Temperature = 22.0 °C

Pressure = 750.0 mmHg = (750/760) atm

The vapor pressure of water at 22.0°C is 19.8 mm Hg = 0.02605 atm

Step 2: Calculate mass of NaNO2

(40/100)*1.05 = 0.42 grams

Step 3: Calculate moles of NaNO2

Moles NaNO2 = 0.42 grams / 69.01 g/mol

Moles NaNO2 = 0.00608 moles

Step 4: Calculate moles of N2

For 2 moles of NaNO2 we'll get 1 mol of N2

For 0.00608 moles of NaNO2 we'll get 0.00608/2 = 0.00304 moles

Step 5: Calculate pressure of N2

P = 750.0 - 19.8 = 730.2 mmHg = (730.2/760)atm = 0.96079 atm = 97352 Pa

Step 6: Calculate volume of N2

PV = nRT

⇒ P = the pressure of N2 = 0.96079

⇒ V = the volume of N2 = TO BE DETERMINED

⇒ n = moles of N2 = 0.00304 moles

⇒ R = the gas constant = 0.08206 L*atm/K*mol

⇒ T = the temperature = 22°C = 295 Kelvin

V = (0.00304*0.08206*295)/0.96079

V = 0.0766 L = 76.6 mL

There is 76.6 mL of nitrogen collected

4 0

3 years ago

How many grams of nano3 remain in solution at 20 ∘c?

Amanda [17]

the table say that at 20 degree celcius 88.0g of NANO3 will remain dissolved in 100 gm of H2O so at 20 degree celcius 80.0g of H20 will dissolve (88.0g)x(80g/100g)=70.4g of NANO3 so at 20 degree celcius 86.3g-70.4g= 15.9 gram of NANO3 will come out of solution.

8 0

2 years ago

1. 100gm of a 55% (M/M) nitric acid solution is to be diluted to 20% (M/M) nitric acid.

stealth61 [152]

Answer:

The volume of water to be added is 0.175 liters of water

Explanation:

The given concentration of the nitric acid = 55% (M/M)

The mass of the nitric acid solution = 100 gm

The concentration solution is to diluted to = 20% (M/M)

The 100 g 55%(M/M) nitric acid solution gives 55g nitric acid in 100 g of solution

Therefore, to have 20% (M/M) nitric acid solution with the 55 g nitric acid, we get

Let "x" represent the volume of the resulting solution, we have;

20% of x = 55 g of nitric acid

∴ 20/100 × x = 55 g

x = 55 g × 100/20 = 275 g

The mass of extra water to be added = The mass of the 20%(M/M) solution solution of nitric acid - The current mass of the 55%(M/M) solution of nitric acid

The mass of extra water to be added = 275 g - 100 g = 175 g

Volume = Mass/Density

The density of water ≈ 1 g/ml

∴ The volume of water to be added that gives 175 g of water = 175 g/(1 g/ml) = 175 ml. = 0.175 l

The volume of water to be added = 0.175 liters of water.

3 0

2 years ago

A chemist prepares a solution of mercury(I) chloride Hg2Cl2 by measuring out

Genrish500 [490]

Answer:

1.26 × 10^-8 M

Explanation:

We are given;

Number of moles of mercury (i) chloride as 0.000126 μmol

Volume is 100 mL

We are required to calculate the concentration of the solution.

We need to know that;

Concentration is also known as molarity is given by;

Molarity = Number of moles ÷ Volume

Number of moles = 1.26 × 10^-10 Moles

Volume = 0.01 L

Therefore;

Concentration = 1.26 × 10^-10 Moles ÷ 0.01 L

= 1.26 × 10^-8 M

Thus, the molarity of the solution is 1.26 × 10^-8 M

6 0

3 years ago