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3 years ago

What makes carbon monoxide

2 answers:

5 0

Carbon monoxide is a colorless, odorless, tasteless gas produced by burning gasoline, wood, propane, charcoal or other fuel. Improperly ventilated appliances and engines, particularly in a tightly sealed or enclosed space, may allow carbon monoxide to accumulate to dangerous levels.

ZanzabumX [31]3 years ago

3 0

Answer:

it comes from burning charcoal

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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

Calculate how many grams of the product form when 16.7 g of calcium metal completely reacts. Assume that there is more than enou

swat32

39.96 g product form when 16.7 g of calcium metal completely reacts.

<h3>What is the stoichiometric process?</h3>

Stoichiometry is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data.

Equation:

$Ca(s) + Cl_2(g)$ → $CaCl_2(s)$

In this case, for the undergoing reaction, we can compute the grams of the formed calcium chloride by noticing the 1:1 molar ratio between calcium and it (stoichiometric coefficients) and using their molar mass of 40 g/mol and 111 g/mol by using the following stoichiometric process:

$m_{ca_C_l_2}$ = 16.7 g Ca x $\frac{1 mol \;of \;Ca}{40g Ca}$ x $\frac{1 mol \;of \;CaCl_2}{1 mol \;Ca}$ x $\frac{111g of \;CaCl_2}{1 mol \;CaCl_2}$