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4 years ago

Which of the following words means "the thing from which something else comes"?

Chemistry

2 answers:

swat324 years ago

7 0

Answer:

b. source

Explanation:

The source is where something comes from. i.e. The source of sunlight is the sun.

grin007 [14]4 years ago

7 0

Answer:

PLZ HELP ME. This is my second time posting my question because the first time a person put a random answer. If you do not know the question please do not answer it and leave it to someone else. Thank you, and my question is on the attached image below

Explanation:

ng from which something else comes"?
a. fusion; b. source; c. attraction; d. force

hehehhehehehheheh teehee

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How much energy must be added to a 2-kg piece of wood with a specific heat

lesya692 [45]

Answer:

A. 140,800 J

Explanation:

Specific heat of a material is defined as the amount of energy required to increase in 1°C 1kg of the material. For wood there are required 1760J perkg per °C.

The equation is:

Q = C*m*ΔT

Where Q is heat in Joules,

C is specific heat (1760J/kg°C),

m is mass (2kg)

ΔT is change in temperature (50°C - 10°C = 40°C)

Replacing:

Q = 1760J/kg°C*2kg*40°C

Q = 140800J

Right option is:

6 0

3 years ago

A 29.00 mL sample of an unknown H3PO4 solution is titrated with a 0.130 M NaOH solution. The equivalence point is reached when 2

beks73 [17]

I am first assuming that you want the molarity of the acid. Use the formula $M_{a} V_{a} = M_{b}V_{b}$ which is the titration formula. M is molarity (a is of acid and b is of base) and V is volume in mL (a is of acid and b is of base). Plugging in gives us $(29 mL)(M_{a}) = (0.130 M)(27.73 mL)$ . Solving gives us $M_{a} = 0.124 M$

6 0

3 years ago

1. 17.0 grams of xenon hexafluoride is in a solid container. How many milliliters of that gas

BlackZzzverrR [31]

Answer: The volume of gas is 3020 ml

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 821.4 torr = 1.08 atm (760 torr = 1atm)

V = Volume of gas in L = ?

n = number of moles = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{17.0g}{245.28g/mol}=0.069mol$

R = gas constant = $0.0821Latm/Kmol$

T =temperature = $302.7^0C=(302.7+273)K=575.7K$

$V=\frac{nRT}{P}$

$V=\frac{0.069mol\times 0.0821Latm/K mol\times 575.7K}{1.08atm}=3.02L=3020ml$

Thus volume of gas is 3020 ml

4 0

3 years ago

What is the chemical name for F5N3

Alexandra [31]

Answer:

Fluorine azide or triazadienyl fluoride (FN3) is a yellow green gas composed of nitrogen and fluorine with formula FN3. It is counted as an interhalogen compound, as the azide functional group is termed a pseudohalogen. It resembles ClN3, BrN3, and IN3 in this respect.

Explanation:

3 0

3 years ago

Read 2 more answers

Calculate the volume of carbon dioxide at 20.0°C and 0.941 atm produced from the complete combustion of 4.00 kg of methane. Comp

tankabanditka [31]

Answer:

The volume of carbon dioxide at 20.0°C and 0.941 atm produced was 6390.89 Liters.

More volume of carbon dioxide gas was released on combustion of 4.00 kg of propane in comparison to the volume of carbon dioxide released on combustion of 4.00 kg of methane.

Explanation:

Methane

$CH_4+2O_2\rightarrow CO_2+2H_2O$

Mass of methane = 4.00 kg = 4000 g (1 kg = 1000 g)

Moles of methane = $\frac{4000 g}{16 g/mol}=250 mol$

According to reaction, 1 mole of methane gives 1 mole of carbon dioxide gas,then 250 moles of methane will give :

$\frac{1}{1}\times 250 mol=250 mol$ of carbon dioxide gas

Moles of carbon dioxide gas = n = 250 mol

Pressure of carbon dioxide gas = P = 0.941 atm

Temperature of carbon dioxide gas = T = 20.0°C = 20.0+273 K = 293 K

Volume of carbon dioxide gas = V

$PV=nRT$ (Ideal gas equation)

$V=\frac{nRT}{P}=\frac{250 mol\times 0.0821 atm L/mol K\times 293 K}{0.941 atm}=6390.89 L$

The volume of carbon dioxide at 20.0°C and 0.941 atm produced was 6390.89 Liters.

Propane

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

Mass of propane = 4.00 kg = 4000 g (1 kg = 1000 g)

Moles of propane = $\frac{4000 g}{44 g/mol}=90.91 mol$

According to reaction, 1 mole of propane gives 3 mole of carbon dioxide gas,then 90.91 moles of propane will give :

$\frac{3}{1}\times 90.91 mol=272.73 mol$ of carbon dioxide gas

Moles of carbon dioxide gas = n = 272.73 mol

Pressure of carbon dioxide gas = P = 0.941 atm

Temperature of carbon dioxide gas = T = 20.0°C = 20.0+273 K = 293 K

Volume of carbon dioxide gas = V

$PV=nRT$ (Ideal gas equation)

$V=\frac{nRT}{P}=\frac{272.73 mol\times 0.0821 atm L/mol K\times 293 K}{0.941 atm}=6,971.95 L$

The volume of carbon dioxide at 20.0°C and 0.941 atm produced was 6,971.95 Liters.

Volume of carbon dioxide given by combustion of 4.00 kg of methane = 6390.89 Liters.

Volume of carbon dioxide given by combustion of 4.00 kg of propane = 6,971.95 Liters.

6390.89 Liters < 6,971.95 Liters

More volume of carbon dioxide gas was released on combustion of 4.00 kg of propane in comparison to the volume of carbon dioxide released on combustion of 4.00 kg of methane.

7 0

3 years ago