3.Write the chemical equation for the reaction when methane burns in (2)

How are independent and dependent variables related?

shtirl [24]

They are pretty much the cause and effect. The independent variable that is being changed in the experiment, so it is the cause. The dependant variable is the result of change the independant variable, so the effect.

3 0

3 years ago

2

stiks02 [169]

Answer:

An organ is a part of the body that carries out its main function. Most organs are made out of tissues.

Explanation:

6 0

3 years ago

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consider a 1000-ml graduated cylinder with marks every 100 ml. a student records the volume of liquid in the cylinder as 750 ml.

Ludmilka [50]

Answer:

Explanation:

The 1000-ml graduated cylinder with marks every 100 ml has least count of 100 ml . It means it can not measure a volume less than 100 ml . It can also measure volume in the integral multiple of 100 ml like 200 ml , 300 ml , 400 ml etc perfectly . All these have significant figure of one .

It can not measure volume like 150 , 250 , 255 , 760 etc because these measurements are not the integral multiple of 100 . 750 is not a integral multiple of 100 so it can not measure volume of 750 ml.

4 0

3 years ago

During the following chemical reaction, 46.3 grams of C3H6O react with 73.2 grams of O2

ra1l [238]

Answer:

a) O2 is the limiting reactant

b) 75.70 grams CO2 (theoretical yield)

c) There remains 12.81 grams of C3H6O

d) The actual yield CO2 is 34.29 grams

Explanation:

Step 1: Data given

Mass of C3H6O = 46.3 grams

Mass of O2 = 73.2 grams

Molar mass of C3H6O = 58.08 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

C3H6O + 4O2 → 3 CO2 + 3H2O

Step 3: Calculate moles C3H6O

Moles C3H6O = mass C3H6O / molar mass C3H6O

Moles C3H6O = 46.3 grams / 58.08 g/mol

Moles C3H6O = 0.793 moles

Step 4: Calculate moles O2

Moles O2 = 73.2 grams / 32 g/mol

Moles O2 = 2.29 moles

Step 5: Calculate limiting reactant

For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O

O2 is the limiting reactant. It will completely be consumed. (2.29 moles).

C3H6O is in excess. There will react 2.29/4 = 0.5725 moles C3H6O

There will remain 0.793 - 0.5725 = 0.2205 moles C3H6O

This is 0.2205 moles * 58.08 g/mol = 12.81 grams

Step 6: Calculate moles of CO2

For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O

For 2.29 moles O2 we need 3/4 * 2.29 = 1.72 moles CO2

This is 1,72 moles * 44.01 g/mol = 75.70 grams CO2

Step 7: Calculate actual yield

% yield = 45.3 % = 0.453 = (actual yield / theoretical yield)

actual yield = 0.453 * 75.70 = 34.29 grams

3 0

3 years ago

Solve the following question

Andrew [12]

Answer:

140. J/g*K

Explanation:

To find the specific heat capacity, you need to use the following equation:

Q = mcΔT

In this equation,

-----> Q = energy/heat (J)

-----> m = mass (g)

-----> c = specific heat (J/mole*K)

-----> ΔT = change in temperature (K)

Before you can use the equation above, you need to (1) convert kg to grams, then (2) convert grams to moles (via molar mass), and then (3) convert Celsius to Kelvin. The final answer should have 3 significant figures.

1.11 kg C₄H₈O₂ x 1,000 = 1110 g

Molar Mass (C₄H₈O₂): 4(12.01 g/mol) + 8(1.008 g/mol) + 2(16.00 g/mol)

Molar Mass (C₄H₈O₂): 88.104 g/mol

1110 grams C₄H₈O₂ 1 mole
------------------------------ x ------------------------- = 12.6 moles C₄H₈O₂
88.104 grams

34.5 °C + 273 = 307.5 K

52.3 °C + 273 = 325.3 K

Q = mcΔT <----- Equation

3.14 x 10⁴ J = (12.6 moles)c(325.3 K - 307.5 K) <----- Insert values

3.14 x 10⁴ J = (12.6 moles)c(17.8) <----- Subtract

3.14 x 10⁴ J = (224.28)c <----- Multiply 12.6 and 17.8

140. = c <----- Divide both sides by 224.28

**this answer may be slightly off due to using different molar masses/Kelvin conversions**

4 0

1 year ago