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3 years ago

10

Isopropanol burns in oxygen to yield carbon dioxide, water, and heat. Which word equation best represents this chemical reaction

?

2 answers:

melomori [17]3 years ago

8 0

C3H8O + O2 → CO2 + H2O

emmainna [20.7K]3 years ago

7 0

Answer:

isopropanol + oxygen (arrow) carbon dioxide + water + heat

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A pan containing 40 grams of water was allowed to cool from a temperature of 91.0C. If the amount of heat repressed is 1,300 jou

Vinvika [58]

Answer:

83°C

Explanation:

The following were obtained from the question:

M = 40g

C = 4.2J/g°C

T1 = 91°C

T2 =?

Q = 1300J

Q = MCΔT

ΔT = Q/CM

ΔT = 1300/(4.2x40)

ΔT = 8°C

But ΔT = T1 — T2 (since the reaction involves cooling)

ΔT = T1 — T2

8 = 91 — T2

Collect like terms

8 — 91 = —T2

— 83 = —T2

Multiply through by —1

T2 = 83°C

The final temperature is 83°C

3 0

3 years ago

ou will prepare 250-mL of this solution using a 30% (m/v) NaOH stock solution. How many mL of the NaOH stock solution will you n

ArbitrLikvidat [17]

Answer:

$\boxed{\text{3.3 mL}}$

Explanation:

You must convert 30 % (m/v) to a molar concentration.

Assume 1 L of solution.

1. Mass of NaOH

$\text{Mass of NaOH} = \text{1000 mL solution } \times \dfrac{\text{30 g NaOH}}{\text{100 mL solution}} = \text{300 g NaOH}$

2. Moles of NaOH

$\text{Moles of NaOH} = \text{300 g NaOH} \times \dfrac{\text{1 mol NaOH}}{\text{40.00 g NaOH}} = \text{7.50 mol NaOH}$

3. Molar concentration of NaOH

$c= \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{7.50 mol}}{\text{1 L}} = \text{7.50 mol/L}$

4. Volume of NaOH

Now that you know the concentration, you can use the dilution formula .

$c_{1}V_{1} = c_{2}V_{2}$

to calculate the volume of stock solution.

Data:

c₁ = 7.50 mol·L⁻¹; V₁ = ?

c₂ = 0.1 mol·L⁻¹; V₂ = 250 mL

Calculations:

(a) Convert millilitres to litres

$V = \text{250 mL} \times \dfrac{ \text{1 L}}{\text{1000 mL}} = \text{0.250 L}$

(b) Calculate the volume of dilute solution

$\begin{array}{rcl}7.50V_{1} & = & 0.1 \times 0.250\\7.50V_{1} &= & 0.0250\\V_{1} & = & \text{0.0033 L}\\& = & \textbf{3.3 mL}\\\end{array}$

$\text{You will need $\boxed{\textbf{3.3 mL}}$ of the stock solution.}$

4 0

3 years ago

How is the composition of a meteorite relevant to finding out the composition of earth’s core?

boyakko [2]

Explanation:

Since scientists think planets and meteorites were made at the same time and in the same place, it seems logical that whatever a meteorite is made of is also what planets are made of.

5 0

2 years ago

Read 2 more answers

Give an example of experimental bias

ale4655 [162]

Answer:

chongus because he's the only good one

3 0

3 years ago

A sealed 1.0 L flask is charged with 0.500 mol of I2 and 0.500 mol of Br2. An equilibrium reaction ensues: I2 (g) + Br2 (g) ↔ 2I

Daniel [21]

Answer: Thus the value of $K_{eq}$ is 110.25

Explanation:

Initial moles of $I_2$ = 0.500 mole

Initial moles of $Br_2$ = 0.500 mole

Volume of container = 1 L

Initial concentration of $I_2=\frac{moles}{volume}=\frac{0.500moles}{1L}=0.500M$

Initial concentration of $Br_2=\frac{moles}{volume}=\frac{0.500moles}{1L}=0.500M$

equilibrium concentration of $IBr=\frac{moles}{volume}=\frac{0.84mole}{1L}=0.84M$ [/tex]

The given balanced equilibrium reaction is,

$I_2(g)+Br_2(g)\rightleftharpoons 2IBr(g)$

Initial conc. 0.500 M 0.500 M 0 M

At eqm. conc. (0.500-x) M (0.500-x) M (2x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[IBr]^2}{[Br_2]\times [I_2]}$

$K_c=\frac{(2x)^2}{(0.500-x)\times (0.500-x)}$

we are given : 2x = 0.84 M

x= 0.42

Now put all the given values in this expression, we get :

$K_c=\frac{(0.84)^2}{(0.500-0.42)\times (0.500-0.42)}$

$K_c=110.25$

Thus the value of the equilibrium constant is 110.25

8 0

3 years ago