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2 years ago

How many times do you have to lick a tooties roll pop to get to the middle?

Chemistry

2 answers:

amid [387]2 years ago

8 0

Answer:

about 364 licks!

Explanation:

pls mark brainliest :)

Debora [2.8K]2 years ago

7 0

Answer:

eso no es un ejercicio. pero igual gracias por los puntos.☺

Explanation:

an average of 364 licks to get to the center of a Tootsie Pop.

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A major component of gasoline is octane (C8H18). When octane is burned in air, it chemically reacts with oxygen gas (O2) to prod

Olin [163]

Answer:

The answer is 6.25g.

Explanation:

First create your balanced equation. This will give you the stoich ratios needed to answer the question:

2C8H18 + 25O2 → 16CO2 + 18H2O

Remember, we need to work in terms of NUMBERS, but the question gives us MASS. Therefore the next step is to convert the mass of O2 into moles of O2 by dividing by the molar mass:

7.72 g / 16 g/mol = 0.482 mol

Now we can use the stoich ratio from the equation to determine how many moles of H2O are produced:

x mol H2O / 0.482 mol O2 = 18 H2O / 25 O2

x = 0.347 mol H2O

The question wants the mass of water, so convert moles back into mass by multiplying by the molar mass of water:

0.347 mol x 18 g/mol = 6.25g

8 0

3 years ago

Read 2 more answers

An inverted pyramid is being filled with water at a constant rate of 45 cubic centimeters per second. The pyramid, at the top, h

vaieri [72.5K]

Answer:

13.20 cm/s is the rate at which the water level is rising when the water level is 4 cm.

Explanation:

Length of the base = l

Width of the base = w

Height of the pyramid = h

Volume of the pyramid = $V=\frac{1}{3}lwh$

We have:

Rate at which water is filled in cube = $\frac{dV}{dt}= 45 cm^3/s$

Square based pyramid:

l = 6 cm, w = 6 cm, h = 13 cm

Volume of the square based pyramid = V

$V=\frac{1}{3}\times l^2\times h$

$\frac{l}{h}=\frac{6}{13}$

$l=\frac{6h}{13}$

$V=\frac{1}{3}\times (\frac{6h}{13})^2\times h$

$V=\frac{12}{169}h^3$

Differentiating V with respect to dt:

$\frac{dV}{dt}=\frac{d(\frac{12}{169}h^3)}{dt}$

$\frac{dV}{dt}=3\times \frac{12}{169}h^2\times \frac{dh}{dt}$

$45 cm^3/s=3\times \frac{12}{169}h^2\times \frac{dh}{dt}$

$\frac{dh}{dt}=\frac{45 cm^3/s\times 169}{3\times 12\times h^2}$

Putting, h = 4 cm

$\frac{dh}{dt}=\frac{45 cm^3/s\times 169}{3\times 12\times (4 cm)^2}$

$=13.20 cm/s$

13.20 cm/s is the rate at which the water level is rising when the water level is 4 cm.

3 0

3 years ago

How is data not actually obtained from the experiment represented in a line graph?

zavuch27 [327]

A colored line, as long as it is one single piece, not broken

5 0

2 years ago

Read 2 more answers

How does biodiversity help sustain a population in an area

Firdavs [7]

Having a variety is always good you never want to have too much of one thing. It helps with the food chain its basically survival of the fittest

3 0

3 years ago

If you combine 230.0 mL 230.0 mL of water at 25.00 ∘ C 25.00 ∘C and 120.0 mL 120.0 mL of water at 95.00 ∘ C, 95.00 ∘C, what is t

Thepotemich [5.8K]

Answer: The final temperature of the mixture is 49°C

Explanation:

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

For cold water:

Density of cold water = 1 g/mL

Volume of cold water = 230.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{230.0mL}\\\\\text{Mass of water}=(1g/mL\times 230.0mL)=230g$

For hot water:

Density of hot water = 1 g/mL

Volume of hot water = 120.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{120.0mL}\\\\\text{Mass of water}=(1g/mL\times 120.0mL)=120g$

When hot water is mixed with cold water, the amount of heat released by hot water will be equal to the amount of heat absorbed by cold water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of hot water = 120 g

$m_2$ = mass of cold water = 230 g

$T_{final}$ = final temperature = ?°C

$T_1$ = initial temperature of hot water = 95°C

$T_2$ = initial temperature of cold water = 25°C

c = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

$120\times 4.186\times (T_{final}-95)=-[230\times 4.186\times (T_{final}-25)]$

$T_{final}=49^oC$

Hence, the final temperature of the mixture is 49°C

4 0

3 years ago