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2 years ago

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A fuel was burned for 5 min, increasing the temperature of 10.0 g of water with a density of 1.00 g/ml by 9.0 oC. The fuel relea

sed _____ ca

1 answer:

jekas [21]2 years ago

7 0

The fuel released 90 calories of heat.

Let suppose that water experiments an entirely <em>sensible</em> heating. Hence, the heat released by the fuel is equal to the heat <em>absorbed</em> by the water because of principle of energy conservation. The heat <em>released</em> by the fuel is expressed by the following formula:

$Q = m\cdot c \cdot \Delta T$ (1)

Where:

$m$ - Mass of the sample, in grams.
$c$ - Specific heat of water, in calories per gram-degree Celsius.
$\Delta T$ - Temperature change, in degrees Celsius.

If we know that $m = 10\,g$ , $c = 1\,\frac{cal}{g\cdot ^{\circ}C}$ and $\Delta T = 9\,^{\circ}C$ , then the heat released by the fuel is:

$Q = (10\,g)\cdot \left(1\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (9\,^{\circ}C)$

The fuel released 90 calories of heat.

We kindly invite to check this question on sensible heat: brainly.com/question/11325154

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Calculate the grams of solute in each of the following solution: 278 mL of 0.038 M Fe2(SO4)3

Goryan [66]

Answer: 4.22 grams of solute is there in 278 ml of 0.038 M $Fe_2(SO_4)_3$

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L

Now put all the given values in the formula of molality, we get

$0.038M=\frac{n}{0.278L}$

$n=0.0105mol$

mass of $Fe_2(SO_4)_3$ = $moles\times {\text {Molar Mass}}=0.0105\times 399.88g/mol=4.22g$

Thus 4.22 grams of solute is there in 278 ml of 0.038 M $Fe_2(SO_4)_3$

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3 years ago

Classify each change as physical or chemical.

trasher [3.6K]

A chemical

B physical

C physical

D Chemical

4 0

4 years ago

At room temperature (20°C} and pressure, the density of air is 1.189 g/L. An object will float in air if its density is less tha

Alekssandra [29.7K]

Explanation:

$Density =\frac{Mass}[Volume}$

Density of the air ,d= 1.189 g/L

(a) Density of the evacuated ball

Mass of the ball ,m = 0.12 g

Volume of the ball = $V=560 cm^3=560 ml=0.560 L$

$D =\frac{0.12 g}{0.560 L}=0.214 g/L$

D<d, teh evacuated ball will flaot in air.

(b) Density of the evacuated ball D = 0.214 g/L

Density of carbon dioxide gas = $d_1=1.830 g/L$

Mass of the carbon dioxide gas :

$1.830 g/L\times 0.560 L=1.0248 g$

Total density of filled ball with carbon dioxide gas:

$\frac{0.12 g+1.0248 g}{0.560 L}==2.044 g/L$

The ball filled with carbon dioxide will not float in the air because total density of filled ball is greater than the density of an air.

(c) Density of the evacuated ball D = 0.214 g/L

Density of hydrogen gas = $d_2=0.0899 g/L$

Mass of the hydrogen gas :

$1.830 g/L\times 0.560 L=0.050344 g$

Total density of filled ball with hydrogen gas:

$\frac{0.12 g+0.050344 g}{0.560 L}==0.3041 g/L$

The ball filled with hydrogen will float in the air because total density of filled ball is lessor than the density of an air.

(d) Density of the evacuated ball D = 0.214 g/L

Density of oxygen gas = $d_3=1.330 g/L$

Mass of the oxygen gas :

$1.330 g/L\times 0.560 L=1.7448 g$

Total density of filled ball with oxygen gas:

$\frac{0.12 g+1.7448 g}{0.560 L}=1.5442 g/L$

The ball filled with oxygen will not float in the air because total density of filled ball is greater than the density of an air.

(e) Density of the evacuated ball D = 0.214 g/L

Density of nitrogen gas = $d_4=1.165 g/L$

Mass of the nitrogen gas :

$1.165 g/L\times 0.560 L=0.6524 g$

Total density of filled ball with nitrogen gas:

$\frac{0.12 g+0.6524 g}{0.560 L}==1.3792 g/L$

The ball filled with nitrogen will not float in the air because total density of filled ball is greater than the density of an air.

f) Mass must be added to sink the ball = m

Density of ball > Density of the air ; to sink the ball.

$\frac{0.12g +m}{0.560L}>1.189 g/L$

m > 0.54584 g

For any case weight added to ball to make it sink in an air should be grater than the value of 0.54584 grams.

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3 years ago

How much does REAL carbon fiber cost ? lets say as big as a piece of paper

coldgirl [10]

Answer:

Today, the average total production cost of “standard modulus” carbon fiber is in the range of $7-9 per pound.

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3 years ago

The organic compound di‑n‑butyl phthalate, C 16 H 22 O 4 ( l ) , C16H22O4(l), is sometimes used as a low‑density ( 1.046 g ⋅ mL

damaskus [11]

Answer:

36.63 Torr

Explanation:

You need to use two expressions, one for pressure and the other with the relation of density and height of the column.

For the pressure:

P = h * d * g (1)

h is height.

d density

g gravity

The second expression put a relation between the densities and height of the column so:

d1/d2 = h1/h2 (2)

let 1 be the phthalate, and 2 the mercury.

Let's calculate first the relation of density:

d1/d2 = 13.53 / 1.046 = 12.93

Now with the first expression, we can calculate the pressure so:

P = hdg

We have two compounds so,

h1d1g = h2d2g ---> gravity cancels out

From here, we can solve for h2:

h2 = h1*(d1/d2)

replacing:

h2 = 459 / 12.53

h2 = 36.63 mm

1 mmHg is 1 torr, therefore the pressure of the gas in Torr would be 36.63 Torr

8 0

3 years ago