In the figure shown, AD, CE, and FB intersect at point F.

A sample of an unknown substance has a mass of 0.158 kg. If 2,510.0 J of heat is required to heat the substance from 32.0°C to 6

JulijaS [17]

Specific heat is the amount of heat absorb or released by a substance to change the temperature to one degree Celsius. To determine the specific heat, we use the expression for the heat absorbed by the system. Heat gained or absorbed in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:
Heat = mC(T2-T1)
By substituting the given values, we can calculate for C which is the specific heat of the material.
2510 J = .158 kg ( 1000 g / 1 kg) (C) ( 61.0 - 32.0 °C)C = 0.5478 J / g °C

8 0

2 years ago

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A student is given the question: "what is the mass of a gold bar that is 7.379 × 10–4 m3 in volume? the density of gold is 19.3

IRINA_888 [86]

<h3><em>Further Explanation</em></h3>

Density is a quantity derived from the mass and volume

density is the ratio of mass per unit volume

With the same mass, the volume of objects that have a high density will be smaller than type

The unit of density can be expressed in g / cm3 or kg / m3

Density formula:

$\large{\boxed{{\bold{\rho~=~\frac{m}{V}}}}$

ρ = density

m = mass

v = volume

A common example is the water density of 1 gr / cm3

The mass itself is often equated with weight, even though it is different. Mass is the amount of matter in the matter while weight is related to the gravitational force

<em>Known variable</em>

volume gold bar 7.379 × 10⁻⁴ m³

a density of 19.3 g/cm³

<em>Asked</em>

the mass of a gold bar

<em>Answer</em>

volume is known to be 7.379 × 10⁻⁴ m³, then we change it first to units of cm³ adjusting to units of density

7.379 × 10⁻⁴ m³ = 7.379 × 10² cm³

then:

mass = volume x density

mass = 7.379 × 10² cm³ x 19.3 g/cm³

mass = 1.424 x 10⁴ gram

<h3><em>Learn more </em></h3>

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Keywords: density, mass, volume, a gold bar

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2 years ago

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Three measurements of 34.5m, 38.4m, and 35.3m are taken. If the accepted value of the measurement is 36.7m, what is the percent

8_murik_8 [283]

Answer:

A

Explanation:

% error in 1

36.7-34.5/36.7*100=5.99%

% error in 2

38.4-36.7/36.7*100=4.63℅

% error in 3

36.7-35.3/36.7*100=3.81%

7 0

2 years ago

2.1. The lithium ion in a 250,00 mL sample of mineral water was

juin [17]

Answer:

11482 ppt of Li

Explanation:

The lithium is extracted by precipitation with B(C₆H₄)₄. That means moles of Lithium = Moles B(C₆H₄)₄. Now, 1 mole of B(C₆H₄)₄ produce the liberation of 4 moles of EDTA. The reaction of EDTA with Mg²⁺ is 1:1. Thus, mass of lithium ion is:

<em>Moles Mg²⁺:</em>

0.02964L * (0.05581mol / L) = 0.00165 moles Mg²⁺ = moles EDTA

<em>Moles B(C₆H₄)₄ = Moles Lithium:</em>

0.00165 moles EDTA * (1mol B(C₆H₄)₄ / 4mol EDTA) = 4.1355x10⁻⁴ mol B(C₆H₄)₄ = Moles Lithium

That means mass of lithium is (Molar mass Li=6.941g/mol):

4.1355x10⁻⁴ moles Lithium * (6.941g/mol) = 0.00287g. In μg:

0.00287g * (1000000μg / g) = 2870μg of Li

As ppt is μg of solute / Liter of solution, ppt of the solution is:

2870μg of Li / 0.250L =

4 0

2 years ago

Anna lives in a city that experiences high precipitation, with an average annual rainfall of 524 millimeters. It is warm all yea

gavmur [86]

Anna lives in a city that is part of the tropical climate types. It has a constantly warm weather, and thus higher humidity, and according to the annual rainfall, it is most probably a rainfall that appears seasonally, not throughout the whole year.

Tim, on the other hand, lives in a city that is part of the dry climate types. It is most probably a place that is deep into the mainland, like the cold deserts of Central Asia, where the temperatures in the summer are high, and in winter are very low. Because of the distance from the sea, the rainfall doesn't reach this places, so they are very dry, and only have symbolic amount of annual rainfall.

6 0

2 years ago