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Oxana [17]
3 years ago
7

What does the specific heat for a substance indicate? A. the heat released when 1 gram of the substance burns B. the amount of h

eat required to increase the temperature of 1 gram by 1oC C. the amount of heat required to increase the temperature of 100 grams by 1oC D. the amount of heat required to increase the temperature of 1 gram by 100oC E. the heat released when 100 grams of the substance burns
Chemistry
2 answers:
Viktor [21]3 years ago
4 0
B. the amount of heat required to increase the temperature of 1 gram by 1oC
Paha777 [63]3 years ago
4 0

Answer:

B. the amount of heat required to increase the temperature of 1 gram by 1°C

Explanation:

The heat required by a substance having mass m to raise the temperature of it by ΔT.

Q = m c ΔT

where c is the specific heat of the substance.

c = \frac{Q}{m\Delta T}

Specific heat for a substance is the amount of heat required to increase the temperature of 1 gram by 1°C.

For example, the specific heat capacity of water is 4.186 J/g°C. This means 4.186 J is required to raise the temperature of 1 g of water by 1°C.

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An ion of oxygen- 16 contains 8 protons and has a 2- charge. How many electrons does it have?
Verdich [7]

Answer:

i would say 10, so the anser is A.

Explanation:

because there are the same number of protons and electrons, therefore for a regular O, you are supposed to have only 8 protons, but it is charged, thus, whatever the charge is will be taken into consideration into how much the proton and electron doe it have. Thus, for this case, it has 10, because the charge is negative and you have 8 electron plus 2 = 10.

3 0
3 years ago
Is the total disappearance of all members of a species.
OLga [1]

Answer: EXTINCTION

Explanation:

4 0
3 years ago
Which of the following methods is not used to locate underground oil reserves?
Ugo [173]
C.) Electrical Signals are not used <span>to locate underground oil reserves

Hope this helps!</span>
6 0
3 years ago
g An oxidized silicon (111) wafer has an initial field oxide thickness of d0. Wet oxidation at 950 °C is then used to grow a thi
77julia77 [94]

Answer:

Explanation:

From the information given:

oxidation of oxidized solution takes place at 950° C

For wet oxidation:

The linear and parabolic coefficient can be computed as:

\dfrac{B}{B/A} = D_o \ exp \Big [\dfrac{-\varepsilon a}{k_BT} \Big]

Using D_o and E_a values obtained from the graph:

Thus;

\dfrac{B}{A} = 1.63 \times 10^8 exp \Big [ \dfrac{-2.05}{8.617 \times 10^{_-5}\times 1173}\Big] \\ \\ = 0.2535 \ \ \mu m/hr

B= 386 \  exp \Big [-\dfrac{0.78}{8.617 \times 10^{-3} \times 1173} \Big] \\ \\  = 0.1719 \ \mu m^2/hr

So, the initial time required to grow oxidation is expressed as:

t_{ox} = \dfrac{x}{B/A}+ \dfrac{x^2}{B} - t_o (initial)

where; \\ \\ t_{ox} = 2 \ hrs;\\ \\ x = 0.5 \\ \\ B/A = 0.2535 \\ \\  B = 0.1719

∴

2= \dfrac{0.5}{0.2535}+ \dfrac{0.5^2}{0.1719} - t_o (initial)

2 = 3.4267 - t_o (initial) \\ \\ t_o(initial) = 3.4267 - 2  \\ \\ t_o(initial) = 1.4267 \ hr

NOW;

1.4267 = \dfrac{d_o}{0.2535} + \dfrac{d_o^2}{0.1719} \\ \\  1.4267 = 3.9448 \ d_o + 5.8173 \ d_o^2 \\ \\ d_o^2 + 0.6781 \ d_o = 0.2453

d_o = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}

d_o = \dfrac{-(0.6781) \pm \sqrt{(0.6781)^2-4(1)(-0.245)}}{2(10)}

d_o = \dfrac{-(0.6781) \pm \sqrt{0.45981961+0.98}}{20}

d_o = \dfrac{-(0.6781) \pm \sqrt{1.43981961}}{20}

d_o = \dfrac{-(0.6781) + \sqrt{1.43981961}}{20} \ OR \   \dfrac{-(0.6781) - \sqrt{1.43981961}}{20}

d_o =0.02609 \ OR \   -0.0939

Thus; since we will consider the positive sign, the initial thickness d_o is ;

≅ 0.261 μm

3 0
3 years ago
Kind of protists can live in groups with different individuals performing different jobs (site 1)
zlopas [31]

Answer:

The plasmodial slime molds are the kind of protist that can live in groups in which different individuals perform different jobs. Some kinds of bacterial cell walls also have other functions.

8 0
3 years ago
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