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3 years ago

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A student thinks that scientific theories are not reliable sources of information, since scientists tend to argue about the deta

ils of them. What is the flaw in the student's reasoning?

1 answer:

Nadusha1986 [10]3 years ago

7 0

A theories are information

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Sound travels faster in: steel water sir

alukav5142 [94]

Sound travels better in water

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3 years ago

In Lesson 20, a magnesium strip was used to ignite the thermite reaction. When magnesium is placed in a flame from a small blow

nadezda [96]

Answer:

2 electrons will be needed by unbound oxygen in order to fill its 2nd shell.

Explanation:

The chemical reaction between magnesium and oxygen gives magnesium oxide as a product.The reaction is chemically represented as:

$2Mg(s)+O_2(g)\rightarrow 2MgO(s)$

Magnesium is a metal of group-2 with 2 valence electrons.It has atomic number of 12.

$[Mg]=1s^22s^22p^63s^2$

In order to attain noble gas configuration it will loose two electrons.

$[Mg]^{2+}=1s^22s^22p^6$

$Mg\rightarrow Mg^{2+}+2e^-$ ...[1]

Oxygen is a non metal of group-16 with 6 valence electrons..It has atomic number of 8.

$[O]=1s^22s^22p^4$

In order to attain noble gas configuration it will gain two electrons.

$[O]^{2-}=1s^22s^22p^6$

$O+2e^-\rightarrow O^{2-}$ ..[2]

2 electrons will be needed by unbound oxygen in order to fill its 2nd shell.

6 0

3 years ago

What is the number of electrons that move past a point in a wire carrying 500 A of current in 4.0 minutes

mr Goodwill [35]

The current is defined as the amount of charge Q that passes through a given point of a wire in a time $\Delta t$ :
$I= \frac{Q}{\Delta t}$
Since I=500 A and the time interval is
$\Delta t=4.0 min=240 s$
the charge is
$Q=I \Delta t=(500 A)(240 s)=1.2 \cdot 10^5 C$

One electron has a charge of $q=1.6 \cdot 10^{-19}C$ , therefore the number of electrons that pass a point in the wire during 4 minutes is
$N= \frac{Q}{q}= \frac{1.2 \cdot 10^5 C}{1.6 \cdot 10^{-19}C}=7.5 \cdot 10^{23}$ electrons

3 0

3 years ago

What is the main fuel consumed in the core of a red giant?<br> a. H<br> b. C<br> c. Fe<br> d. He

vladimir1956 [14]

<span>What is the main fuel consumed in the core of a red giant?
The </span><span>main fuel consumed in the core of a red giant is He or helium. The answer is letter D.</span>

4 0

3 years ago

The rate (in mg carbon/m3/h) at which photosynthesis takes place for a species of phytoplankton is modeled by the function P = 9

Ivanshal [37]

Answer:

At light intensity I = 3, is P a maximum

Explanation:

Given:

$P=\frac{90I}{I^2+I+9}$

now differentiating the above equation with respect to Intensity 'I' we get

$\frac{dp}{dI}=\frac{(I^2+I+9).\frac{d(90I)}{dI}-90I.\frac{d((I^2+I+9)}{dI}}{(I^2+I+9)^2}$

or

$\frac{dp}{dI}=\frac{(I^2+I+9).90-90I.(2I+1)}{(I^2+I+9)^2}$

or

$\frac{dp}{dI}=\frac{90I^2+90I+810)-(180I^2+90I)}{(I^2+I+9)^2}$

or

$\frac{dp}{dI}=\frac{-90I^2+810)}{(I^2+I+9)^2}$

Now for the maxima $\frac{dP}{dI}=0$

thus,

$0=\frac{-90I^2+810)}{(I^2+I+9)^2}$

or

$-90I^2+810=0$

or

$I^2=\frac{810}{90}$

or

$I^2=9$

or

I = 3

thus, <u>for the value of intensity I = 3, the P is maximum</u>

at I = 3

$P=\frac{90\times3}{3^2+3+9}$

or

$P=\frac{270}{21}$

or

$P=12.85$

5 0

3 years ago