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2 years ago

Balance the equationP + 02 → P205

Physics

2 answers:

AlladinOne [14]2 years ago

6 0

Answer:

4P + 502 → 2P205

Explanation:

Hence the above equation is balanced.

ivolga24 [154]2 years ago

3 0

Explanation:

$\underline{ \underline{ \large{ \text{Given \: chemical \: equation}}}} :$

$\tt{P \: + O _{2} \: \: \: \: ➞ \: \: P_{2} \: O_{5}}$

Here , Multiply P2O5 by 2 and O2 by 5 to equalize oxygen :

$\tt{P+ 5O_{2} \: \: ➞\: \: 2P_{2} \:O_{5}}$

Now , To equalize P atoms , Multiply P by 4

$\boxed{ \large{\tt{ 4P + 5O _{2} \: ➞ \: \: \: \: 2P_{2} \: O_{5}}}}$

Here , On the left side we have 4P and On the right side ,we have 4P. Again , On the left side , we have 10 O and On the right side , we have 10 O. And yippie , we got the balanced chemical equation !!!

Hope I helped ! ♡

Have a wonderful day / night ! ツ

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Can sugar molecules produced from photosynthesis be rearranged into different compounds carry out life process

Volgvan

Answer:

Yes, the sugar molecules produced from photosynthesis can be rearranged into different compounds to carry out life processes such energy producing compounds, energy storage compounds, body building compounds, and compounds of other types of food

Explanation:

a) The sugar can be broken down in a plant cell to produce ATP and then energy through cellular respiration as follows;

C₆H₁₂O₆ + O₂ 6CO₂ + 6H₂O + ATP

The produced energy can be used as power for cellular activities, such as the synthesis of protein and cell division

b) The produced sugar can be linked together by enzymes to form starch which is a form of chemical storage of energy as follows;

Sugar C₆H₁₂O₆ transformed into starch (C₆H₁₀O₅)ₙ

c) The sugar can be used in building the plant body by forming cellulose which is a form of starch

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2 years ago

With a thunderstorm brewing, an electric field of magnitude 2.0 × 102 newtons/coulomb exists at a certain point in the earth’s a

Aleks04 [339]

Answer:

The correct option is (b).

Explanation:

Given that,

Electric field, $E=2\times 10^2\ N/C$

We need to find the magnitude of the force on the electron as a result of the electric field.

We know that, the electric force is given by :

$F=qE\\\\=1.6\times 10^{-19}\times 2\times 10^2\\\\F=3.2\times 10^{-17}\ N$

So, the required force on the electron is equal to $3.2\times 10^{-17}\ N$ .

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2 years ago

U need help so can some one help me

GrogVix [38]

Answer:

are sure this is a question

7 0

2 years ago

A small omnidirectional stereo speaker produces waves in all directions that have an intensity of 8.00 at a distance of 4.00 fro

slavikrds [6]

Answer:

A. We have that radius r = 4.00m intensity I = 8.00 W/m^

total power = power/ Area ( 4πr2)= 8.00 w/m^2( 4π ( 4.00 m)2=1607.68 W

b) I = total power/ 4πr2= 8.00 W/m2 ( 4.00 m/ 9.5 m)2= 1.418 W/m2

c) E = total power x time= 1607 . 68 W x 1s= 1607.68 J

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2 years ago

A proton is projected toward a fixed nucleus of charge Ze with velocity vo. Initially the two particles are very far apart. When

11111nata11111 [884]

Answer:

The value is $R_f = \frac{4}{5} R$

Explanation:

From the question we are told that

The initial velocity of the proton is $v_o$

At a distance R from the nucleus the velocity is $v_1 = \frac{1}{2} v_o$

The velocity considered is $v_2 = \frac{1}{4} v_o$

Generally considering from initial position to a position of distance R from the nucleus

Generally from the law of energy conservation we have that

$\Delta K = \Delta P$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K = K__{R}} - K_i$

=> $\Delta K = \frac{1}{2} * m * v_1^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * (\frac{1}{2} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta P = P_f - P_i$

Here $P_i$ is zero because the electric potential energy at the initial stage is zero so

$\Delta P = k * \frac{q_1 * q_2 }{R} - 0$

$\frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R} - 0$

=> $\frac{1}{2} * m *v_0^2 [ \frac{1}{4} -1 ] = k * \frac{q_1 * q_2 }{R}$

=> $- \frac{3}{8} * m *v_0^2 = k * \frac{q_1 * q_2 }{R} ---(1 )$

Generally considering from initial position to a position of distance $R_f$ from the nucleus

Here $R_f$ represented the distance of the proton from the nucleus where the velocity is $\frac{1}{4} v_o$

Generally from the law of energy conservation we have that

$\Delta K_f = \Delta P_f$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K_f = K_f - K_i$

=> $\Delta K_f = \frac{1}{2} * m * v_2^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * (\frac{1}{4} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * \frac{1}{16} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance $R_f$ from the nucleus , this is mathematically represented as

$\Delta P_f = P_f - P_i$