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3 years ago

6

Consider the equation S + O2 → SO2. What type of reaction is this? a. synthesis b. decomposition c. single displacement d. doubl

e displacement

2 answers:

arlik [135]3 years ago

7 0

Answer:

It's a synthesis reaction since S and $O_{2}$ combine to form sulphur dioxide.

Explanation:

In this reaction S and $O_{2}$ synthesized to form a single product, $SO_{2}$ ,so it will be a synthesis reaction.

VashaNatasha [74]3 years ago

4 0

This is a synthesis reaction because there are two reactants combining to form one product.

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ExtremeBDS [4]

Explanation:

In the given situation two forces are working. These are:

1) Electric force (acting in the downward direction) = qE

2) weight (acting in the downward direction) = mg

Therefore, work done by all the forces = change in kinetic energy

Hence, $qE \times S + mg \times S = 0.5 \times mv^{2}$

$1.6 \times 10^{-19} \times 1000 + 9.1 \times 10^{-31} \times 9.8 \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2}$

It is known that the weight of electron is far less compared to electric force. Therefore, we can neglect the weight and the above equation will be as follows.

$(1.6 \times 10^{-19} \times 1000) \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2
}$

v = $sqrt{\frac{1.6 \times 10^{-19}}{(0.5 \times 9.1 \times 10^{-31})}$

= 592999 m/s

Since, the electron is travelling downwards it means that it looses the potential energy.

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2 years ago

Doc Brown has calculated his Delorean can accelerate at a rate of 2.52 m/s/s. How

GREYUIT [131]

Answer:

304.89m

Explanation:

Given

acceleration a = 2.52m/s²

final speed v = 39.2m/s

initial speed = 0m/s (car accelerates from rest)

Using the equation of motion below to get the distance of Doc brown from Marty;

v² = u²+2as

substitute the given parameters

39.2² = 0²+2(2.52)s

1536.64 = 0+5.04s

divide both sides by 5.04

1536.64/5.04 = 5.04s/5.04

rearrange the equation

5.04s/5.04 = 1536.64/5.04

s = 304.89m

Hence He and Marty must stand at 304.89m to allow the car to accelerate from rest to a speed of 39.2 m/s?

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3 years ago

A styrofoam container used as a picnic cooler contains a block of ice at 0°C. If 325 g of ice melts in 1 hour, how much heat ene

Sauron [17]

Answer:

30.0625 W

Explanation:

325 g/h x (1h x 1kg)/(3600s x 1000g) x 3,33 x 10^5 J/Kg = 30.0625 J/Kg = 30.0625 W

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3 years ago

Two people are standing on a 1.75-m-long platform, one at each end. The platform floats parallel to the ground on a cushion of a

Fiesta28 [93]

Answer:

0.05312 m

Explanation:

Given:

Length of platform L = 1.75 m

mass of ball m_b = 5.76 kg

mass of (people + platform) m_p = 184 kg

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Initial Velocity of ball (people + platform) V_i,p = 0

Find:

How far does the platform recoils to rest

Solution:

Using the conservation of momentum on ust before and after the ball was thrown P_i = P_f :

Where, P_i = 0 (initially at rest)

P_f = m_p*V_f,p + m_b * V_f,b

0 = m_p*V_f,p + m_b * V_f,b

V_f,p = - (m_b /m_p) * V_f,b

V_f,p = - (5.76 / 184)*V_f,b

V_f,p = - 0.0313*V_f,b ....1

The time the ball is in air:

t = L / (V_f,b - V_f,p) ...2

The distance that the platform moves d:

d = V_f,p *t ....3

Substitute 2 into 3

d = V_f,p*L /(V_f,b - V_f,p) .... 4

Solve 1 and 4 simultaneously :

d = - m_b*L / (m_b + m_p)

d = - 5.76*1.75 / (5.76 + 184)

d = -0.05312 m

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5 0

3 years ago

Four traveling waves are described by the following equations, where all quantities are measured in SI units and y represents th

agasfer [191]

Answer:

$T_1=T_3=\dfrac{2\pi}{21}$

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Explanation:

Wave 1, $y_1=0.12\ cos(3x-21t)$

Wave 2, $y_2=0.15\ sin(6x+42t)$

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Wave 4, $y_4=-0.27\ sin(3x-42t)$

The general equation of travelling wave is given by :

$y=A\ cos(kx\pm \omega t)$

The value of $\omega$ will remain the same if we take phase difference into account.

For first wave,

$\omega_1=21$

$\dfrac{2\pi }{T_1}=21$

$T_1=\dfrac{2\pi}{21}$

For second wave,

$\omega_2=42$

$\dfrac{2\pi }{T_2}=42$

$T_2=\dfrac{2\pi}{42}$

For the third wave,

$\omega_3=21$

$\dfrac{2\pi }{T_3}=21$

$T_3=\dfrac{2\pi}{21}$

For the fourth wave,

$\omega_4=42$

$\dfrac{2\pi }{T_4}=42$

$T_4=\dfrac{2\pi}{42}$

It is clear from above calculations that waves 1 and 3 have same time period. Also, wave 2 and 4 have same time period. Hence, this is the required solution.

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3 years ago