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3 years ago

The amount of oxygen in the reactants is 4 atoms. In the products, the oxygen is distributed to water (H2O) and O2

Physics

2 answers:

Oliga [24]3 years ago

6 0

Answer:

The amount of oxygen in the reactants is 4 atoms. In the products, the oxygen is distributed to water (H2O) and O2 gas. Which coefficients would be placed in front of water and oxygen?

2 in front of water and 1 in front of oxygen

Kiley wants to balance the equation H2 + N2 → NH3. Which would be her first step in balancing an equation?

NOT adding the coefficient 3 in front of H2 and the coefficient 2 in front of NH3

Which is the balanced equation for V2O5 + CaS → CaO + V2S5?

V2O5 + 5CaS → 5CaO + V2S5

One half of a balanced chemical equation is shown.

3Mg(OH)2 + 2H3PO4

Which lists the numbers of each atom the other half of the equation would contain?

3 Mg, 2 P, 14 O, 12 H

Substances A and B react in a ratio of A + B → AB. Colin was combining substance A and substance B to form substance AB. He added 10 g of substance A and 45 g of substance B.

How much of substance AB would Colin make?

55g

Lana balanced an equation so that the result was 2C2H3Br + 5O2 → 4CO2 + 2H2O + 2HBr. Which most likely represents the starting equation?

NOT 2C4H3Br + 5O2 → 4CO2 + 2H2O + 2HBr

NOT C4H3Br + O2 → CO2 + H2O + HBr

Which best describes the law of conservation of mass?

The mass of the reactants and products is equal and is not dependent on the physical state of the substances.

Gnom [1K]3 years ago

3 0

Answer: is D

Explanation:

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3 years ago

At 5.0 minutes, the temperature of the water reaches 100 °C. The volume of the water in the urn

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2 years ago

LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.

Svetradugi [14.3K]

Answer:

i) E = 269 [MJ] ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

$E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.$

$W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]$

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

$E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]$

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3 years ago

You are investigating an elevator accident which happened in a tall building. An elevator in this building is attached to a stro

Advocard [28]

Answer:

a) F = 2250 Ib

b) F = 550 Ib

c) new max force ( F newmax ) = 2850 Ib

Explanation:

A) The force the wall of the elevator shaft exert on the motor if the elevator starts from rest and goes up

max capacity of elevator = 24000 Ibs

counterweight = 1000 Ibs

To calculate the force (F) :

we first calculate the Tension using this relationship

Counterweight (1000) - T = ( 1000 / g ) ( g/4 )

Hence T = 750 Ib

next determine F

750 + F - 2400 = 2400 / 4

hence F = 2250 Ib

B ) calculate Tension first

T - 1000 = ( 1000/g ) ( g/4)

T = 1250 Ib

F = 2400 -1250 - 2400/ 4

F = 550 Ib

C ) determine design limit

Max = 2400 * 1.2 = 2880 Ib

750 + new force - 2880 = 2880 / 4

new max force ( F newmax ) = 2850 Ib

8 0

3 years ago