All categories

4 years ago

What is the velocity of a rocket that goes 700 km north in 25 second

Chemistry

2 answers:

Dmitrij [34]4 years ago

6 0

Answer: 28000m/s

Explanation:

Displacement = 700 km due north = 700000m

Time = 25secs

Velocity =?

Velocity = Displacement / Time

V = 700000 / 25 = 28000m/s

sukhopar [10]4 years ago

6 0

Velocity= m/s or ms^-1
Km-m= multiply by 1000; 700 x 1000= 700000m
Velocity= distance/time
V= 700000\25= 28000m/s OR 28000ms^-1

You might be interested in

Distinguish<br>between<br>matter<br>and a substance

Irina18 [472]

Answer:

Matter is a physical substance that contains mass. A substance is a particular kind of matter with uniform chracteristics.

Explanation:

Remeber that matter is anything that has mass and takes up space :)

8 0

3 years ago

O which type of molecules, polar or nonpolar, dissolves in water

Anna007 [38]

The polar molecules dissolve in water, because water is also a polar molecule.
"Like dissolves like".
:)

7 0

3 years ago

Convert the following names to balanced formulas.

yulyashka [42]

You can reduce the charges and subscripts so it is PbS2

8 0

3 years ago

A homogeneous mixture is:

LUCKY_DIMON [66]

Answer: A homogeneous mixture is a solid, liquid or gaseous mixture that has the same proportions of its components throughout any given sample.

Explanation:

5 0

3 years ago

For the reaction C2H4(g) + H2O(g) --> CH3CH2OH(g)

Dominik [7]

Answer : The value of equilibrium constant for this reaction at 262.0 K is $3.35\times 10^{2}$

Explanation :

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

$\Delta H^o$ = standard enthalpy = -45.6 kJ = -45600 J

$\Delta S^o$ = standard entropy = -125.7 J/K

T = temperature of reaction = 262.0 K

Now put all the given values in the above formula, we get:

$\Delta G^o=(-45600J)-(262.0K\times -125.7J/K)$

$\Delta G^o=-12666.6J=-12.7kJ$

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln k$

where,

$\Delta G^o$ = standard Gibbs free energy = -12666.6 J

R = gas constant = 8.314 J/K.mol

T = temperature = 262.0 K

K = equilibrium constant = ?

Now put all the given values in the above formula, we get:

$-12666.6J=-(8.314J/K.mol)\times (262.0K)\times \ln k$

$k=3.35\times 10^{2}$

Therefore, the value of equilibrium constant for this reaction at 262.0 K is $3.35\times 10^{2}$

3 0

4 years ago