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Misha Larkins [42]

3 years ago

6

Is the equation balanced? 2H2O + O2 = 4MgO + 3Fe

1 answer:

ipn [44]3 years ago

4 0

Answer:

no

Explanation:

the equation can't be balanced because it doesn't have the same elements on each side of the equal sign.

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What is the MAIN reason water from the oceans turns to water vapor, and then evaporates into the air?

GuDViN [60]

This has a two word answer: sun's heat. The faster moving molecules near the ocean's surface are provided with enough energy from the sun to escape the surface they are near.

8 0

3 years ago

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State four advantages of Mercury over alcohol as thermometric liquids

Gemiola [76]

Answer:

Alcohol wets glass while mercury doesn't
Alcohol has a lower boiling point and hence cannot be used to measure very high temperature.
Alcohol is colorless and is very difficult to read without coloration.
Mercury's expansion is uniform.

6 0

3 years ago

Suppose a 32-kg child sits in a playground swing

otez555 [7]

Answer:

48

Explanation:

6 0

3 years ago

A 0.25 kg ball is suspended from a light 0.65 m string as shown. The string makes an angle of 31° with the vertical. Let U = 0 w

steposvetlana [31]

Explanation:

a) The height of the ball h with respect to the reference line is

$h = L - L\cos{31°} = L(1 - \cos{31°})$

so its initial gravitational potential energy $U_0$ is

$U = mgh = mgL(1 - \cos{31°})$

$\:\:\:\:\:=(0.25\:\text{kg})(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31})$

$\:\:\:\:\:=0.23\:\text{J}$

b) To find the speed of the ball at the reference point, let's use the conservation law of energy:

$\Delta{K} + \Delta{U} = 0 \Rightarrow K_0 + U_0 = K + U$

We know that the initial kinetic energy $K_0,$ as well as its final gravitational potential energy $U$ are zero so we can write the conservation law as

$mgL(1 - \cos{31°}) = \frac{1}{2}mv^2$

Note that the mass gets cancelled out and then we solve for the velocity v as

$v = \sqrt{2gL(1 - \cos{31°})}$

$\:\:\:\:\:= \sqrt{2(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31°})}$

$\:\:\:\:\:= 1.3\:\text{m/s}$

5 0

3 years ago

Read 2 more answers

A 100.0 gram sample of Polonium-210 is contained for 552 days. How many half-lives occur during this period of time, if the half

Leto [7]

Answer:

4 half-lives will occur during this period of time.

Explanation:

Formula used :

$a_o=a\times e^{-\lambda t}$

$\lambda =\frac{0.693}{t_{1/2}}$

$a=\frac{a_o}{2^n}$

where,

a = amount of reactant left after n-half lives and time t

$a_o$ = Initial amount of the reactant.

$\lambda =$ decay constant

$t_{1\2}$ = half life of an isotope

n = number of half lives

We have :

$a_o=100.0 g$

a = ?

t = 552 days

$t_{1/2}=138 days$

$a=100.0 g\times e^{-\frac{0.693}{138}\times 552}$

$a=6.254 g$

$6.254 g=\frac{100.0 g}{2^n}$

$2^n=\frac{100.0 g}{6.254 g}$

n = 4

4 half-lives will occur during this period of time.

4 0

4 years ago