Ask question

Ask question

All categories

3 years ago

8

How does the law of conservation of mass apply to this reaction: Mg + HCI + H2 MgClz?

1 answer:

Brums [2.3K]3 years ago

4 0

It’s the first answer - ie the equation needs to be balanced.

You can’t just look at the individual elements. To balance the equation, Magnesium will be part of the balancing and hence you will be adding Magnesium to both sides of the equation etc

You might be interested in

Convert 3.2 kilometers to feet?

Tanzania [10]

Answer:

10498.7 feet

Explanation:

7 0

3 years ago

Read 2 more answers

Person is lifting a 250 N dumbbell. The weight is 30 cm from the pivot point of the elbow. What force must be exerted five from

qwelly [4]

Refer to the diagram shown below.

The force, F, is applied at 5 cm from the elbow.

For dynamic equilibrium, the sum of moments about the elbow is zero.
Take moments about the elbow.
(5 cm)*(F N) - (30 cm)*(250 N) = 0
F = (30*250)/5 = 1500 N

Answer: 1500 N

4 0

3 years ago

An electron is released from rest at a distance of 0.470 m from a large insulating sheet of charge that has uniform surface char

ArbitrLikvidat [17]

Answer:

Part a)

$W = 1.58 \times 10^{-20} J$

Part b)

$v = 1.86 \times 10^5 m/s$

Explanation:

Part a)

Electric field due to large sheet is given as

$E = \frac{\sigma}{2\epsilon_0}$

$\sigma = 4.00 \times 10^{-12} C/m^2$

now the electric field is given as

$E = \frac{4.00 \times 10^{-12}}{2(8.85 \times 10^{-12})}$

$E = 0.225 N/C$

Now acceleration of an electron due to this electric field is given as

$a = \frac{eE}{m}$

$a = \frac{(1.6 \times 10^{-19})(0.225)}{9.1 \times 10^{-31}}$

$a = 3.97 \times 10^{10}$

Now work done on the electron due to this electric field

$W = F.d$

$d = 0.470 - 0.03$

$d = 0.44 m$

So work done is given as

$W = (ma)(0.44)$

$W = (9.11 \times 10^{-31})(3.97 \times 10^{10})(0.44)$

$W = 1.58 \times 10^{-20} J$

Part b)

Now we know that work done by all forces = change in kinetic energy of the electron

so we will have

$W = \frac{1}{2}mv^2 - 0$

$1.58 \times 10^{-20} = \frac{1}{2}(9.1\times 10^{-31})v^2$

$v = 1.86 \times 10^5 m/s$

7 0

3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

bezimeni [28]

solution:

y = v0t + ½at²

1150 = 79t + ½3.9t²

0 = 3.9t² + 158t - 2300

from quadratic equations and eliminating the negative answer

t = (-158 + v158² -4(3.9)(-2300)) / 2(3.9)

t = 11.37 s to engine cut-off

the velocity at that time is

v = v0 + at

v = 79 + 3.9(11.37)

v = 123.3 m/s

it rises for an additional time

v = gt

t = v/g

t = 123.3 / 9.8

t = 12.59 s

gaining more altitude

y = ½vt

y = 123.3(12.59) /2

y = 776 m

for a peak height of

y = 776 + 1150

5 0

3 years ago

Read 2 more answers

g A thin uniform film of oil that can be varied in thickness covers a sheet of glass of refractive index 1.52. The refractive in

Makovka662 [10]

Answer:

The right solution is "84.09 nm".

Explanation:

The given values are:

Refractive index of glass,

= 1.52

Refractive index of oil (n),

= 1.64

Wavelength (λ),

= 555 nm

Now,

The thickness of the film (t) will be:

= $\frac{\lambda}{4n}$

= $\frac{555}{4\times 1.65}$

= $\frac{555}{6.6}$

= $84.09 \ nm$

6 0

3 years ago