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dimulka [17.4K]

3 years ago

8

In the following reaction, which substance is the precipitate? (NH4)2SO4(aq) + Ba(NO3)2(aq) BaSO4(s) + 2NH4NO3(aq) (NH4)2SO4 Ba(

NO3)2 BaSO4 NH4NO3

2 answers:

Arte-miy333 [17]3 years ago

5 0

Answer:

BaSo4

Explanation:

Mariana [72]3 years ago

4 0

I think it's Barium sulfate, the soild and percipitate

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you should begin viewing a bacteria specimen with what objective lens? view available hint(s)for part g you should begin viewing

Sergio039 [100]

Some people view bacteria specimens with a 100x objective lens in order to see the smallest details.
Others may use a 10x objective lens for more general purposes, such as examining stained slides or pictures.
And still others may use a 40x objective lens to gain maximum resolution when viewing images of thick samples.

It is important to choose the appropriate magnification for your needs so that you can properly examine the specimen under study.

<h3>Why is the 100x objective lens necessary to see bacteria?</h3>

Bacteria must, of course, be viewed at the maximum magnification and resolution possible because to their small size.
Due to optical restrictions, this is approximately 1000x in a light microscope.
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2 years ago

A radio has 100J of energy transferred by electricity. 20J are transferred by heating and 10J are transferred by sound. How many

seraphim [82]

Answer:

70 Joules

Explanation:

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3 years ago

How to round numbers and find significant digits in physics?? <br> Simple answer? Please???

Mrrafil [7]

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3 years ago

A tuning fork vibrating at 508 Hz falls from rest and accelerates at 9.80 m/s^2. How far below the point of release is the tunin

JulijaS [17]

Answer:

Explanation:

given,

tuning fork vibration = 508 Hz

accelerates = 9.80 m/s²

speed of sound = 343 m/s

observed frequency = 490 Hz

$f_s = f(\dfrac{v}{v-(-v_s)})$

$f_s = f(\dfrac{v}{v+v_s})$

$v_s = v[\dfrac{f_s}{f_o}-1]$

$= 343[\dfrac{508}{490}-1]$

$v_s=12.6 m/s$

distance the tunning fork has fallen

$y_1=\dfrac{v^2}{2a_y}$

$=\dfrac{12.6^2}{2\times 9.8}$

=8.1 m

now, time required for the observed will be

$t = \dfrac{8.1}{343} = 0.023 s$

now, for the distance calculation

$y_2 = u\ t + \dfrac{1}{2}at^2$

$= 12.6\times 0.023 +\dfrac{1}{2}\times 9.8 \times 0.023^2$

=0.293 m

total distance

= 8.1 + 0.293 = 8.392 m

3 0

3 years ago

calculate the force required to take away a flat corcular plate of radius 0.01m from the surface of water. the surface tention o

leonid [27]

Answer:

$Force = 0.0047175\ N$

Explanation:

Given

$T = 0.075N/m$ --- Surface Tension

$r = 0.01m$ --- Radius

Required

Determine the required force

First, we calculate the circumference (C) of the circular plate

$C= 2\pi r$

$C= 2 * \frac{22}{7} * 0.01m$

$C= \frac{2 * 22 * 0.01}{7}m$

$C= \frac{0.44}{7}m$

$C= 0.0629 m$

The applied force is then calculated using;

$Force = C * T$

$Force = 0.0629m * 0.075N/m$

$Force = 0.0047175\ N$

8 0

2 years ago