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3 years ago

What do the arrows represent in the figure

Chemistry

2 answers:

anyanavicka [17]3 years ago

7 0

The flow/transfer of energy

astraxan [27]3 years ago

6 0

Answer: Flow of energy in ecosystem.

Explanation: The picture here represents the flow of energy from the producers to consumers.

The first picture shows the plants that takes energy from the sun for the production of glucose by the help of the process known as photosynthesis.

The second picture here is caterpillar who is a primary producer feeding on the plant for its energy needs.

The third picture is a bird who is secondary consumer that is eating the caterpillar for its energy needs.

This shows how the energy travels from the sun→producers→ primary consumer→secondary consumers.

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Why will frequency decrease when wave gets longer

amid [387]

The frequency will decrease because the waves are getting farther and farther apart and if they and as that happens with shorter and shorter so they decrease

5 0

4 years ago

Specific Heat Practice

Alexxx [7]

Answer:

See below

Explanation:

Specific Heat of water = 4.186 J/(g-C)

J/(g-C) multiplied by g and C results in J

4.186 * 540 * (95-32) = 142 407 .72 J

3 0

2 years ago

Help me balence: ___ H2 + ___ O2----> _____ H20

luda_lava [24]

Explanation:

2H2+O2------->2H2O

its yr balanced equation.

hope it helps

<h2>stay safe healthy and happy....</h2>

6 0

3 years ago

What mass of agno3 can be dissolved in 250g of water at 20°c?

Mice21 [21]

Hello there,

You should know the solubility of AgNO3 in water at 20°C equals to 2220 g/L.

So we can say that in 1 L of water, 2220 g of AgNO3 can be dissolve.

Now you should know 1L = 1000g.

Which means 1000 g of water can dissolve 2220 g of AgNO3.

Therefore :

250 g --> x
1000 g --> 2220 g

So : $x = \frac{250*2220}{1000} = \frac{555000}{1000}= 555g$ .

In short, 555g of AgNO3 can be dissolved in 250g of water at 20°C.

Hope this helps !

Photon

4 0

4 years ago

Calculate the molality of a 20.0% by mass ammonium sulfate (nh4)2so4 solution. the density of the solution is 1.117 g/ml.

olasank [31]

Hello!

We have the following data:

m1 (solute mass) = 20 % m/m
M1 (Molar mass of solute) (NH4)2 SO4 = ?
m2 (mass of the solvent) = ? (in Kg)

First we find the solute mass (m1), knowing that:

20% m/m = 20g/100mL

20 ------ 100 mL (0,1 L)
y g --------------- 1 L

y = 20/0,1
y = 200 g --> m1 = 200 g

Let's find Solute's Molar Mass, let's see:

M1 of (Nh4)2SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol

We must find the volume of the solvent and therefore its mass (m2), let us see:

d = 1,117 g/mL
m = 200 g
v (volumen of solute) = ?

$d = \dfrac{m}{V} \to V = \dfrac{m}{d}$

$V = \dfrac{200\:\diagup\!\!\!\!g}{1,117\:\diagup\!\!\!\!g/mL} \to V = 179\:mL\:(volumen\:of\:solute)$

The solvent volume will be:

1000 -179 => V = 821 mL (volumen of disolvent)

If: 1 mL = 1g

Then the mass of the solvent is:

m2 (mass of the solvent) = 821 g → m2 (mass of the solvent) = 0,821 Kg

Now, we apply all the data found to the formula of Molality, let us see:

$\omega = \dfrac{m_1}{M_1*m_2}$

$\omega = \dfrac{200}{132*0,821}$

$\omega = \dfrac{200}{108,372}$

$\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark$

_________________________________
_________________________________

Another way to find the answer:

We have the following data:

W (molality) = ? (in molal)
n (number of mols) = ?
m1 (solute mass) = 20 % m/m = 20g/100mL → (in g to 1L) = 200 g
m2 (disolvent mass) the remaining percentage, in the case: 80 % m/m = 800 g → m2 (disolvent mass) = 0,8 Kg
M1 (Molar mass of solute) (NH4)2 SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol

Let's find the number of mols (n), let's see:

 $n = \dfrac{m_1}{M_1}$

$n = \dfrac{200}{132}$

$n \approx 1,5\:mol$

Now, we apply all the data found to the formula of Molality, let us see:

$\omega = \dfrac{n}{m_2}$

$\omega = \dfrac{1,5}{0,8}$

$
\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark$

I hope this helps. =)

7 0

3 years ago