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3 years ago

A rod shaped organism with no nucleus commonly found in Garbage will most likely belong to which kingdom

Chemistry

1 answer:

Dovator [93]3 years ago

8 0

Answer:

Monera

Explanation:

This organism is only bacteria (bacilli bacteria). They are simple unicellular organism; they have no nucleus. They barely have organelles. These organism in the garbage are heterotrophic /saprophytic and cause the composition of the organic molecules (such as the discarded foods).

You might be interested in

The Difference Between an Alloy and an Amalgam

Delicious77 [7]

Mixtures or combinations of various different metals or metallic substances form things called alloys. An alloy composed of mercury and other metal (or metals) forms "amalgam". When a true alloy is created, the component metals are combined together at a temperature which is greater than the melting point of all of them.

Also, it helps to remember the word "amalgamate", which means "to alloy (a metal) with mercury" according to Dictionary.com.

Hope this helped :)

(btw I'm like 3 brainliest answers away from my next rank so could you...you know... :)

6 0

3 years ago

The enzyme, carbonic anhydrase, is a large zinc-containing protein with a molar mass of 3.00 x10^4 g/mol. Zn is 0.218% by mass o

tia_tia [17]

The mathematical expression for mass percent is given by:

$Mass percent = \frac{mass of the compound}{molar mass of the compound}\times 100$

Put the values,

$0.218 percent of zinc = \frac{mass of zinc}{3.00 \times 10^{4} g/mol}\times 100$

mass of zinc = $0.218 \times (3\times 100) g$

= $0.654 \times 100 g$ or 65.4 g

Now, number of moles of zinc = $\frac{given mass in g}{molar mass of zinc}$

= $\frac{65.4 g}{65.38 g/mol}$

= 1.00 mole

Number of atoms of zinc is calculated by the Avogadro number.

Now, according to mole concept

$6.022\times 10^{23}$ molecules of enzyme consists of $6.022\times 10^{23}$ atoms of zinc

So, 1 molecule of enzyme contains = $\frac{6.022\times 10^{23}atoms of zinc}{6.022\times 10^{23} }$

= 1 atom of zinc.

Hence, every carbonic anhydrase molecule consists of 1 atom of zinc.

5 0

3 years ago

Which of the following is the best name for CaF2?

Vladimir79 [104]

Calcium Flouride. It's an ionic bond. Cation + anion with the suffix -ide

8 0

3 years ago

Read 2 more answers

If the volume occupied by the air in a bicycle pump is 525 cm3, and the pressure changes from 73.2 kPa to 122.5 k.Pa as the pist

Musya8 [376]

Answer:

$V_2=313.71\ cm^3$

Explanation:

Given that,

Initial volume, $V_1=525\ cm^3$

The pressure changes from 73.2 kPa to 122.5 k.Pa.

We need to find the new volume occupied by the air. Let it is V₂. It can be calculated using Boyle's law such that,

$P_1V_1=P_2V_2\\\\V_2=\dfrac{P_1V_1}{P_2}\\\\V_2=\dfrac{73.2\times 525}{122.5}\\\\V_2=313.71\ cm^3$

So, the new volume is $313.71\ cm^3$ .

6 0

2 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