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Mashutka [201]
4 years ago
6

When you look at a speck of mold how many microbes are you most likely seeing

Chemistry
2 answers:
Zigmanuir [339]4 years ago
7 0
You will see about 50.
Blababa [14]4 years ago
4 0
Can you even SEE microbes?

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The surface of the earth is divided into more than 50 large plates
Romashka-Z-Leto [24]

Major Plates

Africa Plate
Antarctic Plate
Indo-Australian Plate
Australian Plate
Eurasian Plate
North American Plate
South American Plate
<span>Pacific Plate
Minor Plates
There are dozens of smaller plates, the seven largest of which are:
</span>Arabian Plate
Caribbean Plate
Juan de Fuca Plate
Cocos Plate
Nazca Plate
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<span>Scotia Plate</span>
5 0
3 years ago
What is the force of gravity acting on an object? (3 points)
oee [108]

the answer of these question is weight

7 0
3 years ago
A scientist measures the standard enthalpy change for the following reaction to be -17.2 kJ : Ca(OH)2(aq) 2 HCl(aq)CaCl2(s)
elixir [45]

Answer: \Delta H^{0}=-173.72 kJ/mol

Explanation: <u>Enthalpy</u> <u>Change</u> is the amount of energy in a reaction - absorption or release - at a constant pressure. So, <u>Standard</u> <u>Enthalpy</u> <u>of</u> <u>Formation</u> is how much energy is necessary to form a substance.

The standard enthalpy of formation of HCl is calculated as:

\Delta ^{0}=\Sigma H_{products}-\Sigma H_{reactants}

Ca(OH)_{2}_{(aq)}+2HCl_{(aq)} → CaCl_{2}_{(s)}+2H_{2}O_{(l)}

Standard Enthalpy of formation for the other compounds are:

Calcium Hydroxide: \Delta H^{0}= -1002.82 kJ/mol

Calcium chloride: \Delta H^{0}= -795.8 kJ/mol

Water: \Delta H^{0}= -285.83 kJ/mol

Enthalpy is given per mol, which means we have to multiply by the mols in the balanced equation.

Calculating:

-17.2=[-795.8+2(285.85)]-[-1002.82+2\Delta H]

-17.2=-1367.46+1002.82-2\Delta H

2\Delta H=17.2-364.64

\Delta H=-173.72

So, the standard enthalpy of formation of HCl is -173.72 kJ/mol

8 0
3 years ago
The reaction 2a → a2​​​​​ was experimentally determined to be second order with a rate constant, k, equal to 0.0265 m–1min–1. if
katovenus [111]

The integrated rate law for a second-order reaction is given by:

\frac{1}{[A]t} =   \frac{1}{[A]0} + kt

where, [A]t= the concentration of A at time t,

[A]0= the concentration of A at time t=0

<span>k =</span> the rate constant for the reaction


<u>Given</u>: [A]0= 4 M, k = 0.0265 m–1min–1 and t = 180.0 min


Hence, \frac{1}{[A]t} = \frac{1}{4} + (0.0265 X 180)

<span>                                        = 4.858</span>

<span><span><span>Therefore, [A]</span>t</span>= 0.2058 M.</span>

<span>
</span>

<span>Answer: C</span>oncentration of A, after 180 min, is 0.2058 M

7 0
3 years ago
Read 2 more answers
HELP ME WITH THIS LAST QUESTION FROM PAGE 2
spayn [35]

Carbonates

OPTION A is the correct answer

7 0
3 years ago
Read 2 more answers
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