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podryga [215]
3 years ago
9

The density of mercury is 13.6 g/cm3, What is its density in mg/mm3?

Chemistry
1 answer:
forsale [732]3 years ago
8 0

Answer:

Density of mercury is 13600 kg

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How many atoms are in 5.2 moles of Na ?
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One mole of a substance is defined by Avogadro as consisting of 6.022 x 1023 atoms. This is Avogadro's number. To calculate the number of atoms in two moles of sodium, use dimensional analysis. 2.0 moles Na x 6.022⋅1023g1mol=1.20⋅1024 atoms of Na

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What is dalang puppeteer​
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3 0
3 years ago
In AG calculations, temperature is expressed in a. degrees Celsius. b. kelvins. c. degrees Fahrenheit, d. kilojoules.
schepotkina [342]

Answer:

The correct option is: a. degrees Celsius

Explanation:

The anion gap is the difference in the cations and anions in plasma, serum or urine, calculated from medical lab test results. It can be calculated by measuring the concentration of the anions or cations, which are expressed in millimoles/litre (mmol/L) or milliequivalents/liter (mEq/L).

The temperature in this test is expressed in degrees Celsius (°C).

6 0
3 years ago
Sometimes people do not want to change their belief regardless of the evidence.
bagirrra123 [75]

Answer:

True

Explanation:

Ex. People against vaccines tend to believe crazy theories over facts because they think they'll end up brainwashing them.

6 0
2 years ago
The reaction C 4 H 8 ( g ) ⟶ 2 C 2 H 4 ( g ) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ / mol. 262 kJ/mol. At 600.0 K,
ludmilkaskok [199]

Answer: 4.3\times 10^{-13}s^{-1}

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 600.0K = 6.1\times 10^{-8}s^{-1}

K_2 = rate constant at 775.0 = ?

Ea = activation energy for the reaction = 262 kJ/mol = 262000J/mol

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 600.0K

T_2 = final temperature = 775.0K

Now put all the given values in this formula, we get

\log (\frac{6.1\times 10^{-8}}{K_2})=\frac{262000}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{775.0K}]

\log (\frac{6.1\times 10^{-8}s^}{K_2})=5.150

(\frac{6.1\times 10^{-8}}{K_2})=141253.8

Therefore, the value of the rate constant at 775.0 K is 4.3\times 10^{-13}s^{-1}

5 0
2 years ago
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