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2 years ago

During El Niño years, why is Christchurch, New Zealand’s temperature cooler than usual?

Chemistry

1 answer:

valina [46]2 years ago

3 0

Answer:

It's probably claim 3

Explanation:

It's not claim 1 because the amount of energy from the sun can't change and claim 2 doesn't really make sense.

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Which branch of chemistry would be used by physicians when deciding

weqwewe [10]

Explanation:

Medicinal chemistry and pharmaceutical chemistry are disciplines at the intersection ofchemistry, especially synthetic organicchemistry, and pharmacology and various other biological specialties, where they are involved with design,chemical synthesis and development for market of pharmaceutical agents, or bio-active ...

6 0

3 years ago

Does the energy go from the surrounding to the chemicals or from the chemicals to the surround in an exothermic reaction?

Karo-lina-s [1.5K]

In general terms, exothermic reactions release energy, so the energy goes from the system to the surroundings.

4 0

2 years ago

For a first-order reaction, A → B, the rate coefficient was found to be 3.4 × 10-4 s-1 at 23 °C. After 5.0 h, the concentration

Illusion [34]

Answer:

the original concentration of A = 0.0817092 M

Explanation:

A reaction is considered to be of first order it it strictly obeys the graphical equation method.

$k_1 = \dfrac{2.303}{t}log \dfrac{a}{a-x}$

where;

k = the specific rate coefficient = 3.4 × 10⁻⁴ s⁻¹

t = time = 5.0 h = 5.0 × 3600 = 18000 seconds

a = initial concentration = ???

a - x = remaining concentration of initial concentration at time t = 0.00018 mol L⁻¹

$3.4 \times 10^{-4}= \dfrac{2.303}{18000}log \dfrac{a}{0.00018}$

$3.4 \times 10^{-4}= 1.27944 \times 10^{-4} \times log \dfrac{a}{0.00018}$

$\dfrac{3.4 \times 10^{-4}}{1.27944 \times 10^{-4}}= log \dfrac{a}{0.00018}$

$2.657= log \dfrac{a}{0.00018}$

$10^{2.657}= \dfrac{a}{0.00018}$

$453.94 = \dfrac{a}{0.00018}$

$a =453.94 \times 0.00018$

a = 0.0817092 M

Thus , the original concentration of A = 0.0817092 M

8 0

3 years ago

A liquid-fueled rocket engine has two liquids, usually called the fuel and the oxidizer. Which two liquids are most commonly use

hammer [34]

Answer:

A liquid-fueled rocket has two liquids (liquids are good because of the density, they need less space than a gas to be stored), such that these liquids are called the fuel and the oxidizer.

These liquids are injected into a system that leads to a combustion chamber, where the liquids are mixed (we need to mix the fuel with the oxidizer to enable the combustion of the fuel) and burned to produce thrust.

Some common examples of oxidizers are liquid oxygen, which may be combined with fuels like liquid hydrogen, liquid methane, kerosene and hydrazine.

Other oxidizers are liquid fluorine (which also can be combined with the fuels liquid hydrogen and hydrazine), nitrogen tetroxide (which can be combined whit kerosene, hydrazine and other fuels) and FLOX-70, which can only be combined with kerosene.

The "most commonly used" may depend on the country and the type of liquid propellant ( petroleum, cryogens, and hypergols)

Such that the most common oxidizer may be liquid oxygen, and the most common fuel the kerosene.

8 0

3 years ago

How many molecules of carbon dioxide are dissolved in 0.550 L of water at 25 °C if the pressure of CO2 above the water is 0.250

Grace [21]

<u>Answer:</u> The number of molecules of carbon dioxide gas are $2.815\times 10^{21}$

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{CO_2}=K_H\times p_{CO_2}$

where,

$K_H$ = Henry's constant = $0.034mol/L.atm$

$C_{CO_2}$ = molar solubility of carbon dioxide gas

$p_{CO_2}$ = pressure of carbon dioxide gas = 0.250 atm

Putting values in above equation, we get:

$C_{CO_2}=0.034mol/L.atm\times 0.250atm\\\\C_{CO_2}=8.5\times 10^{-3}M$

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of carbon dioxide = $8.5\times 10^{-5}M$

Volume of solution = 0.550 L

Putting values in above equation, we get:

$8.5\times 10^{-3}M=\frac{\text{Moles of }CO_2}{0.550L}\\\\\text{Moles of }CO_2=(8.5\times 10^{-3}mol/L\times 0.550L)=4.675\times 10^{-3}mol$

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules

So, $4.675\times 10^{-3}$ moles of carbon dioxide will contain = $(6.022\times 10^{23}\times 4.675\times 10^{-3})=2.815\times 10^{21}$ number of molecules

Hence, the number of molecules of carbon dioxide gas are $2.815\times 10^{21}$

3 0

3 years ago

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