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ohaa [14]
3 years ago
10

Which substance can lead to addiction, or drug dependence, if abused?

Chemistry
1 answer:
worty [1.4K]3 years ago
7 0

Answer:

D

Explanation:

All of those can produce happy chemicals which cause your brain to become dependent. The side effects can be life-threatening but if it makes your brain irrationally happy then it doesn't matter

You might be interested in
Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of sodium sulfa
notka56 [123]

Answer:

27%

Explanation:

Hello,

The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

2NaOH+H_2SO_4-->Na_2SO_4+2H_2O

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4

Finally, the percent yield turns out into:

Y=\frac{1.92g}{7.1g} *100

Y=27.0%

Best regards.

6 0
3 years ago
GoOd MoRnInG <br> EvErY OnE <br> BlA BlA BlA BlA
azamat

Answer:

you too ^^

Explanation:

4 0
3 years ago
Read 2 more answers
How many ions are present in iron 2 oxide
krek1111 [17]
Iron(II) oxide also refers to a family of related non-stoichiometric compounds, which are typically iron deficient with compositions ranging from Fe0.84O to Fe0.95O.
8 0
3 years ago
A sample of gas in a balloon has an initial temperature of 20. ∘C and a volume of 1.92×103 L . If the temperature changes to 68
morpeh [17]

Answer:

\boxed{2.23 \times 10^{3} \text{ L}}

Explanation:

The pressure is constant, so we can use Charles' Law.

\dfrac{ V_{1} }{T_{1}} = \dfrac{ V_{2} }{T_{2}}

Data:

V₁ = 1.92 × 10³ L; T₁ = 20 °C  

V₂ = ?;                 T₂ = 68 °C

Calculations:

(a) Convert temperatures to kelvins

T₁ = (20 + 273.15) K = 293.15 K

T₂ = (68 + 273.15) K = 341.15 K

(b) Calculate the volume

\dfrac{ 1.92 \times 10^{3}}{293.15} = \dfrac{ V_{2}}{341.15}\\\\6.550 = \dfrac{ V_{2}}{341.15}\\\\V_{2} = 6.550 \times 341.15 = 2.23 \times 10^{3} \text{ L}

The new volume of the gas is \boxed{2.23 \times 10^{3} \text{ L}}.

6 0
3 years ago
If 5 L of butane is reacted what volume of carbon dioxide is produced ILL GIVE BRAINLIEST
Len [333]

Answer: First, here is the balanced reaction:  2C4H10  +  13O2  ===>  8CO2  +  10H2O.

This says for every mole of butane burned 4 moles of CO2 are produced, in other words a 2:1 ratio.

Next, let's determine how many moles of butane are burned.  This is obtained by

5.50 g / 58.1 g/mole  =  0.0947 moles butane.  As CO2 is produced in a 2:1 ratio, the # moles of CO2 produced is 2 x 0.0947  =  0.1894 moles CO2.

Now we need to figure out the volume.  This depends on the temperature and pressure of the CO2 which is not given, so we will assume standard conditions:  273 K and 1 atmosphere.

We now use the ideal gas law PV = nRT, or V =nRT/P, where n is the # of moles of CO2, T the absolute temperature, R the gas constant (0.082 L-atm/mole degree), and P the pressure in atmospheres ( 1 atm).

V = 0.1894 x 0.082 x 273.0 / 1  =  4.24 Liters.

Explanation:

8 0
3 years ago
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