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2 years ago

5

A drug user who has developed a high tolerance to a drug could take an overdose large enough to cause death.

1 answer:

lidiya [134]2 years ago

4 0

If you're wanting to know true or false, then yes they can.

When they develop a tolerance for it, they don't get the same effects as the did before the tolerance, so then they'd take more of it to get those effects back. If they keep having to increase the amount, they eventually take too much, overdose, and in a lot of cases, die.

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The graphic shows two isomers of a chemical compound with molecular formula C5H11NO2. Which type of isomers are these? A. struct

-BARSIC- [3]

The two compounds shown indeed have tha same molecular formula, C5 H11 NO2. One of the molecules has a group NH2 and a group COOH, the other molecule has a NOO group, that makes that the two isomers have a completely different structure, with the atoms arranged in a completely different order. <span>This kind of isomers fits in the definition of structural isomers, so the answer is structural isomers.</span>

6 0

3 years ago

Read 2 more answers

Read the following summary of the article

True [87]

Answer:

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Explanation:

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6 0

2 years ago

A volume of 90.0 mLmL of aqueous potassium hydroxide (KOHKOH) was titrated against a standard solution of sulfuric acid (H2SO4H2

Alja [10]

Answer:

0.823 M was the molarity of the KOH solution.

Explanation:

$H_2SO_4+KOH\rightarrow K_2SO_4+2H_2O$ (Neutralization reaction)

To calculate the concentration of base , we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is KOH.

We are given:

$n_1=2\\M_1=1.50 M\\V_1=24.7 mL\\n_2=1\\M_2=?\\V_2=90.0 mL$

Putting values in above equation, we get:

$2\times \1.50 M\times 24.7 mL=1\times M_2\times 90.0 mL$

$M_2=\frac{2\times 1.50M\times 24.7 mL}{1\times 90.0 mL}=0.823 M$

0.823 M was the molarity of the KOH solution.

7 0

2 years ago

You are a chemist working in your laboratory. In your storage closet you have a

Nastasia [14]

Ans is B.Bromine. Only two elements are liquid at room temperature ( mercury and Bromine) . Helium and chlorine, arsenic are gases.

8 0

2 years ago

Iodine, I2, is a solid at room temperature but sublimes (converts from a solid into a gas) when warmed. What is the temperature

Stels [109]

<h3>Answer:</h3>

382.63 K

<h3>Explanation:</h3>

We are given;

Volume of Iodine as 71.4 mL
Mass of Iodine as 0.276 g
Pressure of Iodine as 0.478 atm

We are required to calculate the temperature of Iodine

We are going to use the ideal gas equation;

According to the ideal gas equation; PV = nRT, where R is the ideal gas constant, 0.082057 L.atm/mol.K.

Rearranging the formula;

T = PV ÷ nR

But, n, the number of moles = Mass ÷ Molar mass

Molar mass of iodine = 253.8089 g/mol

Thus, n = 0.276 g ÷ 253.8089 g/mol

= 0.001087 moles

Therefore;

T = (0.478 atm × 0.0714 L) ÷ (0.001087 moles × 0.082057)

= 382.63 K

Thus, the temperature of Iodine in Kelvin is 382.63 K

3 0

3 years ago