Can someone please help

1. Which is not a characteristic of background radiation?

siniylev [52]

Answer: C.) It burns and causes lung cancer

5 0

3 years ago

A 41.1 g sample of solid CO2 (dry ice) is added to a container at a temperature of 100 K with a volume of 3.4 L. If the containe

Morgarella [4.7K]

Answer:

The pressure inside the container is 6.7 atm

Explanation:

We have the ideal gas equation: P x V = n x R x T

whereas, P (pressure, atm), V (volume, L), n (mole, mol), R (ideal gas constant, 0.082), T (temperature, Kelvin)

Since the container is evacuated and then sealed, the volume of the body of gas is the volume of the container.

So we can calculate the pressure by

P = n x R x T / V

where as,

n = 41.1 g / 44 g/mol = 0.934 mol

Hence P = 0.934 x 0.082 x 298 / 3.4 L = 6.7 atm

3 0

3 years ago

A compound is determined to have the empirical formula C2OH4. If the molar mass of the compound is 132 g/mol, determine the mole

8_murik_8 [283]

Answer: The molecular formula will be $C_6O_3H_{12}$

Explanation:

Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.

Empirical weight of $C_2OH_{4}$ is $12\times 2+16\times 1+1\times 4=44g$

Molecular mass of compound is = 132 g

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{132}{44}=3$

The molecular formula will be= $3\times C_2OH_4=C_6O_3H_{12}$

5 0

3 years ago

n an experiment, 39.26 mL of 0.1062 M NaOH solution was required to titrate 37.54 mL of \ v unknown acetic acid solution to a ph

olasank [31]

Answer:

Molarity: 0.111M

% (w/w): 0.666

Explanation:

The reaction of NaOH with acetic acid (CH₃COOH) is:

NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O

where 1 mole of NaOH reacts per mole of acetic acid producing 1 mole of water and 1 mole of sodium acetate.

As 39.26mL ≡ 0.03926L of 0.1062M are required to titrate the solution of acetic acid. Moles are:

0.03926L × (0.1062mol / L) = 4.169x10⁻³ moles of NaOH. As 1 mole of NaOH reacts per mole of acetic acid:

4.169x10⁻³ moles of CH₃COOH.

Molarity is defined as ratio between moles of substance and volume of solution in liters. Thus, molarity of acetic acid solution is:

4.169x10⁻³ moles of CH₃COOH / 0.03754L = 0.111M

As molar mass of acetic acid is 60g/mol, 4.169x10⁻³ moles weights:

4.169x10⁻³ moles × (60g / mol) = 0.2501 g of acetic acid

Now, assuming density of solution as 1.00g/mL, 37.54mL weights 37.54g.

Thus, percent by weight is:

0.2501g CH₃COOH / 37.54g × 100 = 0.666% (w/w)

8 0

3 years ago

Respond if you got a min to help me with my problem

grandymaker [24]

Answer:

I mean i got my own problems

Explanation:

3 0

3 years ago

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