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3 years ago

What properties of matter are intensive independent of their size

Chemistry

1 answer:

Paha777 [63]3 years ago

8 0

Answer:

temperature, refractive index, density, and hardness of an object

Explanation:

it is a bulk property, meaning that it is a physical property of a system that does not depend on the system size or the amount of material in the system

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"What volume of" a 0.300 M BaF2 "solution is needed to prepare" 240.0 mL of a 0.100 M F- solution

zhannawk [14.2K]

Use M x V = M' x V'

0.300 x V = 0.100 x 250

V = .......... ml

6 0

4 years ago

A sample of a compound is decomposed in the laboratory and produces 330 g carbon, 69.5 g hydrogen, and 220.2 g oxygen. Calculate

Zarrin [17]

Answer: Empirical formula is $C_2H_5O$

Explanation: We are given the masses of elements present in a sample of compound. To evaluate empirical formula, we will be following some steps.

Step 1 : Converting each of the given masses into their moles by dividing them by Molar masses.

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of Carbon = 12.0 g/mol

Molar mass of Hydrogen = 1.0 g/mol

Molar mass of Oxygen = 16.0 g/mol

Moles of Carbon = $\frac{330g}{12g/mol}=27.5moles$

Moles of Hydrogen = $\frac{69.5g}{1g/mol}=69.5moles$

Moles of Oxygen = $\frac{220.2g}{16g/mol}=13.76moles$

Step 2: Dividing each mole value by the smallest number of moles calculated above and rounding it off to the nearest whole number value

Smallest number of moles = 13.76 moles

$\text{Mole ratio of Carbon}=\frac{27.5moles}{13.76moles}=1.99\approx 2$

$\text{Mole ratio of Hydrogen}=\frac{69.5moles}{13.76moles}=5.05\approx 5$

$\text{Mole ratio of Oxygen}=\frac{13.76moles}{13.76moles}=1$

Step 3: Now, the moles ratio of the elements are represented by the subscripts in the empirical formula

Empirical formula becomes = $C_2H_5O$

7 0

3 years ago

Help me with this please

MrRissso [65]

Answer:

I think B no is correct orterwise Ano

8 0

3 years ago

HI decomposes to H2 and I2 by the following equation: 2HI(g) → H2(g) + I2(g);Kc = 1.6 × 10−3 at 25∘C If 1.0 M HI is placed into

dsp73

Answer: The concentration of hydrogen gas at equilibrium is 0.037 M

Explanation:

We are given:

Initial concentration of HI = 1.0 M

The given chemical equation follows:

$2HI(g)\rightleftharpoons H_2(g)+I_2(g)$

Initial: 1.0

At eqllm: 1.0-2x x x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[H_2][I_2]}{[HI]^2}$

We are given:

$Kc=1.6\times 10^{-3}$

Putting values in above expression, we get:

$1.6\times 10^{-3}=\frac{x\times x}{(1.0-2x)^2}\\\\x=-0.043,0.037$

Neglecting the negative value of 'x' because concentration cannot be negative

So, equilibrium concentration of hydrogen gas = x = 0.037 M

Hence, the concentration of hydrogen gas at equilibrium is 0.037 M

6 0

3 years ago

A nuturing parent

Phantasy [73]

Answer:

Answer choice C

Explanation:

Neglecting either your own or your child's needs is always a bad idea and is bad for the child, meaning that A is not the correct answer. You should also not expect more of and push your child above their natural ability level, so B is also not correct. However, you should try to discipline and teach your child without hurting them physically or mentally. Therefore, C is the correct answer. Hope this helps!

3 0

3 years ago

Read 2 more answers