What kind of mixture is a solution? A suspension? A colloid?

The angle between the horizon and the sun is a.large b.small at both sunrise and sunset

Inessa [10]

Answer: b

Explanation:

6 0

3 years ago

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Which one of these is NOT a sign of chemical change? *

Gekata [30.6K]

I think it’s light
Have fun :)

3 0

3 years ago

Balloon A started with a volume of 1.76 L. The temperature of the room the balloon was in was 295 K. The balloon was heated to a

valentina_108 [34]

Answer:

V₂ = 1.5 L

Explanation:

Given data:

Initial volume of balloon = 1.76 L

Initial temperature = 295 K

Final temperature = 253.15 K

Final volume = ?

Solution:

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 1.76 L ×253.15 K / 295 K

V₂ = 445.54 L.K /295 K

V₂ = 1.5 L

4 0

4 years ago

50.0 mL of 0.200 M HNO2 is titrated to its equivalence point with 1.00 M NaOH. What is the pH at the equivalence point?

yKpoI14uk [10]

Answer:

8.279

Explanation:

The pH can be determined by hydrolysis of a conjugate base of weak acid at the equivalence point.

At the equivalence point, we have

$n_{NaOH}=n_{HNO_2}$

= 25.00 x 0.200

= 5.00 m-mol

= 0.005 mol

Volume of the base that is added to reach the equivalence point is

$\frac{0.005}{1.00} \times 1000= 5.00 \ mL$

Number of moles of $NO^-_{2}=n_{HNO_2}$

= 0.005 mol

Volume at the equivalence point is 25 + 5 = 30.00 mL

Therefore, concentration of $NO^-_{2}= \frac{5}{30}$

= 0.167 M

Now the ICE table :

$NO^-_2 + H_2O \rightarrow HNO_3 + OH^-$

I (M) 0.167 0 0

C (M) -x +x +x

E (M) 0.167-x x x

Now, the value of the base dissociation constant is ,

$K_w=K_a \times K_b$ $(K_w \text{ is the ionic product of water })$

$K_b =\frac{K_w}{K_a}$

$K_b =\frac{1 \times 10^{-14}}{4.6 \times 10^{-4}}$

= $2.174 \times 10^{-11}$

Base ionization constant, $K_b = \frac{\left[HNO_2\right] \left[OH^- \right]}{\left[NO^-_2 \right]}$

$2.174 \times 10^{-11}=\frac{x^2}{0.167 -x}$

$x= 1.9054 \times 10^{-6}$

So, $[OH^-]=1.9054 \times 10^{-6 } \ M$

pOH =- $\log[OH^-]$

= $- \log(1.9054 \times 10^{-6} \ M)$

=5.72

Now, since pH + pOH = 14

pH = 14.00 - 5.72

= 8.279

Therefore the ph is 8.279 at the end of the titration.

8 0

3 years ago

Fill in the missing mass number and atomic number for each of these particles or types of radiation.

Akimi4 [234]

Answer:

Particle / mass number atomic number

type of radiation

alpha / ⁴₂ He 4 2

beta / e⁰ 0 - 1

gamma / γ 0 0

neutron / n 1 0

Explanation:

These are the basic types of radiation: alpha (α), beta (β), gamma (γ), and neutrons (n).

The radiation is emitted by unstable nuclei when undergo radiactive decay or by nuclei that are shooted by other particles.

Alpha radiation (⁴₂ He):

They are nuclei of helium-4 atoms: 2 protons and 2 neutrons.

Hence, the atomic number, which is the number of protons, of these particles, is 2; and the mass number, which is the sum of protons and neutrons, is 2 + 2 = 4.

The symbol of this radiation is ⁴₂ He, where the superscript to the left of the chemical symbol is the mass number and the subscript to the left of the chemical symbol is the atomic number.

Beta (⁰₋ ₁ e)

These are electrons emiited from an unstable nucleus. The symbols used for this particle are either ⁰₋ ₁β or ⁰₋ ₁e.

The superscript 0 indicates that the relative mass of this particle is practically zero and the subscript -1 tells that the emission of electrons increases the atomic number of the nucleus that emits it.

Gamma (⁰₀γ)

The gamma rays are electromagnetic radiation of high (the highest) energy.

The both superscript and subscript are zero, meaning that this radiation does not change either the mass or atomic numbers of the nucleus.

Neutron (¹₀n):

Neutrons are also emiited from the nucleus and so they may be considered a radiation.

The atomic number of neutrons is 0 (since it does not have protons) and its mass number is 1.

4 0

3 years ago

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