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3 years ago

Studying and got stuck if someone can help me.

Chemistry

1 answer:

Lostsunrise [7]3 years ago

3 0

Answer:

Ph level 0-7 is a acid

ph level 7-14 is base

bases turn paper blue

acids dissolve metal

acids contain hydrogen ions

Explanation:

I dont know the last one all I know that they become neutral when acids and bases are mixed.

You might be interested in

Please answer both questions.

Deffense [45]

Answer:

1)Krypton

2)11H

Explanation:

electrons=protons

protons=atomic number

mass number=protons+neutrons

mass number is the superscript

atomic number is the subscript.

1)The answer is Krypton because its atomic number= number of protons=number of electrons is 36.

mass number is 46+36=82.

2)subscript=atomic number=number of protons=number of electrons

i. H = electrons=1

=neutrons=0

ii. Cl=electrons=17

=neutrons=35-17=18

iii. Na=electrons=11

=neutrons=23-11= 12

so the answer is Hydrogen because it has 1 electron and 0 neutron.

I hope this helps.

6 0

3 years ago

Select the correct answer.

Illusion [34]

Answer: the new force will be 1/9 of the original force

Explanation: Please see attachment for explanation

5 0

3 years ago

Write the balanced chemical equation between H2SO4 and KOH in aqueous solution. This is called a neutralization reaction and wil

emmainna [20.7K]

Answer:

0.166M

Explanation:

In a neutralization, the acid, H₂SO₄, reacts with a base, KOH, to produce a salt, K₂SO₄ and water. The reaction is:

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

To solve this problem, we need to determine moles of H2SO4 and moles of KOH that reacts to find the moles of sulfuric acid that remains after the reaction:

<em>Moles H2SO4:</em>

0.650L * (0.430mol /L) = 0.2795moles H2SO4

<em>Moles KOH:</em>

0.600L * (0.240mol / L) = 0.144 moles KOH

Moles of sulfuric acid that reacts with 0.144 moles of KOH are:

0.144 moles KOH * (1mol H2SO4 / 2 mol KOH) = 0.072 moles of H2SO4 react.

And remain:

0.2795moles H2SO4 - 0.072moles H2SO4 = 0.2075 moles of H2SO4 reamains.

In 0.650L + 0.600L = 1.25L:

Molar concentration of sulfuric acid:

0.2075 moles of H2SO4 / 1.25L =

7 0

3 years ago

At 25 °C, what is the hydroxide ion concentration, [OH−] , in an aqueous solution with a hydrogen ion concentration of [H+]=4.8×

algol [13]

[OH⁻]= 1 x 10⁻¹⁴ : 4.8 x 10⁻⁴ = 2.083 x 10⁻¹¹

3 0

2 years ago

You are requested to reduce the size of 50 ton/hr of a given solid. The size of the feed is such 80% passes a 4-in (76.2 mm) scr

defon

Answer:

1) The power needed to process 50 ton/hr is 135.4 HP.

2) The void fraction of the bed is 0.37.

Explanation:

1) For this type of milling operations, we can estimate the power needed for an operation according to the work index (Ei), the passing size of the circuit feed (F80) and the passing size of the product (P80).

We assume the units of Ei are kWh/t.

The equation that relates this parameters and the power is (size of particles in μm):

$W=Ei*(\frac{10}{\sqrt{P80}} -\frac{10}{\sqrt{F80}} )\\\\W=9.45*(\frac{10}{\sqrt{3175\mu m}} -\frac{10}{\sqrt{76200mm}} )\\\\\\W=9.45*(0.1774+0.0362)=2.019 kWh/t$

The power needed to process 50 ton/hor is

$P=2.0194\frac{kWh}{Ton}*\frac{50Ton}{h}*\frac{1.341HP}{1kW}= 135.4 \, HP$

2) The density of the packed bed can be expressed as

$\rho=f_v*\rho_v+f_s*\rho_s$

being f the fraction and ρ the density of every fraction. We know that the density of the void is 0 (ρv=0) and that fv=1-fs (the sum of the fractions ois equal to the total space).

Then we can rearrange

$\rho=f_v*\rho_v+f_s*\rho_s\\\\\rho=f_v*0+(1-f_v)*\rho_s\\\\\rho/\rho_s=1-f_v\\\\f_v=1-\rho/\rho_s=1-990/1570=1-0.63=0.37$

The void fraction of the bed is 0.37.

6 0

4 years ago