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vagabundo [1.1K]
3 years ago
6

This biome has the nutrient poor soil

Chemistry
1 answer:
nikklg [1K]3 years ago
7 0

Answer:

the answer is the tropical rain forest

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A gas that occupies 50.0 liters has its volume increased to 68 liters when the pressure was changed to 3.0 ATM. What was the ori
Yuliya22 [10]

Answer:

4.1 atm = 3,116 mmHg = 415.4 kPa

Explanation:

According to Boyle's law, as volume is increased the pressure of the gas is decreased. That can be expressed as:

P₁ x V₁= P₂ x V₂

Where P₁ and V₁ are the initial pressure and volume respectively, and P₂ and V₂ are final pressure and volume, respectively.

From the problem, we have:

V₁= 50.0 L

V₂= 68.0 L

P₂= 3.0 atm

Thus, we calculate the initial pressure as follows:

P₁= (P₂ x V₂)/V₁= (3.0 atm x 68.0 L)/(50.0 L)= 4.08 atm ≅ 4.1 atm

To transform to mmHg, we know that 1 atm= 760 mmHg:

4.1 atm x 760 mmHg/1 atm = 3,116 mmHg

To transform to kPa we use: 1 atm= 101.325 kPa

4.1 atm x 101.325 kPa = 415.4 kPa

5 0
3 years ago
I need to check some chemistry questions. Help with any question is appreciated! :) Please include an explanation with your conc
finlep [7]

1. 4.67 kg; 2. 4.8 ×10^5 kg; 3. 0.106 cm^3; 4. 1.7 g/cm^3

<em>Q1. Mass of Hg </em>

Mass = 345 mL × (13.53 g/1 mL) = 4670 g = 4.67 kg

<em>Q2. Mass of Pb </em>

<em>Step 1</em>. Calculate the <em>volume of the Pb</em>.

<em>V = lwh</em> = 6.0 m × 3.5 m × 2.0 m = 42.0 m^3

<em>Step 2</em>. Calculate the <em>mass of the Pb</em>.

Mass = 42.0 m^3 × (11 340 kg/1 m^3) = 4.8 × 10^5 kg

<em>Q3. Volume of displaced water </em>

Volume of Ag = 0.987 g × (1 cm^3/9.320 g) = 0.106 cm^3

<em>Archimedes</em>: volume of displaced water = volume of Ag = <em>0.106 cm^3</em>

<em>4. Density of metal </em>

<em>Step 1</em>. Convert <em>ounces to grams </em>

Mass = 3.35 oz × (28.35 g/1 oz) = 94.97 g

<em>Step 2</em>. Calculate the <em>volume in cubic inches </em>

<em>V = lwh</em> = 3.0 in × 2.5 in × 0.45 in = 3.38 in^3

<em>Step 3</em>. Convert <em>cubic inches to cubic centimetres</em><em> </em>

<em>V</em> = 3.38 in^3 × (2.54 cm/1 in)^3 = 55.3 cm^3

<em>Step 4</em>. Calculate the <em>density</em>

ρ = <em>m</em>/<em>V</em> = (94.97 g/55.3 cm^3) = 1.7 g/cm^3 (magnesium?)

4 0
3 years ago
Rank the following salts in order of decreasing pH of their 0.1 M aqueous solutions.(a) FeCl2, FeCl3, MgCl2, KClO2 .(b) NH4Br, N
Firlakuza [10]

Answer:

a) FeCl2, FeCl3, MgCl2, KClO2.

KClO2 --> K+ + ClO2-; ClO2- will hydrolyse to form HClO +OH-

Mg+2, Fe+2 and Fe+3 ions will form acidic solutions, since theyfom slightly amount of

Mg+2 + 2H2O <-> Mg(OH)2 + 2H+

Fe+2 + 2H2O <-> Fe(OH)2 + 2H+

Fe+3 + 3H2O <-> Fe(OH)3 + 3H+

Therefore;

decreasing pH is high pH to low pH:

KClO2 > MgCl2 > FeCl2 > FeCl3

b) NH4Br, NaBrO2, NaBr, NaClO2.

NH4Br is acidic, forms NH4+ and NH4+ dnates H+ to form NH3 andH+

NaBrO2 is basic, forms Na+ + BrO2- then H2O + BrO2- HBrO2

NABr is neutral, NaClO2 is basics, forms Na+ + ClO2-then H2O + ClO2- HClO2

decreasig pH:

NaClO2 > NaBrO2 > NaBr >NH4Br

Note that HClO2 is stronger acid than HBrO2, therefore, expectmore HBrO2 formation

NaBrO2 > NaClO2 > NaBr >NH4Br

5 0
3 years ago
What is the pH of an acid with a [H+] of 1.0 x 10&amp;-4 M?<br><br> 1. 4<br> 2. 10<br> 3. 1<br> 4. 7
True [87]
PH scale is from 1 to 14 and indicates how acidic or basic a solution is. To find pH or pOH we need to know the H⁺ ion concentration or OH⁻ concentration.
pH can be calculated using the following equation;
pH = -log[H⁺]
the H⁺ concentration of the given acid is 1.0 x 10⁻⁴ M. substituting this we can find the pH 
pH = -log[1x10⁻⁴]
pH = 4
answer is 1) 4
3 0
3 years ago
What is an echo? (1 point)
zaharov [31]

Answer:

The reflection of sound waves

Explanation:

An echo is a reflection of sound that bounces of of one thing to another.

4 0
3 years ago
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