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3 years ago

Jen makes a Venn diagram to compare active transport and passive transport.

Chemistry

2 answers:

Oksanka [162]3 years ago

6 0

Answer:

moves molecules

Explanation:

I did this one yesterday! Active transport moves low to high concentration and passive does the opposite so C and D are not an option. Active transport requires energy and Passive Transport does not so it has to be A.!

Bond [772]3 years ago

6 0

Moves Molecules I got this answer right before:)

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How many grams are in 2.3 x 1024 formula units of KNO3?

Harlamova29_29 [7]

<h2>Answer:</h2>

390 g KNO₃

<h2>General Formulas and Concepts:</h2><h3><u>Chemistry</u></h3>

<u>Atomic Structure</u>

Reading a Periodic Table
Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<h2>Explanation:</h2>

<u>Step 1: Define</u>

2.3 × 10²⁴ formula units KNO₃

<u>Step 2: Identify Conversions</u>

Avogadro's Number

Molar Mass of K - 39.10 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of O - 16.00 g.mol

Molar Mass of KNO₃ - 39.10 + 14.01 + 3(16.00) = 101.11 g/mol

<u>Step 3: Convert</u>

<u /> $2.3 \cdot 10^{24} \ formula \ units \ KNO_3(\frac{1 \ mol \ KNO_3}{6.022 \cdot 10^{23} \ formula \ units \ KNO_3} )(\frac{101.11 \ g \ KNO_3}{1 \ mol \ KNO_3} )$ = 386.172 g KNO₃

<u>Step 4: Check</u>

<em>We are given 2 sig figs. Follow sig fig rules and round.</em>

386.172 g KNO₃ ≈ 390 g KNO₃

5 0

3 years ago

Help pleaaasee??????????? can u tell me whether it's physical or chemical change

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5 0

3 years ago

gulaghasi [49]

Taking into account the change of units, the size of the tank is 54.88 liters.

In first place, the rule of three is a tool used to quickly solve problems involving a proportional relationship between two variables.

The rule of three is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them.

If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other), the direct rule of three should be applied.

The rule of three is applied as follows, with A, B and C being known values and D being an unknown value: if A maintains a certain relationship with B, and it is known that C and D have the same relationship, it is possible to calculate the unknown value D using the following expression:

$D=\frac{CxB}{A}$

In this case, the rule of three can be applied as follows: if 1 gallon is equal to 3.785 liters, 14.5 gallons are equal to how many liters?

$D=\frac{14.5 gallonsx3.785 liters}{1 gallon}$

Solving:

D= 54.88 liters

In summary, the size of the tank is 54.88 liters.

Learn more with this example:

7 0

2 years ago

Read 2 more answers

Calculate δg o for each reaction using δg of values:(a) h2(g) + i2(s) → 2hi(g) kj (b) mno2(s) + 2co(g) → mn(s) + 2co2(g) kj (c)

steposvetlana [31]

Part (a) :
H₂(g) + I₂(s) → 2 HI(g)
From given table:
G HI = + 1.3 kJ/mol
G H₂ = 0
G I₂ = 0
ΔG = G(products) - G(reactants) = 2 (1.3) = 2.6 kJ/mol

Part (b):
MnO₂(s) + 2 CO(g) → Mn(s) + 2 CO₂(g)
G MnO₂ = - 465.2
G CO = -137.16
G CO₂ = - 394.39
G Mn = 0
ΔG = G(products) - G(reactants) = (1(0) + 2*-394.39) - (-465.2 + 2*-137.16) = - 49.3 kJ/mol

Part (c):
NH₄Cl(s) → NH₃(g) + HCl(g)
ΔG = ΔH - T ΔS
ΔG = (H(products) - H(reactants)) - 298 * (S(products) - S(reactants))
= (-92.31 - 45.94) - (-314.4) - (298 k) * (192.3 + 186.8 - 94.6) J/K
= 176.15 kJ - 84.78 kJ = 91.38 kJ

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3 years ago

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With the levels closest to the nucleus of

Dmitry_Shevchenko [17]

Answer:

The first energy level is closest to the nucleus. The second energy level is a little farther away than the first. ... The electrons in the energy level farthest from the nucleus are called valence electrons. Atoms in the same column (group) in the periodic table have the same number of valence electrons.

Explanation:

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3 years ago