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3 years ago

Which example is a way that intercellular communication occurs?

Chemistry

2 answers:

mafiozo [28]3 years ago

7 0

Answer: A signal is sent from a nerve cell to a muscle cell.

Explanation: i got the answer right on my test!

harina [27]3 years ago

5 0

Answer: the answer is d

Explanation: i took the test

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SOMEONE ANSWER?!!?!?!,!,EMERGENCY Can someone tell me the answer plz plz I’ll give you brainlist and points plz This is my final

Vera_Pavlovna [14]

Answer: Should be A)

Explanation:

the size of planets effects the amount of gravitational force each planet has, like jupiter, it has the most gravity.

6 0

3 years ago

Consider the reaction between iron (III) oxide, Fe2O3 and carbon monoxide, CO. Fe2O3 + 3CO --> 2Fe + 3CO2 In the process, 213

s344n2d4d5 [400]

To calculate the mass of Fe formed in a) we get first the limiting reactant between Fe2O3 and CO. Given the masses, the ratio of Fe2O3 is 1.33 while that of CO is 1.67. Hence the limiting reagent is Fe2O3. The mass of Fe formed is 148.98 grams. In b, the needed CO is only 112.04 grams. Hence, the excess is 27. 96 grams.

8 0

4 years ago

Calculate the solubilities of the following compounds in a 0.02 M solution of barium nitrate using molar concentrations, first i

Law Incorporation [45]

Answer:

a. 1.7 × 10⁻⁴ mol·L⁻¹; b. 5.5 × 10⁻⁹ mol·L⁻¹

c. 2.3 × 10⁻⁴ mol·L⁻¹; 5.5 × 10⁻⁸ mol·L⁻¹

Explanation:

a. Silver iodate

Let s = the molar solubility.

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq); Ksp = 3.0 × 10⁻⁸

E/mol·L⁻¹: s s

$K_{sp} =\text{[Ag$^{+}$][IO$_{3}$$^{-}$]} = s\times s = s^{2} = 3.0\times 10^{-8}\\s = \sqrt{3.0\times 10^{-8}} \text{ mol/L} = 1.7 \times 10^{-4} \text{ mol/L}$

b. Barium sulfate

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq); Ksp = 1.1 × 10⁻¹⁰

I/mol·L⁻¹: 0.02 0

C/mol·L⁻¹: +s +s

E/mol·L⁻¹: 0.02 + s s

$K_{sp} =\text{[Ba$^{2+}$][SO$_{4}$$^{2-}$]} = (0.02 + s) \times s \approx 0.02s = 1.1\times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{0.02} \text{ mol/L} = 5.5 \times 10^{-9} \text{ mol/L}$

c. Using ionic strength and activities

(i) Calculate the ionic strength of 0.02 mol·L⁻¹ Ba(NO₃)₂

The formula for ionic strength is

$\mu = \dfrac{1}{2} \sum_{i} {c_{i}z_{i}^{2}}\\\\\mu = \dfrac{1}{2} (\text{[Ba$^{2+}$]}\cdot (2+)^{2} + \text{[NO$_{3}$$^{-}$]}\times(-1)^{2}) = \dfrac{1}{2} (\text{0.02}\times 4 + \text{0.04}\times1)= \dfrac{1}{2} (0.08 + 0.04)\\\\= \dfrac{1}{2} \times0.12 = 0.06$

(ii) Silver iodate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^{2}\sqrt{I} = -0.051(1)^{2}\sqrt{0.06} = -0.51\times 0.24 = -0.12\\\gamma = 10^{-0.12} = 0.75$

b. Calculate the solubility

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq)

$K_{sp} =\text{[Ag$^{+}$]$\gamma_{Ag^{+}}$[IO$_{3}$$^{-}$]$\gamma_{IO_{3}^{-}}$} = s\times0.75\times s \times 0.75 =0.56s^{2}= 3.0 \times 10^{-8}\\s^{2} = \dfrac{3.0 \times 10^{-8}}{0.56} = 5.3 \times 10^{-8}\\\\s =2.3 \times 10^{-4}\text{ mol/L}$

(iii) Barium sulfate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^{2}\sqrt{I} = -0.051(2)^{2}\sqrt{0.06} = -0.51\times16\times 0.24 = -0.50\\\gamma = 10^{-0.50} = 0.32$

b. Calculate the solubility

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq

$K_{sp} =\text{[Ba$^{2+}$]$\gamma_{ Ba^{2+}}$[SO$_{4}$$^{2-}$]$\gamma_{ SO_{4}^{2-}}$} = (0.02 + s) \times 0.32\times s\times 0.32 \approx 0.02\times0.10s\\2.0\times 10^{-3}s = 1.1 \times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{2.0 \times 10^{-3}} \text{ mol/L} = 5.5 \times 10^{-8} \text{ mol/L}$

7 0

3 years ago

In the reaction A + B C, doubling the concentration of A doubles the reaction rate and doubling the concentration of B does not

frutty [35]

Answer:

rate = k[A]

Explanation:

The equation that relate reaction rate with reactant concentrations is known as the rate law.

for a reaction:

A + B → C

the rate law can be expressed as:

Rate = k[A]ᵃ[B]ᵇ

The proportionality constant, k, is known as the rate constant, the powers a and b is the reaction order with respect to reactants A and B, respectively.

for this reaction doubling the concentration of A doubles the reaction rate that means

Rate₂ = 2 *Rate₁ and [A]₂ = 2 [A]₁

Rate₁ = k[A]₁ᵃ[B]ᵇ → eq. 1

Rate₂ = k[A]₂ᵃ[B]ᵇ → eq. 2

Dividing eq. 2 by eq. 1 one can get

(Rate₂ / Rate₁) = (k [A]₂ᵃ[B]ᵇ) / (k[A]₁ᵃ[B]ᵇ)

using

Rate₂ = 2 *Rate₁ and [A]₂ = 2 [A]₁

∴ (2 Rate₁ / Rate₁) = ( k [2]ᵃ[B]ᵇ) / (k[1]ᵃ[B]ᵇ)

(2) = (2)ᵃ
taking log of both sides
log (2) = a Log (2)
0.693 = a * 0.693
a =1

∴ order of reaction with respect to A is first (=1) → (1)

Doubling the concentration of B does not affect the reaction rate.

that means

Rate₂ = Rate₁ and [B]₂ = 2 [B]₁

Rate₁ = k[A]ᵃ[B]₁ᵇ → eq. 1

Rate₂ = k[A]ᵃ[B]₂ᵇ → eq. 2

Dividing eq. 2 by eq. 1 one can get

(Rate₂ / Rate₁) = (k [A]ᵃ[B]₂ᵇ) / (k[A]ᵃ[B]₁ᵇ)

using

Rate₂ = Rate₁ and [B]₂ = 2 [B]₁

∴ (Rate₁ / Rate₁) = ( k [A]ᵃ[2]ᵇ) / (k[A]ᵃ[1]ᵇ)

(1) = (2)ᵇ
taking log of both sides
log (1) = b Log (2)
0 = 0.693 * b
b = 0

∴ order of reaction with respect to B is zero → (2)

So, from 1 and 2 the right choice is rate = k[A]¹[B]⁰= k[A]

6 0

3 years ago

Determine the maximum number of moles of product that can be produced from 7.0 mol al and 8.0 mol cl2 according to the equation

Vilka [71]

The balanced chemical reaction is

<span>2al + 3cl2 = 2alcl3

To determine the maximum amount of product, we need to determine which is the limiting reactant. Then, use the initial amount of that reactant to calculate the amount of the product that would be produced. We do as follows:

7 mol Al (3 mol Cl2 / 2 mol Al) = 10.5 mol Cl2
8 mol Cl2 ( 2 mol Al / 3 mol Cl2) = 5.3 mol Al
Thus, it is Cl2 that is the limiting reactant.

8 mol Cl2 ( 2 mol AlCl3 / 3 mol Cl2) = 5.3 moles of AlCl3 is produced</span>

5 0

4 years ago