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3 years ago

Plants have functions that make them a(n) ?????????? part of the world. ^Insignificant or vital

1 answer:

5 0

I don't fully understand your question, but I believe that plants have functions that are vital to the planet. They take sunlight, make it into food for themselves, and release oxygen that we need to survive.

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In experiment 1, how many moles of benzoic acid are present? how many moles of sodium bicarbonate are contained in 1 ml of a 10%

nexus9112 [7]

First, let us calculate the moles of solute or sodium bicarbonate is in the 1 ml solution.

moles = 1 mL * (1 g / 9 mL) = 0.11 moles

The molar mass of sodium bicarbonate is 84 g/mol, therefore the mass is:

mass = 0.11 moles * 84 g/mol

mass = 9.33 g

6 0

3 years ago

Match each item with the correct statement below.

adell [148]

1. Formula

2. Compound

3. Properties

4. Solute

5. Mass

6. Matter

7. Alloy

8. aqueous solution

9. quantitative

5 0

4 years ago

Read 2 more answers

Un lote de 250 pijamas tiene un de 20.000, ¿cuál es el costo de cada pijama?

zysi [14]

Answer in file below

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6 0

3 years ago

HELPPPPP ME ASAP I'll give you a thanks and mark you as brainliest and give u points

yawa3891 [41]

Answer: 2nd one

Explanation:

PLATO

3 0

3 years ago

At a particular temperature, K = 4.1 ✕ 10−6 for the following reaction. 2 CO2(g) 2 CO(g) + O2(g) If 2.3 moles of CO2 is initiall

mestny [16]

Answer:

concentration of $[O_2]$ = 0.0124 = 12.4 ×10⁻³ M

concentration of $[CO]$ = 0.0248 = 2.48 ×10⁻² M

concentration of $[CO_2]$ = 0.4442 M

Explanation:

Equation for the reaction:

$2CO_2_{(g)$ ⇄ $2CO_{(g)$ + $O_2_{(g)$

Concentration of $CO_2_{(g)$ = $\frac{2.3}{4.9}$ = 0.469

For our ICE Table; we have:

$2CO_2_{(g)$ ⇄ $2CO_{(g)$ + $O_2_{(g)$

Initial 0.469 0 0

Change - 2x +2x +x

Equilibrium (0.469-2x) 2x x

K = $\frac{[CO]^2[O]}{[CO_2]^2}$

K = $\frac{[2x]^2[x]}{[0.469-2x]^2}$

$4.1*10^{-6}=\frac{2x^3}{(0.469-2x)^2}$

Since the value pf K is very small, only little small of reactant goes into product; so (0.469-2x)² = (0.469)²

$4.1*10^{-6} = \frac{2x^3}{(0.938)}$

$2x^3 =3.8458*10^{-6$

$x^3 =\frac{3.8458*10^{-6}}{2}$

$x^3=1.9229*10^{-6$

$x=\sqrt[3]{1.9929*10^{-6}}$

x = 0.0124

∴ at equilibrium; concentration of $[O_2]$ = 0.0124 = 12.4 ×10⁻³ M

concentration of $[CO]$ = 2x = 2 ( 0.0124)

= 0.0248

= 2.48 ×10⁻² M

concentration of $[CO_2]$ = 0.469-2x

= 0.469-2(0.0124)

= 0.469 - 0.0248

= 0.4442 M

3 0

3 years ago