How are water and minerals transported and used in vascular plants?

Why Is Any Chemical Reaction always Balanced? Give reasons and Explain the Easiest way to solve the Balancing Problems in Chemic

erastova [34]

Answer:

The chemical equation needs to be balanced so that it follows the law of conservation of mass. A balanced chemical equation occurs when the number of the different atoms of elements in the reactants side is equal to that of the products side.

8 0

3 years ago

What is the concentration of the barium hydroxide solution if 50.0 mL of a 0.425 M HNO3 solution is required to neutralize a 36.

fomenos

The molarity of Barium Hydroxide is 0.289 M.

<u>Explanation:</u>

We have to write the balanced equation as,

Ba(OH)₂ + 2 HNO₃ → Ba(NO₃)₂ + 2 H₂O

We need 2 moles of nitric acid to react with a mole of Barium hydroxide, so we can write the law of volumetric analysis as,

V1M1 = 2 V2M2

Here V1 and M1 are the volume and molarity of nitric acid

V2 and M2 are the volume and molarity of Barium hydroxide.

So the molarity of Ba(OH)₂, can be found as,

$$ M2 = \frac{V1 \times M1}{2 \times V2}$

$$M2 = \frac{50 \times 0.425 }{2 \times 36.8}$

= 0.289 M

5 0

3 years ago

Water covers over 70% of planet Earth and most of the water is salt water. Which

Serga [27]

Answer:

D. 4

Explanation:

6 0

3 years ago

If 16.00 g of O₂ reacts with 80.00 g NO, how many the excess reactant are left over? (enter only the value, round to whole numbe

pishuonlain [190]

Answer:

50

Explanation:

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

Mᵣ: 30.01 32.00 46.01

2NO + O₂ ⟶ 2NO₂

Mass/g: 80.00 16.00

2. Calculate the moles of each reactant

$\text{moles of NO} = \text{80.00 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{2.666 mol NO}\\\\\text{moles of O}_{2} = \text{16.00 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.5000 mol O}_{2}$

3. Calculate the moles of NO₂ we can obtain from each reactant

From NO:

The molar ratio is 2 mol NO₂:2 mol NO

$\text{Moles of NO}_{2} = \text{2.333 mol NO} \times \dfrac{\text{2 mol NO}_{2}}{\text{2 mol NO}} = \text{2.333 mol NO}_{2}$

From O₂:

The molar ratio is 2 mol NO₂:1 mol O₂

$\text{Moles of NO}_{2} = \text{0.5000 mol O}_{2}\times \dfrac{\text{2 mol NO}_{2}}{\text{1 mol Cl}_{2}} = \text{1.000 mol NO}_{2}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO₂.

The excess reactant is NO.

5. Mass of excess reactant

(a) Moles of NO reacted

The molar ratio is 2 mol NO:1 mol O₂

$\text{Moles reacted} = \text{0.500 mol O}_{2} \times \dfrac{\text{2 mol NO}}{\text{1 mol O}_{2}} = \text{1.000 mol NO}$

(b) Mass of NO reacted

$\text{Mass reacted} = \text{1.000 mol NO} \times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \text{30.01 g NO}$

(c) Mass of NO remaining

Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO

5 0

3 years ago

HURRY HELP 10 POINTS ILL GOVE BRAINLIEST I HAVE TO PASS PICTURE PROVIDED

kramer

Answer:

C₃H₄O₄

Explanation:

In order to get the empirical formula of a compound, we have to follow a series of steps.

Step 1: Divide the percent by mass of each element by its atomic mass.

C: 34.6/12.01 = 2.88

H: 3.9/1.01 = 3.86

O: 61.5/16.00 = 3.84

Step 2: Divide all the numbers by the smallest one, i.e., 2.88

C: 2.88/2.88 = 1

H: 3.86/2.88 ≈ 1.34

O: 3.84/2.88 ≈ 1.33

Step 3: Multiply all the numbers by a number that makes all of them integer

C: 1 × 3 = 3

H: 1.34 × 3 = 4

O: 1.33 × 3 = 4

The empirical formula is C₃H₄O₄.

5 0

2 years ago