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Virty [35]
2 years ago
12

Which statements describe the relationship between the area and the bioversity

Chemistry
1 answer:
AveGali [126]2 years ago
3 0

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To a 25.00 mL volumetric flask, a lab technician adds a 0.150 g sample of a weak monoprotic acid, HA , and dilutes to the mark w
Elis [28]

<u>Answer:</u> The number of moles of weak acid is 4.24\times 10^{-3} moles.

<u>Explanation:</u>

To calculate the moles of KOH, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}\text{Volume of solution (in L)}}

We are given:

Volume of solution = 43.81 mL = 0.04381 L      (Conversion factor: 1L = 1000 mL)

Molarity of the solution = 0.0969 moles/ L

Putting values in above equation, we get:

0.0969mol/L=\frac{\text{Moles of KOH}}{0.04381}\\\\\text{Moles of KOH}=4.24\times 10^{-3}mol

The chemical reaction of weak monoprotic acid and KOH follows the equation:

HA+KOH\rightarrow KA+H_2O

By Stoichiometry of the reaction:

1 mole of KOH reacts with 1 mole of weak monoprotic acid.

So, 4.24\times 10^{-3}mol of KOH will react with = \frac{1}{1}\times 4.24\times 10^{-3}=4.24\times 10^{-3}mol of weak monoprotic acid.

Hence, the number of moles of weak acid is 4.24\times 10^{-3} moles.

6 0
3 years ago
2. How many moles of salt are present in 1.5L of a 5.OM salt water solution?
madam [21]

Answer:

There are 7.5 moles of salt

Explanation:

5.0M means that in every liter of solution, there are 5 moles of salt. So, 1.5L of solution times 5 moles per liter equals 7.5 moles

8 0
3 years ago
What would you need to do to calculate the molality of 10 g of NaCl in 2 kg of
jasenka [17]

Answer:

O B. Convert the 10 g of NaCl to moles of NaCl.

Explanation:

The formula for finding the molality is m=moles of solute/kg of solvent. The solute for this question is NaCl and the solvent is water.

(10g NaCl)(1 mol NaCl/58.44g NaCl)=0.1711 mol NaCl

58.44 is the molar mass of NaCl

m=0.1711 mol NaCl/2 kg H2O

m=0.085557837

7 0
3 years ago
Select the correct hybridization for the central atom based on the electron geometry cocl2 (carbon is the central atom).
Rainbow [258]
In this compound (Phosgene) the central atom (carbon is Sp² Hybridized).

Sp, Sp² and Sp³ can be calculated very simply by doing three steps, 

Step 1:
           Assume triple bond and double bond as one bond and assign s or p to it. In this example carbon double bond oxygen is considered once and let suppose it is s. Now we are having our s.

Step 2:
          Count lone pair of electron, each lone pair counts for s and p. In this case there is no lone pair of electron on carbon, so not included.

Step 3:
          Count single bonds for s and p. As we have already assigned s to the double bond, now one p for one single bond, and other p for the other single bond.

Result:
          So, we counted 1 s for double bond, 1 p for one single and other p for second single bond. As a whole we got,

                                                       Sp²

Practice:
You can practice for hybridization of Oxygen in this molecule. Oxygen has 2 lone pair of electrons. (Hint: Sp² Hybridization)
7 0
2 years ago
A particle with a charge of 9.40 nC is in a uniform electric field directed to the left. Another force, in addition to the elect
juin [17]

Answer:

a. Work done by the electric force = -2.85 * ×10⁻⁵ J

b. The potential of the starting point with respect to the end point = -3.03 * 10³ V

c. The magnitude of the electric field is 33.7kV/m

Explanation:

Given.

Charge = Q = 9.40 nC

Distance = d = 9.00 cm = 0.09m

Amount of work = 7.10×10⁻⁵ J

Kinetic energy = K = 4.25×10⁻⁵ J

a. What work was done by the electric force?

This is calculated by; change in Kinetic Energy i.e. ∆KE

∆KE = ∆K2 - ∆Kæ

Where K2 = 4.25×10⁻⁵ J

The body is released at rest, so the initial velocity is 0.

So, K1 = 0

Also, total work done = W1 + W2

Where W2 = 7.10×10⁻⁵J

So, W1 + W2 = W = K2

W1 + 7.10×10⁻⁵ = 4.25×10⁻⁵

W1 = 4.25×10⁻⁵ - 7.10×10⁻⁵

W = -2.85 * ×10⁻⁵ J

Work done by the electric force = -2.85 * ×10⁻⁵ J

b. What is the potential of the starting point with respect to the end point?

The change in potential energy is given as

W = ∆U

W = Q|V2 - V1| where V1 = 0 because the body starts from rest

So, W = QV2

Make V the Subject of the formula

V2 = W/Q

V2 = -2.85 * ×10⁻⁵ J / 9.40 nC

V2 = -2.85 * ×10⁻⁵ J / 9.40 * 10^-9C

V2 = −3031.9148936170212765957V

V2 = -3.03 * 10³ V

The potential of the starting point with respect to the end point = -3.03 * 10³ V

c. What is the magnitude of the electric field?

The magnitude of the electric field is calculated as follows;

W = -Fd = -QEd

And E = V/d

E = -3.03 * 10³ V / 0.09 m

E = −33687.943262411347517730 V/m

E = -33.7kV/m

The magnitude of the electric field is 33.7kV/m

4 0
3 years ago
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