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3 years ago

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A titration is completed using 10.0 mL of 0.50 M HCl and an unknown concentration of NaOH solution. If the equivalence point is

reached when exactly 20.0 mL of the NaOH solution is added, what must the concentration of the NaOH solution be?

1 answer:

Drupady [299]3 years ago

7 0

Answer:

0.172 M

Explanation:

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Complete the following equations : H3PO4 = 3H+ --------------- CaSO4 = Ca 2+ ---------------- (2) b) Write the formulae of the s

jeka94

Answer:

H₃PO₄ → 3H⁺ + PO₄³⁻

CaSO₄ → Ca²⁺ + SO₄²⁻

b. CaCl₂

Explanation:

When H₃PO₄ is dissolved in water, there are produced the H⁺ and PO₄³⁻ ions. The equation is:

H₃PO₄ → 3H⁺ + PO₄³⁻

In the same way, CaSO₄ is dissolved in:

CaSO₄ → Ca²⁺ + SO₄²⁻

b. Now, in a reaction of an acid (HCl) and a base (Ca(OH)₂), water, H₂O and a salt are produced:

2 HCl + Ca(OH)₂ → 2H₂O + Salt

The ions that are not present in the reaction are Cl⁻ and Ca²⁺, the salt is CaCl₂ and the balanced reaction is:

2 HCl + Ca(OH)₂ → 2H₂O + CaCl₂

5 0

2 years ago

A 1.50-kilogram ball is attached to the end of a 0.520-meter string and swung in a circle. The velocity of the ball is 9.78 m/s.

Sedaia [141]

Answer:

F centripetal force (tension) = 275.9 N

Explanation:

Given data:

Mass = 1.50 kg

Radius = 0.520 m

Velocity of ball = 9.78 m/s

Tension = ?

Solution:

F centripetal force (tension) = m.v² / R

F centripetal force (tension) = 1.50 kg . (9.78 m/s)² / 0.520 m

F centripetal force (tension) = 1.50 kg . 95.65 m²/s² / 0.520 m

F centripetal force (tension) = 143.5 kg. m²/s² / 0.520 m

F centripetal force (tension) = 275.9 N

7 0

2 years ago

What is not a product of modern chemistry ?

Doss [256]

Answer:

Modern chemistry describes the composition, structure, and physical and chemical properties of substances. ... The main aim of modern chemistry is to improve the communication and conversation among the scientists, researchers, engineers, and policymakers, who are working under the area of modern chemistry

Explanation:

I hope this helps you!!

7 0

2 years ago

An ideal gas sample is confined to 3.0 L and kept at 27 °C. If the temperature is raised to 77 °C and the initial pressure was 1

Nadusha1986 [10]

The gas is confined in 3.0 L container ( rigid container) ⇒ the volume remains constant when the temperature is increased from from 27oC to 77oC and therefore V1=V2 .

<h2>Hope it helps you please mark as brainlist</h2>

6 0

2 years ago

The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151

vitfil [10]

Answer:- The natural abundance of $^1^5^1_E_u$ is 0.478 or 47.8% and $^1^5^3_E_u$ is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for $^1^5^1_E_u$ and $^1^5^3_E_u$ as 150.919860 and 152.921243 amu, respectively. Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of $^1^5^1_E_u$ as n then the abundance of $^1^5^3_E_u$ would be 1-n .

Let's plug in the values in the formula:

$151.964=150.919860(n)+152.921243(1-n)$

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

$151.964-152.921243=150.919860n-152.921243n$

$-0.957243=-2.001383n$

negative sign is on both sides so it is canceled:

$0.957243=2.001383n$

$n=\frac{0.957243}{2.001383}$

$n=0.478$

The abundance of $^1^5^1_E_u$ is 0.478 which is 47.8%.

The abundance of $^1^5^3_E_u$ is = $1-0.478$

= 0.522 which is 52.2%

Hence, the natural abundance of $^1^5^1_E_u$ is 0.478 or 47.8% and $^1^5^3_E_u$ is 0.522 or 52.2% .

3 0

2 years ago