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RoseWind [281]
2 years ago
11

as the number of neutrons in the nucleus of a given atom of an element increases the atomic number of that element

Chemistry
1 answer:
algol [13]2 years ago
4 0
There is no change to the atomic number. 
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Consider a solution containing 0.100 m fluoride ions and 0.126 m hydrogen fluoride. the concentration of hydrogen fluoride after
Nataly_w [17]
Fluorine ions reacts with Hydrogen chloride to form more hydrogen fluoride.
Therefore, moles of HCl = 0.005 l × 0.01 M = 5 ×10^-5 moles
The initial moles of Hydrogen fluoride will be;
 = 0.0126 M× 0.0250 = 0.00315 Moles
Moles of hydrogen fluoride after the addition of HCl 
= 0.00315 + 5.0× 10^-5 = 0.0032 moles 
Therefore, the concentration of Hydrogen chloride 
            = 0.0032 moles/ 0.030 L 
            = 0.107 M
3 0
3 years ago
Read 2 more answers
Which section of the reaction represents the products?<br> A) a <br> B) b <br> C) c <br> D) d
vladimir2022 [97]
The answer will be D because the reaction is what's first(A) and then it is the product that comes out.
6 0
3 years ago
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How many atoms are in 3.5 moles of He?
Volgvan

Answer:

The answer is 2.107 × 10²⁴ He atoms

Explanation:

To find the number of atoms given the number moles we use the formula

<h3>N = n × L</h3>

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

We have

N = 3.5 × 6.02 × 10²³

We have the final answer as

<h3>2.107 × 10²⁴ He atoms</h3>

Hope this helps you

4 0
2 years ago
What is the conjugate base<br> of H2O?
kolbaska11 [484]

Answer:

A conjugate acid, is a species formed by the reception of a proton (H+) by a base—in other words, it is a base with a hydrogen ion added to it.

OH- is the conjugate base of H2O..

6 0
2 years ago
A nitric acid solution flows at a constant rate of 5L/min into a large tank that initially held 200L of a 0.5% nitric acid solut
leonid [27]

Answer:

x(t) = −39e

−0.03t + 40.

Explanation:

Let V (t) be the volume of solution (water and

nitric acid) measured in liters after t minutes. Let x(t) be the volume of nitric acid

measured in liters after t minutes, and let c(t) be the concentration (by volume) of

nitric acid in solution after t minutes.

The volume of solution V (t) doesn’t change over time since the inflow and outflow

of solution is equal. Thus V = 200 L. The concentration of nitric acid c(t) is

c(t) = x(t)

V (t)

=

x(t)

200

.

We model this problem as

dx

dt = I(t) − O(t),

where I(t) is the input rate of nitric acid and O(t) is the output rate of nitric acid,

both measured in liters of nitric acid per minute. The input rate is

I(t) = 6 Lsol.

1 min

·

20 Lnit.

100 Lsol.

=

120 Lnit.

100 min

= 1.2 Lnit./min.

The output rate is

O(t) = (6 Lsol./min)c(t) = 6 Lsol.

1 min

·

x(t) Lnit.

200 Lsol.

=

3x(t) Lnit.

100 min

= 0.03 x(t) Lnit./min.

The equation is then

dx

dt = 1.2 − 0.03x,

or

dx

dt + 0.03x = 1.2, (1)

which is a linear equation. The initial condition condition is found in the following

way:

c(0) = 0.5% = 5 Lnit.

1000 Lsol.

=

x(0) Lnit.

200 Lsol.

.

Thus x(0) = 1.

In Eq. (1) we let P(t) = 0.03 and Q(t) = 1.2. The integrating factor for Eq. (1) is

µ(t) = exp Z

P(t) dt

= exp

0.03 Z

dt

= e

0.03t

.

The solution is

x(t) = 1

µ(t)

Z

µ(t)Q(t) dt + C

= Ce−0.03t + 1.2e

−0.03t

Z

e

0.03t

dt

= Ce−0.03t +

1.2

0.03

e

−0.03t

e

0.03t

= Ce−0.03t +

1.2

0.03

= Ce−0.03t + 40.

The constant is found using x(t) = 1:

x(0) = Ce−0.03(0) + 40 = C + 40 = 1.

Thus C = −39, and the solution is

x(t) = −39e

−0.03t + 40.

3 0
2 years ago
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