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3 years ago

Which chemical is added to many municipal water supplies to help prevent tooth decay?

Chemistry

2 answers:

romanna [79]3 years ago

6 0

Fluoride I believe is the answer

Debora [2.8K]3 years ago

3 0

flouride gradpoint..

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A disk of radius 2.0 cm has a surface charge density of 6.3 μC/m2 on its upper face. What is the magnitude of the electric field

maksim [4K]

Answer:

the electric field at Z = 12 cm is E = 9.68 × 10³ N/C = 9.68 kN/C

Explanation:

Given: radius of disk, R = 2.0 cm = 2 × 10⁻² cm, surface charge density,σ = 6.3 μC/m² = 6.3 × 10⁻⁶ C/m², distance on central axis, z = 12 cm = 12 × 10⁻² cm.

The electric field, E at a point on the central axis of a charged disk is given by E = σ/ε₀( $1 - \frac{z}{\sqrt{z^{2} + R^{2} } }$ )

Substituting the values into the equation, it becomes

E = σ/ε₀( $1 - \frac{z}{\sqrt{z^{2} + R^{2} } }$ ) = 6.3 × 10⁻⁶/8.854 × 10⁻¹²( $1 - \frac{0.12}{\sqrt{0.12^{2} + 0.02^{2} } }$ ) = 7.12 × 10⁵( $1 - \frac{0.12}{0.1216}$ ) = 7.12 × 10⁵(1 - 0.9864) = 7.12 × 10⁵ × 0.0136 = 0.0968 × 10⁵ = 9.68 × 10³ N/C = 9.68 kN/C

Therefore, the electric field at Z = 12 cm is E = 9.68 × 10³ N/C = 9.68 kN/C

7 0

3 years ago

Express 1340000 in scientific notation

Katen [24]

1.34*10^6
Move the decimal 6 times to the left, it is a number 1-10

8 0

3 years ago

Read 2 more answers

Would isotopes of the same element have the same number of electrons?

posledela

No, isotopes would have a different number of electrons

8 0

3 years ago

Calculate the pH for the following weak acid. A solution of HCOOH has 0.12M HCOOH at equilibrium. The Ka for HCOOH is 1.8×10−4.

Shtirlitz [24]

Answer:

the pH of HCOOH solution is 2.33

Explanation:

The ionization equation for the given acid is written as:

$HCOOH\leftrightarrow H^++HCOO^-$

Let's say the initial concentration of the acid is c and the change in concentration x.

Then, equilibrium concentration of acid = (c-x)

and the equilibrium concentration for each of the product would be x

Equilibrium expression for the above equation would be:

$\Ka= \frac{[H^+][HCOO^-]}{[HCOOH]}$

$1.8*10^-^4=\frac{x^2}{c-x}$

From given info, equilibrium concentration of the acid is 0.12

So, (c-x) = 0.12

hence,

$1.8*10^-^4=\frac{x^2}{0.12}$

Let's solve this for x. Multiply both sides by 0.12

$2.16*10^-^5=x^2$

taking square root to both sides:

$x=0.00465$

Now, we have got the concentration of $[H^+] .$

$[H^+] = 0.00465 M$

We know that, $pH=-log[H^+]$

pH = -log(0.00465)

pH = 2.33

Hence, the pH of HCOOH solution is 2.33.

7 0

3 years ago

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I need helppppppp plssssss

Irina-Kira [14]

Answer:

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4 0

2 years ago