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4 years ago

What is a base used in toothpaste?

Chemistry

2 answers:

stich3 [128]4 years ago

5 0

Sodium fluoride is used in a lot of toothpastes<span />

Svetlanka [38]4 years ago

3 0

Most of toothpastes<span> includes sodium fluoride,silica,sodium lauryl phosphate,sorbitol,water etc. It may neutralise the acid formation in mouth cavity as well as at teeth gaps.So tooth paste is basic in nature.</span>

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Which interaction is responsible for the force of attraction between the particles in a salt crystal? proton-proton proton-elect

Mars2501 [29]

The answer is B. proton-electron.

3 0

4 years ago

Read 2 more answers

Suppose you start with a solution of red dye #40 that is 2.3 ✕ 10−5 M. If you do three successive volumetric dilutions pipetting

larisa [96]

Answer: Therefore, the concentration of final solution is $4.0\times 10^{-8}M$

Explanation:

According to the neutralization law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $2.3\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$2.3\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=0.28\times 10^{-5}M$

Therefore, the concentration of diluted solution is $0.28\times 10^{-5}M$

2) On further dilution

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $0.28\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$0.28\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=0.034\times 10^{-5}M$

Therefore, the concentration of diluted solution is $0.034\times 10^{-5}M$

3) On further dilution

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $0.034\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$0.034\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=4.0\times 10^{-8}M$

Therefore, the concentration of final solution is $4.0\times 10^{-8}M$

5 0

3 years ago

. The density of gold is 19.32 g/mL. A U.S. $5.00 gold coin has a diameter of 1.62 cm and a thickness of 0.085 cm. The price of

Lostsunrise [7]

Answer:

$182 is the value of the gold in the coin

Explanation:

Diameter is 2r, the ratio of the coin is = 1.62cm / 2 = 0.81cm

Following the formula of the volume of the coin:

V = π r² h

V = π*(0.81cm)²*0.085cm

V = 0.172cm³ = mL

As the density of the gold is 19.32g/mL, the mass of 0.1752mL of gold is:

0.1752mL * (19.32g / mL) =

3.385g is the mass of the coin

As the price of the gold is $53.90 / g, the value of the gold in the coin is:

3.385g * ($53.90 / g) =

<h3>$182 is the value of the gold in the coin</h3>

7 0

3 years ago

1 poin

8090 [49]

Answer:

It's called Independent variable

Explanation:

....

4 0

3 years ago

The purification of hydrogen gas by diffusion through a palladium sheet was discussed in Section 5.3. Compute the number of kilo

Lynna [10]

Answer:

NA = 6.8 E-12 Kg H2(g) / hour

Explanation:

steady-state diffusion of A through non-diffuser B:

NA = (DAB/RTz)(p*A1 - p*A2)

∴ (A): H2(g)

∴ (B): Pd

∴ DAB = 1.7 E-8 m²/s

∴ p*A1 = 2.0 Kg H2 / m³ Pd

∴ p*A2 = 0.4 Kg H2 / m³ Pd

∴ z = 6 mm = 6 E-3 m

∴ T = 600°C ≅ 873 K

∴ R = 8.314 J/mol.K = 8.314 N.m/mol.K

⇒ NA = ((1.7 E-8)/(8.314)(873)(6 E-3))(2.0 - 0.4)

⇒ NA = 6.246 E-10 mol/s.m³

for A = 0.25 m²

⇒ volume (v) = A×z = (0.25)(6 E-3) = 1.5 E-3 m³

∴ Mw H2(g) = 2.016 g/mol

⇒ NA = (6.246 E-10 mol/s.m³)(1.5 E-3 m³)(2.016 g/mol)(Kg/1000 g)(3600 s/h)

⇒ NA = 6.8 E-12 Kg H2(g)/h

4 0

3 years ago