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Sloan
2 months ago
12

El teñido de una camiseta blanca con pintura es un cambio fisico o quimico

Chemistry
1 answer:
Anarel [2.9K]2 months ago
4 0

respuesta: cambio fisico

explicacion: Un cambio químico ocurre cuando se altera la estructura atómica, en este caso, al teñir la camiseta, esta sigue siendo una camiseta sin perder su identidad.

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Why is it important to have regular supervision of the weight and measurements in the market
lions [2927]

Answer:

Oversight of weights and measures ensures correct evaluations of goods and services so that everyone receives a fair exchange in the marketplace. It also acts as a deterrent, promoting honesty among traders.

Explanation:

4 0
3 months ago
Calculate the molarity of 48.0 mL of 6.00 M H2SO4 diluted to 0.250 L
lorasvet [2795]

Answer:

The molality is 1.15 m.

Molality is calculated by dividing the number of moles of solute by the kilograms of solvent, which in this case is water.

Calculate moles of H₂SO₄ from molarity:

C = n/V → n = C × V = 6.00 mol/L × 0.048 L = 0.288 moles

Mass of solvent (water) based on density:

m = ρ × V = 1.00 kg/L × 0.250 L = 0.250 kg

Therefore, molality is:

m = moles/solvent mass = 0.288 moles / 0.250 kg = 1.15 m

4 0
2 months ago
Read 2 more answers
A scientist measures the speed of sound in a monatomic gas to be 449 m/s at 20∘C. What is the molar mass of this gas?
Tems11 [2777]

Answer:

The molar mass of the gas is 36.25 g/mol.

Explanation:

  • To determine this, we utilize the mathematical relationship:

ν = \sqrt{3RT/M}

Where, ν represents the speed of light in a gas (ν = 449 m/s),

R denotes the universal gas constant (R = 8.314 J/mol.K),

T stands for the temperature of the gas in Kelvin (T = 20 °C + 273 = 293 K),

M is the molar mass of the gas in (Kg/mol).

ν = \sqrt{3RT/M}

(449 m/s) = √(3(8.314 J/mol.K)(293 K)/M,

by squaring both sides:

(449 m/s)² = (3(8.314 J/mol.K)(293 K))/M,

thus M = (3(8.314 J/mol.K)(293 K)/(449 m/s)² = 7308.006/201601 = 0.03625 Kg/mol.

Thus, the molar mass of the gas is 36.25 g/mol.


7 0
1 month ago
(a) calculate the %ic of the interatomic bond for the intermetallic compound tial3. (b) on the basis of this result, what type o
Tems11 [2777]

Answer :

The percentage ionic character (%IC) equals 10%, indicating the bond is mostly covalent with slight polarity.

Percent Ionic Character:

This reflects the fraction of ionic nature within a polar covalent bond. The formula for %IC (% ionic character) is:

Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^X^a^-^X^b^) * 100

Here, Xa is the electronegativity of atom A and Xb is that of atom B.

Given: The compound is TiAl₃.

Electronegativity of Ti = 2.0

Electronegativity of Al = 1.6 (as shown in the provided image)

Substitute these values into the formula:

Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^2^.^0^-^1^.^6^) * 100

Percent Ionic character = 1 - e^(^-^0^.^2^5 ^*^0^.^4^) * 100

Percent Ionic character = 1 - e^(^-^0^.^1^) * 100

The value of e⁻¹ equals 0.90.

Therefore, percent ionic character = (1 - 0.90) × 100

Percent Ionic Character = 10%

Because the % IC is only 10%, which is relatively low, the bond is classified as covalent with minimal polarity.

8 0
3 months ago
5 grams of NaHCO3 was added to 50 cm3 of 1.5 mol dm-3 hydrochloric acid, causing the temperature to decrease by 6.8 K. What is t
Anarel [2989]

The balanced chemical equation for the neutralization of HCl with NaHCO_{3} is:

HCl (aq) + NaHCO_{3}(aq)--> NaCl (aq) + H_{2}O(l)+CO_{2} (g)

Given weight of NaHCO_{3} = 5g

Moles of NaHCO_{3} = 5 g *\frac{1 mol}{84 g} = 0.05952 mol NaHCO_{3}

Volume of HCl solution = 50 cm^{3} * \frac{1 mL}{1cm^{3}} = 50 mL

Assuming the density of the solution is 1.0 g/mL

Mass of HCl solution = 50 g

Overall mass of the solution = 50 g + 5 g = 55 g

To find the heat of neutralization, we calculate:

Q = m C ΔT

where m equals the mass of the solution = 55 g

C represents the specific heat capacity of the solution = 4.184\frac{J}{g. ^{0}C}

ΔT signifies the temperature change = 6.8 K = (6.8 - 273) C = -266.2^{0}C

Q = 55 g * 4.184 \frac{J}{g. K}(6.8K) = 1565 J

The enthalpy of neutralization per mole of NaHCO_{3}

= \frac{1565J}{0.05952 mol} = 26294 \frac{J}{mol}*\frac{1 kJ}{1000J} =26.294kJ/mol

3 0
2 months ago
Read 2 more answers
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