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3 years ago

60 grams of ice will require _____ calories to raise the temperature 1°C.

Chemistry

1 answer:

guajiro [1.7K]3 years ago

7 0

Answer:

60 grams of ice will require 30.26 calories to raise the temperature 1°C.

Explanation:

The amount of heat (Q) to raise the temperature of 60.0 g of ice by 1°C can be calculated from:

Q = m.c.ΔT,

where, Q is the amount of heat released or absorbed by the system.

m is the mass of the ice (m = 60.0 g).

c is the specific heat capacity of ice (c = 2.108 J/g.°C).

ΔT is the temperature difference (ΔT = 1.0 °C).

∴ Q = m.c.ΔT = (60.0 g)(2.108 J/g.°C)(1.0 °C) = 126.48 J.

It is known that 1.0 cal = 4.18 J.

∴ Q = (126.48 J)(1.0 cal / 4.18 J) = 30.26 cal.

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What did aristotle believe about the nature of matter

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Explanation:

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3 years ago

A reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by O 3 ( g ) + NO ( g )

Aleksandr [31]

Answer:

(a) 7.11x10⁻⁴ M/s

(b) 2.56 mol.L⁻¹.h⁻¹

Explanation:

(a) The reaction is:

O₃(g) + NO(g) → O₂(g) + NO₂(g) (1)

The reaction rate of equation (1) is given by:

$rate = k*[O_{3}][NO]$ (2)

We have:

k: is the rate constant of reaction = 3.91x10⁶ M⁻¹.s⁻¹

[O₃]₀ = 2.35x10⁻⁶ M

[NO]₀ = 7.74x10⁻⁵ M

Hence, to find the inital reacion rate we will use equation (2):

$rate = k*[O_{3}]_{0}[NO]_{0} = 3.91 \cdot 10^{6} M^{-1}s^{-1}*2.35\cdot 10^{-6} M*7.74 \cdot 10^{-5} M = 7.11 \cdot 10^{-4} M/s$

Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s

(b) The number of moles of NO₂(g) produced per hour per liter of air is:

t = 1 h

V = 1 L

$\frac{\Delta[NO_{2}]}{\Delta t} = rate$

$\frac{\Delta[NO_{2}]}{\Delta t} = 7.11 \cdot 10^{-4} M/s*\frac{3600 s}{1 h} = 2.56 mol.L^{-1}.h{-1}$

Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹

I hope it helps you!

5 0

2 years ago

A gas of 3.4 moles occupies a volume of 0.046 L at 298 K. What is the pressure in kPa?

Phantasy [73]

Answer : The correct option is, $2.1\times 10^5kPa$

Explanation :

To calculate the pressure of gas we are using ideal gas equation as:

$PV=nRT$

where,

P = pressure of gas = ?

V = volume of gas = 0.046 L

n = number of moles of gas = 3.4

R = gas constant = 8.314 L.kPa/mol.K

T = temperature of gas = 298 K

Now put all the given values in the above formula, we get:

$P\times (0.046L)=(3.4mol)\times (8.314L.kPa/mol.K)\times (298K)$

$P=1.83\times 10^5kPa\approx 2.1\times 10^5kPa$

Therefore, the pressure of gas is, $2.1\times 10^5kPa$

3 0

3 years ago