If 1.2 kilograms of rust forms on a bridge in 5 days, what should be the rate of reaction in grams per hour?

1 answer:

8 0

We have 1.2 kg of rust whithin 5 days.
the rate of reaction in grams per hour:

1.2 kg/5 day * 1000 g/kg * 1 day/24 hours = 10.0 grams/hour

hope this help

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Question 2: Heat (5 points)

Ronch [10]

Answer:

(i) specific heat

(ii) latent heat of vaporization

(iii) latent heat of fusion

Explanation:

i. Q = mcΔT; identify c.

Here, Q is heat, m is the mass, c is the specific heat and ΔT is the change in temperature.

The amount of heat required to raise the temperature of substance of mass 1 kg by 1 degree C is known as the specific heat.

ii. Q = mLvapor; identify Lvapor

Here, Q is the heat, m is the mass and L is the latent heat of vaporization.

The amount of heat required to convert the 1 kg liquid into 1 kg vapor at constant temperature.

iii. Q = mLfusion; identify Lfusion

Here, Q is the heat, m is the mass and L is the latent heat of fusion.

Here, Q is the heat, m is the mass and L is the latent heat of vaporization.

The amount of heat required to convert the 1 kg solid into 1 kg liquid at constant temperature.

8 0

3 years ago

Rick has taken a loan of $3,200 from the bank to buy new appliances. His first loan payment is due at the end of this month. He

marysya [2.9K]

Answer:

140

Explanation:

4 0

3 years ago

Read 2 more answers

Calculate the number of grams of oxygen gas in a 24L container at STP

romanna [79]

There are 34 g of oxygen in the container.

We can use the<em> Ideal Gas Law</em> to solve this problem.

$pV = nRT$

But $n = \frac{m}{M}$ , so

$pV = \frac{m}{M}RT$ and

$m = \frac{pVM}{RT}\\$

STP is 0 °C and 1 bar, so

$m = \frac{\text{1 bar} \times \text{24 L} \times 32.00 \text{ g}\cdot\text{mol}^{-1}}{\text{0.083 14 } \text{bar}\cdot\text{L}\cdot\text{K}^{-1}\text{mol}^{-1}\times\text{273.15 K} } = \textbf{34 g}\\$