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SVETLANKA909090 [29]
3 years ago
10

How many grams are in 4.39 mol of copper

Chemistry
1 answer:
I am Lyosha [343]3 years ago
5 0

Answer: 278.966

Explanation: You can also approximate the number to 278.97

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If 25 mol of C8H18 are available, how many mol of CO2 can be produced
Nady [450]

Answer:

200 mol

Explanation:

2C8H18+25O2=16CO2+18H2O

2 mol → 16 mol

25 mol → X mol

X=25×16÷2=200

7 0
3 years ago
How is nitric acid classified?
Darina [25.2K]
It's classified as an acid
4 0
3 years ago
Why is water necessary for rusting to happen?
Triss [41]
You need oxygen and water
5 0
3 years ago
As an athlete exercises, sweat is produced and evaporated to help maintain a proper body temperature. On average, an athlete los
dmitriy555 [2]

<u>Answer:</u> For the given amount of sweat lost, the amount of energy required will be 692,899 Joules.

<u>Explanation:</u>

We are given:

Heat of vaporization for water = 2257 J/g

Amount of sweat lost = 307 grams

Applying unitary method:

For 1 g of sweat lost, the energy required is 2257 Joules

So, for 307 grams of sweat lost, the energy required will be = \frac{2257J}{1g}\times 307g=692,899J

Hence, for the given amount of sweat lost, the amount of energy required will be 692,899 Joules.

7 0
2 years ago
Promethium-147 is sometimes used in luminescent paint. It has a half-life of 956.3 days. If 250 grams (g) of promethium-147 is u
Fantom [35]

Answer:

% = 76.75%

Explanation:

To solve this problem, we just need to use the expressions of half life and it's relation with the concentration or mass of a compound. That expression is the following:

A = A₀ e^(-kt)   (1)

Where:

A and A₀: concentrations or mass of the compounds, (final and initial)

k: constant decay of the compound

t: given time

Now to get the value of k, we should use the following expression:

k = ln2 / t₁/₂   (2)

You should note that this expression is valid when the reaction is of order 1 or first order. In this kind of exercises, we can assume it's a first order because we are not using the isotope for a reaction.

Now, let's calculate k:

k = ln2 / 956.3

k = 7.25x10⁻⁴ d⁻¹

With this value, we just replace it in (1) to get the final mass of the isotope. The given time is 1 year or 365 days so:

A = 250 e^(-7.25x10⁻⁴ * 365)

A = 250 e^(-0.7675)

A = 191.87 g

However, the question is the percentage left after 1 year so:

% = (191.87 / 250) * 100

<h2>% = 76.75%</h2><h2>And this is the % of isotope after 1 year</h2>
3 0
3 years ago
Read 2 more answers
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