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Oxana [17]
3 years ago
13

Use a graph to solve the equation on the interval

Chemistry
1 answer:
Fofino [41]3 years ago
8 0
In order to solve this graphically, you would construct two lines:
The first of y = √2
The second of y = sec(x)

The solutions will be the intersections over the given domain.
Solving mathematically:
sec x = √2
1/cos(x) = √2
cos (x) = 1/√2

x = -π/4 and π/4
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19)Calculate the number of moles of Al2O3 that are produced when 15 mol of Fe is produced
Margarita [4]

Answer:

C

Explanation:

Calculate the number of moles of Al2O3 that are produced when 15 mol of Fe is produced

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6 0
2 years ago
For each of the following units and concentration values, mention if they are parts per million (ppm), parts per billion (ppb) o
arsen [322]

Answer:

a) ppm

b) ppm

c) ppb

d) ppt

e) ppb

Explanation:

a) You know that 1000 g are 1 kg, and 1000 kg are 1 ton, so (1000)*(1000) g are 1 ton, so 1,000,000 grams are one ton.

b) 1000 mg are 1 g, and 1000 g are 1 liter, so 1,000,000 grams are one liter.

c) You know that 1000 ug are 1 mg, so with the b), we just need to multiply the answer by 1000, so 1,000,000,000 ug are 1 liter.

d) The same as c, 1000 ng are 1 mg. So we are talkinf of ppt.

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7 0
3 years ago
Which one is correct
anyanavicka [17]

Answer:

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Explanation:

4 0
3 years ago
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Calculate the number of calories needed to increase the temperature of 50.0 g of copper metal from 21.0 degrees C to 75.0 degree
KonstantinChe [14]
<h3>Answer:</h3>

1031.4 Calories.

<h3>Explanation:</h3>

We are given;

Mass of the copper metal = 50.0 g

Initial temperature = 21.0 °C

Final temperature, = 75°C

Change in temperature = 54°C

Specific heat capacity of copper = 0.382 Cal/g°C

We are required to calculate the amount of heat in calories required to raise the temperature of the copper metal;

Quantity of heat is given by the formula,

Q = Mass × specific heat capacity × change in temperature

   = 50.0 g × 0.382 Cal/g°C × 54 °C

   = 1031.4 Calories

Thus, the amount of heat energy required is 1031.4 Calories.

4 0
3 years ago
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