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2 years ago

15

What is the total number of grams of mg consumed when 0.50 mole of h2 is produced

1 answer:

rosijanka [135]2 years ago

4 0

Answer:

12 grams

<em>Please mark as Brainliest :)</em>

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10.0 grams of water are heated during the preparation of a cup of coffee 1.0x 103 j of the heat are added to the water. which is

katovenus [111]

<u>Answer:</u> The final temperature of the coffee is 43.9°C

<u>Explanation:</u>

To calculate the final temperature, we use the equation:

$q=mC(T_2-T_1)$

where,

q = heat released = $1.0\times 10^3J=1000J$

m = mass of water = 10.0 grams

C = specific heat capacity of water = 4.184 J/g°C

$T_2$ = final temperature = ?

$T_1$ = initial temperature = 20°C

Putting values in above equation, we get:

$1000J=10.0g\times 4.184J/g^oC\times (T_2-20)\\\\T_2=43.9^oC$

Hence, the final temperature of the coffee is 43.9°C

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3 years ago

Which substance is a pure substance? <br> Cereal <br> Silver <br> Jelly bean <br> Shampoo

Tju [1.3M]

Silver?

Fun FACT : An apple, potato, and onion all taste the same if you eat them with your nose plugged

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3 years ago

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A gas with a volume of 500. mL at 75°C is heated to 225 °C. What is the new volume of the gas?

mihalych1998 [28]

Using v1/t1=v2/t2
v1=500
v2=?
t1=75=368k
t2=225=498
500/368=v2/498
1.4x498=v2
v2=697.2ml

6 0

2 years ago

PLEASE ANSWER <br> ARE THESE CORRECTTTTTT

Naddik [55]

1. The answer is E.

2. I agree with your answers.

3. I also agree with your answers

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3 years ago

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Calculate the percent of each component in the mixture. Show your calculations. Circle final answers.

Colt1911 [192]

Answer:

See Explanation

Explanation:

The question is incomplete; as the mixtures are not given.

However, I'll give a general explanation on how to go about it and I'll also give an example.

The percentage of a component in a mixture is calculated as:

$\%C_E = \frac{E}{T} * 100\%$

Where

E = Amount of element/component

T = Amount of all elements/components

Take for instance:

In $(Ca(OH)_2)$

The amount of all elements is: (i.e formula mass of $(Ca(OH)_2)$ )

$T = 1 * Ca + 2 * H + 2 * O$

$T = 1 * 40 + 2 * 1 + 2 * 16$

$T = 74$

The amount of calcium is: (i.e formula mass of calcium)

$E = 1 * Ca$

$E = 1 * 40$

$E = 40$

So, the percentage component of calcium is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{40}{74} * 100\%$

$\%C_E = \frac{4000}{74}\%$

$\%C_E = 54.05\%$

The amount of hydrogen is:

$E = 2 * H$

$E = 2 * 1$

$E = 2$

So, the percentage component of hydrogen is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{2}{74} * 100\%$

$\%C_E = \frac{200}{74}\%$

$\%C_E = 2.70\%$

Similarly, for oxygen:

The amount of oxygen is:

$E = 2 * O$

$E = 2 * 16$

$E = 32$

So, the percentage component of oxygen is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{32}{74} * 100\%$

$\%C_E = \frac{3200}{74}\%$

$\%C_E = 43.24\%$

5 0

2 years ago