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3 years ago

13

nadira is using her laptop for about 6 hours straight,yet her laptops barley hot or warm. Does her laptop have high or low speci

fic heat?

1 answer:

makkiz [27]3 years ago

8 0

Her computer will start to heat up, and the temp. would be 60 degrees celcius.

I hope this helps you ᕕ( ᐛ )ᕗ

You might be interested in

For the following example, list the given and unknown information (including gratis or moles)

Setler79 [48]

Answer:

9.6 moles O2

Explanation:

I'll assume it is 345 grams, not gratis, of water. Hydrogen's molar mass is 1.01, not 101.

The molar mass of water is 18.0 grams/mole.

Therefore: (345g)/(18.0 g/mole) = 19.17 or 19.2 moles water (3 sig figs).

The balanced equation states that: 2H20 ⇒ 2H2 +02

It promises that we'll get 1 mole of oxygen for every 2 moles of H2O, a molar ratio of 1/2.

get (1 mole O2/2 moles H2O)*(19.2 moles H2O) or 9.6 moles O2

6 0

2 years ago

What is the balanced form of the following equation? Br 2 + S 2 O 3 2– + H 2 O → Br 1– + SO 4 2– + H +

nikitadnepr [17]

Answer:

4Br₂+ 5H₂O+ S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8Br⁻

Explanation:

Br₂ + S₂O₃²⁻ + H₂O → Br⁻ + SO₄²⁻ + H⁺

This is a redox reaction:

Br₂ changes the oxidation state from 0 to -1, so it was reduced

In the S₂O₃⁻² anion S changes the oxidation state from +2 to +6 in sulfate anion. (S₂O₃⁻², it is called thiosulfate)

We have protons in the main equation, so we assume we are in acidic medium:

Br₂ + 2e⁻ → 2Br⁻ Reduction

We balanced the bromide with 2, so the bromine has gained 2 electrons.

<u>5H₂O</u> + S₂O₃²⁻ → 2SO₄²⁻ + <u>10H⁺</u> + <em>8e</em>- Oxidation

First of all, we add 2 to the sulfate anion in the product side, in order to balance the S.

As we have 8 O in right side, and 3 O in left side, we must add 5 O. We add 5 water in the place where the O are lower (reactant side).

Now, we have 10 H, in the reactant side, so we balance the product side with protons (10 H⁺).

Sulfur changed the oxidation state from +2 to +6, so it released 4 electrons, but, if you see thiosulfate anion you have 2 sulfurs so finally it has released 8 electrons.

Electrons are unbalanced so we multiply reduction x4, and oxidation x1.

(Br₂ + 2e⁻ → 2Br⁻) . 4 = 4Br₂ + 8e⁻ → 8Br⁻

(5H₂O + S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + <em>8e</em>-) . 1 = STAYS THE SAME.

We sum both half reactions, to cancel the elecetrons:

4Br₂ + 8e⁻ + 5H₂O + S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + <em>8e</em>- + 8Br⁻

Finally the balanced reaction is: 4Br₂+ 5H₂O+ S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8Br⁻

5 0

3 years ago

1. A gas having the following composition is burnt under a boiler with 50% excess air.

jeka94

The composition of the stack gas are :

$CH_4$ = 0.8713

$C_3H_8$ = 0.0202

CO = 0.107

<h3 /><h3>What is a mole fraction?</h3>

The ratio of the number of moles of one component of a solution or other mixture to the total number of moles representing all of the components.

Assuming 100 g of the stack gas. Calculate the mass of each species in this sample according to their percentages.

Mass of $CH_4$ : 70% of 100 g = 70 g

Mass of $C_3H_8$ : 15% of 100 g = 15 g

Mass of CO : 15% of 100 g = 15 g

Now calculate the number of moles of each species:

Number of moles of $CH_4$ : $\frac{70 g}{16.04 g/mol}$ = 4.3 mole

Number of moles of $C_3H_8$ : $\frac{15 g}{144.1 g/mol}$ = 0.10 mole

Mass of CO : $\frac{15 g}{28.01 g/mol}$ = 0.53 mole

Now to calculate the mole fraction of each we use the formula:

Mole fraction of $CH_4$ : $\frac{4.3}{4.935}$ = 0.8713

Mole fraction of $C_3H_8$ : $\frac{0.10}{4.935}$ = 0.0202

Mole fraction of CO : $\frac{0.53}{4.935}$ = 0.107

Hence, composition of the stack gas are:

$CH_4$ = 0.8713

$C_3H_8$ = 0.0202

CO = 0.107

Learn more about mole fraction here:

brainly.com/question/13135950

#SPJ1

8 0

2 years ago

1. Analysis of an unknown substance formerly used in rocket fuel reveals a composition of 93.28% nitrogen and 6.72% hydrogen by

Nitella [24]

Answer:

The formula of the compound is:

N2H2

Explanation:

Data obtained from the question:

Nitrogen (N) = 93.28%

Hydrogen (H) = 6.72%

Next, we shall determine the empirical formula for the unknown compound. This is illustrated below:

N = 93.28%

H = 6.72%

Divide by their molar mass

N = 93.28 /14 = 6.663

H = 6.72 /1 = 6.7

Divide by the smallest

N = 6.663 / 6.663 = 1

H = 6.72 /6.663 = 1

Therefore, the empirical formula is NH.

Now, we can obtain the formula of the compound as follow:

The formula of a compound is simply a multiple of the empirical formula.

[NH]n = 30.04

[14 + 1]n = 30.04

15n = 30.04

Divide both side by 15

n = 30.04/15

n = 2

Therefore, the formula of the compound is:

[NH]n => [NH]2 => N2H2

6 0

3 years ago

0.00001011 kilograms is a larger mass than 0.001011 grams. True or False

valentina_108 [34]

True...............................

6 0

3 years ago

Read 2 more answers