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Usimov [2.4K]
3 years ago
14

Help pleasesHow do you maximize heat/energy transfer?​

Chemistry
1 answer:
jok3333 [9.3K]3 years ago
3 0
You increase the activators
You might be interested in
What is the mass of a 5.00 cm 3 piece of copper having a density of 8.96 g/cm3?
skad [1K]

Mass/volume = density. We know the volume (5 cm cubed) and the density (8.96). We plug these into the equation to find the mass:

M/5 = 8.96

M = 44.8 grams

4 0
4 years ago
Consider the addition of HBr to 2-pentene. Indicate the re
lana [24]

Answer:

This addition reaction yields 3-BromoPentane and 2-BromoPentane.

Explanation: The reaction is an addition reaction that follows the Markonikoff's principle engaging the electrophillic addition mechnism with electrophile having no lone pair so rearrangement of carbonation is possible. It yields two possible products.

5 0
4 years ago
A metal X from two oxide A and B .3.oogm of A and B contain 0.72 and 1.16g of oxygen respectively.calculate the maases of metal
devlian [24]

Answer:

Explanation:

Firstly, we have to determine the mass of metal X. We can do that by interpreting the first and second statement mathematically.

Metal X can form 2 oxides (A and B).

A + B = 3g

The mass of oxygen in A is 0.72g and the mass of oxygen in B is 1.16g.

The mass of metal X in the two oxides will be the same because it's the same metal.

Thus, we represent the mass of the metal in the two oxides as 2X.

2X + 0.72 + 1.16 = 3

2X + 1.88 = 3

2X = 3 - 1.88

2X = 1.12

X = 0.56

<u>Thus, 0.56 g of the metal combines with 0.72g of oxygen in A and 1.16 g of oxygen in B.</u>

Thus, mass of metal (X) in 1g of oxygen in A is

0.56g ⇒ 0.72g

X ⇒ 1

X = 1 × 0.56/0.72

X = 0.78 g

Hence, 0.78g of the metal will combine with 1g of oxygen for A

Also, mass of metal (X) in 1g of oxygen in B is

0.56g ⇒ 1.16g

X ⇒ 1g

X = 1×0.56/1.16

X = 0.48 g

Thus, 0.48g of the metal will combine with 1g of oxygen for B

6 0
3 years ago
Complete combustion of 2.60 g of a hydrocarbon produced 8.46 g of co2 and 2.60 g of h2o. what is the empirical formula for the h
zubka84 [21]
1 mole of carbon dioxide contains a mass of 44 g, out of which 12 g are carbon. 
Hence, in this case the mass of carbon in 8.46 g of CO2:
 (12/44) × 8.46 = 2.3073 g
 1 mole of water contains 18 g, out of which 2 g is hydrogen;
Therefore, 2.6 g of water contains;
 (2/18) × 2.6 = 0.2889 g of hydrogen.
Therefore, with the amount of carbon and hydrogen from the hydrocarbon we can calculate the empirical formula.
We first calculate the number of moles of each,
Carbon = 2.3073/12  = 0.1923 moles
Hydrogen = 0.2889/1 = 0.2889 moles
Then, we calculate the ratio of Carbon to hydrogen by dividing with the smallest number value;
             Carbon : Hydrogen
  0.1923/0.1923 : 0.2889/0.1923
                       1 :  1.5
                      (1 : 1.5) 2
                     = 2 : 3
Hence, the empirical formula of the hydrocarbon is C2H3
5 0
4 years ago
Good morning, I have a question..
matrenka [14]

Answer:

<h3>The answer is 5.24 mL</h3>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density} \\

From the question

mass = 152 g

density = 29 g/cm³

We have

volume =  \frac{152}{29}  \\  = 5.24137931...

We have the final answer as

<h3>5.24 mL</h3>

Hope this helps you

7 0
3 years ago
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