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sveta [45]
2 years ago
14

What is the name of Br6F10 ?

Chemistry
1 answer:
astraxan [27]2 years ago
3 0

Answer:

idfk

Explanation:

You might be interested in
Alexandra decides to climb Mt. Krumpett, which is 5000 m high. She determines that this will require a total of 1350 kcal of ene
Kisachek [45]
Alexandra requires a total energy of 1350 kcal for the climb
by eating proteins, fats and carbohydrates the amount of calories per gram contributed varies.
Proteins and carbohydrates - 4 calories per gram 
fats - 9 calories and gram
This means that by eating the same mass of fats and proteins/ carbohydrats the calories gained from fats is higher.
each bar contains;
<span>50 g of carbohydrates - 4 calories/g x 50 g = 200 calories
10 g of fat - 9 calories/g x 10 g = 90 calories 
40 g of protein - 4 calories/g x 40 g = 160 calories 
total amount of calories from 1 bar = 200 + 90 + 160 = 450 calories 
energy required = 1 350 000 calories 
bars required = 1 350 000/450 = 3000
alexandra should consume 3000 bars </span>
8 0
2 years ago
HELP! Ethanol has a density of 0.8 g/cm3. a. What is the mass of 225 cm3 of ethanol? b. What is the volume of 75.0 g of ethanol?
IRINA_888 [86]

Answer:

The correct answers are:

a) 180 g

b) 93.7 cm³

Explanation:

The density of a substance is the mass of the substance per unit of volume. So, it is calculated as follows:

density= mass/volume

From the data provided in the problem:

density = 0.8 g/cm³

a) Given: volume= 225 cm³

mass= density x volume = 0.8 g/cm³ x 225 cm³ = 180 g

b) Given: mass= 75.0 g

volume = mass/density = 75.0 g/(0.8 g/cm³)= 93.75 cm³≅ 93.7 cm³

5 0
1 year ago
How many moles of oxygen atoms are in 132.2 g of MgSO4?
zzz [600]

4.4moles of oxygen atoms

Explanation:

Given parameters:

Mass of MgSO₄ = 132.2g

Unknown:

Number of moles of oxygen atoms = ?

Solution:

The number of moles is the quantity of substance that contains the avogadro's number of particles.

 To solve for this;

 Number of moles = \frac{mass}{molar mass}

Molar mass of MgSO₄ = 24 + 32 + 4(16) = 120g/mole

  Number of moles = \frac{132.2}{120} = 1.1 moles

In

     1 moles of MgSO₄ we have 4 moles of oxygen atoms

    1.1 moles of MgSO₄ contains 4 x 1.1 moles = 4.4moles of oxygen atoms

learn more:

number of moles  brainly.com/question/1841136

#learnwithBrainly

8 0
2 years ago
Assume the atomic mass of element X is 22.99 amu. A 17.15−g sample of X combines with 14.17 g of another element Y to form a com
Alenkasestr [34]
The formula of the compound is XY. This means that the relation between the moles is 1: 1. One mole of X per one mole of Y.

From the information about the element X you can determine the number of moles of X (which is the same that the number of moles of Y).

# of moles of X = weigth of X / atomic mass of X = 17.15 g / 22.9 g/mol = 0.74598

Now the atomic mass of Y = weight of Y / # of moles of Y = 14.17 g / 0.74598 mol = 19 amu
8 0
1 year ago
Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equation N2O4(g) 2N
Amanda [17]

Answer : The correct option is, (a) 0.44

Explanation :

First we have to calculate the concentration of N_2O_4.

\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}

\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M

Now we have to calculate the dissociated concentration of N_2O_4.

The balanced equilibrium reaction is,

                             N_2O_4(g)\rightleftharpoons 2NO_2(aq)

Initial conc.           1.0 M          0

At eqm. conc.     (1.0-x) M    (2x) M

As we are given,

The percent of dissociation of N_2O_4 = \alpha = 28.0 %

So, the dissociate concentration of N_2O_4 = C\alpha=1.0M\times \frac{28.0}{100}=0.28M

The value of x = C\alpha = 0.28 M

Now we have to calculate the concentration of N_2O_4\text{ and }NO_2 at equilibrium.

Concentration of N_2O_4 = 1.0 - x  = 1.0 - 0.28 = 0.72 M

Concentration of NO_2 = 2x = 2 × 0.28 = 0.56 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be:

K_c=\frac{[NO_2]^2}{[N_2O_4]}

Now put all the values in this expression, we get :

K_c=\frac{(0.56)^2}{0.72}=0.44

Therefore, the equilibrium constant K_c for the reaction is, 0.44

8 0
2 years ago
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