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3 years ago

Which atoms combine together during fusion reaction on the sun?

Chemistry

2 answers:

PtichkaEL [24]3 years ago

8 0

Explanation:

its hydrogen I took the test and got it right

shusha [124]3 years ago

5 0

Answer:

Hydrogen

Explanation:

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A binary compound of boron and hydrogen has the following percentage composition: 78.14% boron, 21.86% hydrogen. If the molar ma

algol [13]

Answer:

Empirical formula: BH3

Molecular Formula: B2H6

Explanation:

To solve the exercise, we need to know how many boron atoms and how many hydrogen atoms the compound has. We know that of the total weight of the compound, 78.14% correspond to boron and 21.86% to hydrogen. As the weight of the compound is between 27 g and 28 g, using the above percentages we can solve that the compound has between 21.1 g and 21.8 g of boron, and between 5.9 g and 6.1 g of hydrogen:

100% _____ 27 g

78.14% _____ x = 78.14% * 27g / 100% = 21.1 g boron

100% ______27 g

21.86% ______ x = 21.86% * 27g / 100% = 5.9 g hydrogen

100% _____ 28 g

78.14% _____ x = 78.14% * 28g / 100% = 21.8 g boron

100% _____ 28g

21.86% _____ x = 21.86% * 28g / 100% = 6.1 g hydrogen

So, if the atomic weight of boron is 10.8 g, there must be two boron atoms in the compound that sum 21.6 g. The weight of hydrogen is 1 g, so the compound must have six hydrogen atoms.

The molecular formula represents the real amount of atoms that form a compound. Therefore, the molecular formula of the compound is B2H6.

The empirical formula is the minimum expression that represents the proportion of atoms in a compound. For example, ethane has 2 carbon atoms and 6 hydrogen atoms, so its molecular formula is C2H6, however, its empirical formula is CH3. Therefore, the empirical formula of the boron compound is BH3.

8 0

3 years ago

The molar mass of V2O3 is A. 66.94 B. 133.88 C. 149.88 D. 200.82

enot [183]

B. 133.88
Sorry po kung mali

4 0

3 years ago

Tim wants to store 0.122 moles of helium gas at 203 kPa and 401 K. Which is the volume of the helium gas needed for storage?

algol13

The volume of Helium gas needed for storage is 2.00 L (answer C)

calculation

The volume of Helium is calculated using ideal gas equation

That is Pv =nRT

where;

P( pressure) = 203 KPa

V(volume)=?

n(number of moles) = 0.122 moles

R(gas constant) = 8.314 L.Kpa/mol.K

T(temperature)= 401 K

make V the subject of the formula by diving both side by P

V=nRT/p

V={[0.122 moles x 8.314 L. KPa/mol.K x 401 K] / 203 KPa} = 2.00 L

3 0

3 years ago

I NEED TOUR HELP, PLEASE.

kvasek [131]

It would be 5.0 more in i did this

8 0

2 years ago

What mass of butane in grams is necessary to produce 1.5×103 kj of heat what mass of co2 is produced?

kari74 [83]

The heat of reaction (i.e. combustion) of butane ( $C_{4} H_{10}$ ) when reacted with oxygen ( $O_{2}$ )  is -2658 kJ/mol butane, and the chemical reaction is given by:

$C_{4} H_{10}$ + $\frac{13}{2} O_{2}$ ---> $4 CO_{2}$ + $5 H_{2}O$

The mass of butane required in the reaction is based on the heat produced by the reaction, which is given to be -1,500 kJ. The minus sign is added because the reaction releases heat (exothermic), which means that the products are in a "lower energy state" than the reactants.

Dividing this with the heat of reaction per mole of butane reacted would give the number of moles butane required. Then, multiplying the answer with the molar mass of butane which is 58 grams/mole, will give the mass of butane required.

Moles of butane = [(-1,500 kJ)/(-2658 kJ/mol butane)]
Moles of butane = 0.5643 moles butane

Mass of butane  = 0.5643 moles butane * 58 grams/mol butane
Mass of butane = 32.73 grams butane

The mass of carbon dioxide ( $CO_{2}$ ) can be determined by multiplying the moles of butane ( $C_{4} H_{10}$ ) with the mole ratio of ( $CO_{2}$ ) produced to the ( $C_{4} H_{10}$ ) reacted, and then with the molar mass of ( $CO_{2}$ ), which is 44 grams/mole.

Mass of carbon dioxide produced
= 0.5643 moles butane * [4 moles $CO_{2}$ / 1 mole $C_{4} H_{10}$ ] * 44 grams/mole $CO_{2}$

Mass of carbon dioxide produced
= 99.32 grams $CO_{2}$

Thus, the mass of butane required is 32.73 grams, and the mass of carbon dioxide produced from the reaction of this amount of butane is 99.32 grams.

4 0

3 years ago

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