A soccer balll is travelling at a velcrocity of 50ms the kinetic energy of the ball is 500 what is the mass of the soccer ball

If 2.27 moles of bromine react with excess phosphorus, how many moles of phosphorus tribromide will be produced

zalisa [80]

Answer:

The answer to your question is 1.51 moles of PBr₃

Explanation:

Data

moles of Br = 2.27

moles of PBr₃ = x

Balanced chemical reaction

2P + 3Br₂ ⇒ 2PBr₃

Reactant Element Products

2 P 2

6 Br 6

-Use proportions to find the answer

3 moles of Br₂ ------------------ 2 moles of PBr₃

2.27 moles of Br₂ ------------------ x

x = (2.27 x 2) / 3

x = 4.54 / 3

x = 1.51 moles

7 0

2 years ago

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A critical reaction in the production of energy in biological systems is the hydrolysis of adenosine triphosphate (ATP) to adeno

Archy [21]

Answer:

ΔG° of reaction = -47.3 x $10^{3}$ J/mol

Explanation:

As we can see, we have been a particular reaction and Energy values as well.

ΔG° of reaction = -30.5 kJ/mol

Temperature = 37°C.

And we have to calculat the ΔG° of reaction in the biological cell which contains ATP, ADP and HPO4-2:

The first step is to calculate the equilibrium constant for the reaction:

Equilibrium Constant K = $\frac{[HPO4-2] x [ADP]}{ATP}$

And we have values given for these quantities in the biological cell:

[HP04-2] = 2.1 x $10^{-3}$ M

[ATP] = 1.2 x $10^{-2}$ M

[ADP] = 8.4 x $10^{-3}$ M

Let's plug in these values in the above equation for equilibrium constant:

K = $\frac{[2.1x10^{-3}] x [8.4x10^{-3}] }{[1.2 x 10^{-2}] }$

K = 1.47 x $10^{-3}$ M

Now, we have to calculate the ΔG° of reaction for the biological cell:

But first we have to convert the temperature in Kelvin scale.

Temp = 37°C

Temp = 37 + 273

Temp = 310 K

ΔG° of reaction = (-30.5 $10^{3}$ ) + (8.314)x (310K)xln(0.00147)

Where 8.314 = value of Gas Constant

ΔG° of reaction = (-30.5 x $10^{3}$ ) + (-16810.68)

ΔG° of reaction = -47.3 x $10^{3}$ J/mol

5 0

2 years ago

What is the final temperature of the solution formed when 1.52 g of NaOH is added to 35.5 g of water at 20.1 °C in a calorimeter

Inessa [10]

Answer : The final temperature of the solution in the calorimeter is, $31.0^oC$

Explanation :

First we have to calculate the heat produced.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = -44.5 kJ/mol

q = heat released = ?

m = mass of $NaOH$ = 1.52 g

Molar mass of $NaOH$ = 40 g/mol

$\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{1.52g}{40g/mole}=0.038mole$

Now put all the given values in the above formula, we get:

$44.5kJ/mol=\frac{q}{0.038mol}$

$q=1.691kJ$

Now we have to calculate the final temperature of solution in the calorimeter.

$q=m\times c\times (T_2-T_1)$

where,

q = heat produced = 1.691 kJ = 1691 J

m = mass of solution = 1.52 + 35.5 = 37.02 g

c = specific heat capacity of water = $4.18J/g^oC$

$T_1$ = initial temperature = $20.1^oC$

$T_2$ = final temperature = ?

Now put all the given values in the above formula, we get:

$1691J=37.02g\times 4.18J/g^oC\times (T_2-20.1)$

$T_2=31.0^oC$

Thus, the final temperature of the solution in the calorimeter is, $31.0^oC$

4 0

2 years ago

A tablet of one antacid contains citric acid, H3C6H5O7, and sodium hydrogen carbonate, NaHCO3. When the tablet dissolves in wate

Nadusha1986 [10]

H2O
This equation is a double displacement reaction, and it forms H2CO3, which is very unstable and separates into H2O and CO2.

8 0

2 years ago

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What do all atoms in a period (horizontal rows) have in common?

scoray [572]

Answer:

the same number of atomic orbitals.

Explanation:

5 0

1 year ago

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