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S_A_V [24]
3 years ago
15

Write an expression that can be used to check the quotient of 646÷3

Chemistry
1 answer:
iVinArrow [24]3 years ago
5 0

Answer:

215.33

Explanation:

A mathematical expression is an equation that connect one or more variables with constants.

The Quotient is the result of carrying out the division operation.let "y" be the quotient of dividing 646 by 3

therefore 646÷3 =y

\frac{646}{3}=y\\ 3y=646\\

y=\frac{646}{3}\\ y=215.33

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A solution is prepared by adding 1.43 mol of potassium chloride (kcl) to 889 g of water. The concentration of kcl is 1.61  molal.
mol of Kcl (potassium chloride)= 1.43
water = 889 g
the formula for calculating molality is:
molality = moles of solute/kilograms of solvent
1kg = 1000g so, 889g = 0.889kg

m = 1.43/0.889 = 1.61 molal
7 0
3 years ago
A person walking covers 5.20 m in 10.4 s. How fast is the person moving?
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Answer:

<h2>0.50 m/s</h2><h2></h2>

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What does a scientific law do ?
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Which of the following metals may tarnish to a greenish-blue color called verdigris
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Read 2 more answers
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
3 years ago
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