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3 years ago

What does the M in Newton's law equation stand for?

1 answer:

5 0

In Newton's 2nd law, written in terms of force, mass and acceleration, m stands for the mass in the original Newton's work, i.e. the quantity of the substance which makes up a material body.

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What would be the final volume (in mL) of the aqueous potassium fluoride which was prepared by dissolving 8.0 g of potassium flu

zhuklara [117]

Answer:

0.15 L

Explanation:

Step 1. Calculate the moles of KF.

Molar mass = 58.10 g/mol

Moles of KF = 8.0 × 1/58.10

Moles of KF = 0.138 mol KF

===============

Step 2. Calculate the volume of KF

c = n/V Multiply both sides by V

V = 0.138 × 1/0.89

V = 0.15 L

5 0

3 years ago

Ter the symbol for the isotope used in blood flow analyses.

blondinia [14]

Answer:

The symbol of isotopes used for blood flow analysis is $\begin{array}{c}{141} \\ {58}\end{array} C e.$

Explanation:

Isotopes are the substances that exhibit the same atomic number but has a different mass number of an element.
The atomic number explains the number of protons present in the element and mass number explains the number of neutrons available in the element.
For blood flow analysis, the isotope element is cerium-141 and it is used in the chemical examination of blood flow particles.
Symbol used for this isotope is $\begin{array}{c}{141} \\ {58}\end{array} C e$ , where 141 indicates the amount of mass present and 58 indicates the proton number and 83 indicates neutron number present in that element.
The amount of mass in an atom is calculated by the sum of protons and neutrons present in it. Thus mass of isotope is 141 obtained by the sum of 58 protons and 83 neutrons present in that isotope.

8 0

4 years ago

Which example best shows conservation of resources?

zaharov [31]

Answer:

Explanation:

I took this test as a freshman and this was the only question of 4 that I can remember

6 0

4 years ago

Read 2 more answers

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max2010maxim [7]

Answer: permanent removal of hair by energy or heat

Explanation:

8 0

3 years ago

Consider the Gibbs energies at 25 ∘C.

zysi [14]

Answer:

A)ΔGrxn∘=55.7kJ/mol

B)Ksp=1.75×10^−10

C)ΔGrxn∘=70.0kJ/mol

D)Ksp=5.45×10^−13

Explanation:

ΔGrxn∘=ΔGf,products∘−ΔGf,reactants∘

To calculate for the

Ksp

of the dissolution reaction can be claculated

ΔGrxn∘=−RTlnKsp

where R is the proportionality constant equal to 8.3145 J/molK.

ΔGrxn∘=[ΔGf,Ag(aq)+∘+ΔGf,Cl(aq)−∘]−ΔGf,

AgCl(s)⇌Ag(aq)++Cl(aq)−

ΔG∘rxn=[77.1kJ/mol+(−131.2kJ/mol)]−(−109.8kJ/mol)

ΔGrxn∘=55.7kJ/mol

b) Calculate the solubility-product constant of AgCI.

ΔGrxn∘=−RTlnKsp

55.7kJ/mol=−(8.3145×10−3J/molK)(298.15K)InKsp

Ksp=1.75×10^−10

c) Calculate

To calculate ΔG°rxn

for the dissolution of AgBr(s).

ΔGrxn∘=[ΔGf,Ag(aq)+∘+ΔGf,Br(aq)−∘]−ΔGf,

ΔGrxn∘=[77.1kJ/mol+(−104.0kJ/mol)]−(−96.90kj/mol

ΔGrxn∘=70.0kJ/mol

d)To Calculate the solubility-product constant of AgBr.

ΔGrxn∘=−RTlnKsp

70.0kJ/mol=−(8.3145×10−3J/molK)(298.15K)lnKsp

70.0kJ/mol=−(8.3145×10−3J/molK)(298.15K

Ksp=5.45×10^−13

8 0

3 years ago