The formula for solving current given with resistance and power source or voltage is shown below:
I = V/R
When two 5 ohms resistors are in series, we have:
I = 9 volts / (5+5 ohms)
I = 0.9 amperes
When it is being added with another 7.5 resistors, we have:
I = 9 volts / (5+5+7.5 ohms)
I = 0.529 ampere
The answer to the question is the letter "D. decrease; 0.51 amps".
Here's a short answer
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Answer:
- Question 19: the three are molecular compounds.
Explanation:
<em>Question 19.</em>
All of them are the combination of two kinds of different atoms in fixed proportions.
- C₂H₄: two carbon atoms per four hydrogen atoms
- HF: one hydrogen atom per one fluorine atom
- H₂O₂: two hydrogen atoms per two oxygent atoms
Thus, they all meet the definition of compund: a pure substance formed by two or more different elements with a definite composition.
Molecular compounds are formed by covalent bonds and ionic compounds are formed by ionic bonds.
Two non-metal elements, like H-F, C - C, C - H, H-O, H - H, and O - O will share electrons forming covalent bonds to complete their valence shell. Thus, the three compounds are molecular and not ionic.
<em>Question 20. </em>Formula of copper(II) sulfate hydrate with 36.0% water.
Copper(II) sulfate is CuSO₄. Its molar mass is 159.609g/mol
Water is H₂O. Its molar mass is 18.015g/mol
Calling x the number of water molecules in the hydrate, the percentage of water is:

From which we can solve for x:

Thus, there are 5 molecules of water per each unit of CuSO₄, and the formula is:
Answer:
Explanation:
1) a, b) A <em>solution</em><em> is a homogeneous mixture of two or more substances</em>. The <em>solute</em><em> is the substance present in a smaller amount</em>, and the <em>solvent</em><em> is the substance present in a larger amount. </em>
c) <em>A </em><em>saturated solution</em><em> contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature. </em>
2) See picture in attachment.
Answer:
10043.225 J
Explanation:
We'll begin by calculating the amount of heat needed to change ice to water since water at 0°C is ice. This is illustrated below:
Mass (m) = 15.5g
Latent heat of fussion of water (L) = 334J/g
Heat (Q1) =..?
Q1 = mL
Q1 = 15.5 x 334
Q1 = 5177 J
Next, we shall calculate the amount of heat needed to raise the temperature of water from 0°C to 75°C.
This is illustrated below:
Mass = 15.5g
Initial temperature (T1) = 0°C
Final temperature (T2) = 75°C
Change in temperature (ΔT) = T2 – T1 = 75 – 0 = 75°C
Specific heat capacity (C) of water = 4.186J/g°C
Heat (Q2) =?
Q2 = MCΔT
Q2 = 15.5 x 4.186 x 75
Q2 = 4866.225 J
The overall heat energy needed is given by:
QT = Q1 + Q2
QT = 5177 + 4866.225
QT = 10043.225 J
Therefore, the amount of energy required is 10043.225 J