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Masteriza [31]
3 years ago
15

A series circuit has two 5.0-ohm resistors and a power source of 9.0 volts. How will the current be affected and what will it be

if another resistor (with a resistance of 7.5 ohms) is added to the circuit?
A.increase; 0.51 amps
B.increase; 0.72 amps
C.increase; 3.6 amps
D.decrease; 0.51 amps
E.decrease; 0.72 amps
Chemistry
1 answer:
miskamm [114]3 years ago
3 0
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".
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What is the mass of 1.2 mol IrI3?​
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Answer:

690 g IrI₃

Explanation:

To convert from moles to grams, you have to use the molar mass of the compound. The molar mass of IrI₃ is 572.92 g/mol. You use this as the unit converter in this equation:

1.2molIrI3 * \frac{572.92g}{1mol} = 687.504 g IrI3

Round to the lowest number of significant figures which is 2 to get 690 g IrI₃.

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What is the ratio of carbon, hydrogen, and oxygen in most carbohydrates?
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 In carbohydrates the C:H:O is 1:2:1
7 0
3 years ago
Read 2 more answers
Iron has a density of 7.87mL. How much mass would a 12.3mL sample contain?
algol [13]

The correct answer is 96.80 grams

Explanation:

The density of a substance shows the total mass the substance contains in 1 mL or 1 cm3 as density is calculated by using the formula D= M (mass) / V (volume). This implies, Iron contains 7.87 grams per milliliter. Moreover, this value and formula can be used to calculate the mass or volume of any other sample. The process to calculate the mass of an iron sample with a volume of 12.3 mL is shown below.

D = M ÷ V

7.87 mL = x ÷ 12.3 mL - x represents the missing value. Now find the value of x by solving the equation

x = 7.87 · 12.3  

x = 96.80

This means a sample of 12.3 mL contains a mass of 96.80 grams. Also, you can know this value is correct because if you divide the mass by the value the density is the same (96.80 grams ÷ 12.3 mL = 7.87 g/mL)

4 0
3 years ago
For the reaction IO3–(aq) + 5 I–(aq) + 6 H+(aq) → 3 I2(aq) + 3 H2O(l) the rate of disappearance of I–(aq) at a particular time a
scoray [572]

Answer: The rate of appearance of I_2(aq) is 1.44\times 10^{-3}mol/Ls

Explanation:

Rate of a reaction is defined as the rate of change of concentration per unit time.

Thus for reaction:

IO_3^-(aq)+5I^-(aq)+6H^+(aq)\rightarrow 3I_2(aq)+3H_2O(l)

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate=-\frac{1d[I^-]}{5dt}=+\frac{d[I_2]}{3dt}

Given: -\frac{d[I^-]}{dt}] = 2.4\times 10^{-3}mol/Ls

+\frac{d[I_2]}{dt}=-\frac{3d[I^-]}{5dt}=-\frac{3}{5}\times 2.4\times 10^{-3}mol/Ls=1.44\times 10^{-3}mol/Ls

The rate of appearance of I_2(aq) is 1.44\times 10^{-3}mol/Ls

6 0
3 years ago
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