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3 years ago

Does adding resistors and a serious increase or decrease the overall resistance of a circuit

Chemistry

1 answer:

Oksanka [162]3 years ago

4 0

It decreases while in a parallel circuit.

In a series circuit it will increase the resistance.

mark brainliest :)

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What does an atom become when it gains an electron in an ionic bond?

Elden [556K]

It becomes a acceptor because in an ionic bonding the element who gives out is a donor while the atom which accept is a acceptor

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2 years ago

How does the oceanic crust compare to continental crust?

dimulka [17.4K]

I think the answer is number (4)

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3 years ago

Draw the structures of the 3 isomers of C8H18 that contain 3 methyl branches on the main chain, 2 of which are on the same carbo

Vlad1618 [11]

Answer:

Please see the attachments

Explanation:

Please see the attachments below structures of the 3 isomers of C8H18 that contain 3 methyl branches on the main chain, 2 of which are on the same carbon.

3 0

3 years ago

A 60 and N net force is applied to a 12 kg box which is the acceleration of the box?

GarryVolchara [31]

The relation between force and mass and acceleration is

$Force = mass X acceleration$

The SI unit will be

Force = Newton

mass =kg

acceleration= $\frac{m}{s^{2} }$

thus

$acceleration = \frac{Force}{mass}$

putting values

$acceleration=\frac{60 N}{12kg}=5\frac{m}{s^{2} }$

Thus the acceleration will be

$5\frac{m}{s^{2} }$

6 0

3 years ago

Carbon-14 is a radioactive isotope that decays according to first-order kinetics in a process that has a half-life of 5730 years

Sliva [168]

Answer : The time passed in years is $2.83\times 10^3\text{ years}$

Explanation :

Half-life = 5730 years

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{5730\text{ years}}$

$k=1.21\times 10^{-4}\text{ years}^{-1}$

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $1.21\times 10^{-4}\text{ years}^{-1}$

t = time passed by the sample = ?

a = let initial amount of the reactant = X g

a - x = amount left after decay process = $71\% \times (x)=\frac{71}{100}\times (X)=0.71Xg$

Now put all the given values in above equation, we get

$t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{X}{0.71X}$

$t=2831.00\text{ years}=2.83\times 10^3\text{ years}$

Therefore, the time passed in years is $2.83\times 10^3\text{ years}$

8 0

2 years ago