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3 years ago

A SERIES CIRCUIT is a closed circuit with one path. a. True b. False

2 answers:

5 0

A series circuit is a closed circuit where the current follows one path. In a series circuit, the devices along the circuit loop are connected in a continuous row, so that if one device fails or is disconnected, the entire circuit is interrupted.

lisov135 [29]3 years ago

4 0

TRUE the devices in a series circuit are connected in a continuous row so if one device fails they all fail

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If you have 0.045 L of 0.465 M potassium bromide. How many moles of potassium bromide are present?

Mekhanik [1.2K]

The answer to your question is letter A! Hope that helps

7 0

3 years ago

1. What type of reaction is listed below? Prove that it is balanced by listing the number of elements on both side of the equati

ohaa [14]

Is there a mistake or am i doing something wrong

6 0

3 years ago

Read 2 more answers

Balance <br> Na2O+H2O → NaOH

AleksAgata [21]

Answer:
Na2O+H2O=2NaOH
Step by step exp.
Given:
Equation Na2O+H2O=NaOH
To find: Balance the equation
Solution:
Taking LHS of the equation
LHS=Na2O+H2O
There is 2 sodium, 2 oxygen,& 2 hydrogen
To balance the equation we have equal number of atom so we multply 2 to the RHS=2NaOH
There fore the equation form is
Na2O+H2O=2NaOH

8 0

3 years ago

How many molecules are there in 4dm³ of nitrogen gas

daser333 [38]

There are 1.078 x 10²³ molecules

<h3>Further explanation</h3>

Given

4 dm³ = 4 L Nitrogen gas

Required

Number of molecules

Solution

Assumptions on STP (1 atm, 273 K), 1 mol gas = 22.4 L, so for 4 L :

mol = 4 : 22.4

mol = 0.179

1 mol = 6.02 x 10²³ particles(molecules, atoms)

For 0.179 :

= 0.179 x 6.02 x 10²³

= 1.078 x 10²³

5 0

2 years ago

PLEASE HELP FAST IM ABOUT TO FAIL PLEASE HELP PLEASE HELP FAST HELP HELP THIS IS DESPRATE

MariettaO [177]

1. 0.240 liters of water would be needed to dissolve 21.6 g of lithium nitrate to make a 1.3 M (molar) solution.

2. 2.9 M is the molarity of a solution made of 215.1 g of HCl is dissolved to make 2.0 L of solution.

3.83.3 ml of concentrated 18 M H2SO4 is needed to prepare 250.0 mL of a 6.0 M solution.

4. 135 ml of stock HBr will be required to dilute the solution.

5. 150 ml of water should be added to 50.0 mL of 12 M hydrochloric acid to make a 4.0 M solution

6. The pH of the resulting solution is 13.89

Explanation:

The formula used in solving the problems is

number of moles= $\frac{mass}{atomic mass of one mole}$ 1st equation

molarity = $\frac{number of moles}{volume}$ 2nd equation

Dilution formula

M1V1 = M2V2 3rd equation

1. Data given

mass of Lithium nitrate = 21.6 grams

atomic mass of on emole lithium nitrate = 68.946 gram/mole

Molarity is given as 1.3 M

VOLUME=?

Calculate the number of moles using equation 1

n = $\frac{21.6}{68.946}$

= 0.313 moles of lithium nitrate.

volume is calculated by applying equation 2.

volume = $\frac{0.313}{1.3}$

= 0.240 litres of water will be used.

2. Data given:

mass of HCl = 215.1 gram

atomic mass of HCl = 36.46 gram/mole

volume = 2 litres

molarity = ?

using equation 1 number of moles calculated

number of moles = $\frac{215.1}{36.46}$

number of moles of HCl = 5.899 moles

molarity is calculated by using equation 2

M = $\frac{5.899}{2}$

= 2.9 M is the molarity of the solution of 2 litre HCl.

3. data given:

molarity of H2SO4 = 18 M

Solution to be made 250 ml of 6 M

USING EQUATION 3

18 x V1= 250 x 6

V1 = 83.3 ml of concentrated 18 M H2SO4 will be required.

4. data given:

M1= 10M, V1 =?, M2= 3 ,V2= 450 ml

applying the equation 3

10 x VI = 3x 450

V1 = 135 ml of stock HBr will be required.

5. Data given:

V1 = 50 ml

M1= 12 M

V2=?

M2= 4

applying the equation 3

50 x 12 = 4 x v2

V2 = 150 ml.

6. data given:

HCl + NaOH ⇒ NaCl + H20

molarity of NaOH = 0.525 M

volume of NaOH = 25 ml

molarity of acid HCl= 75 ml

volume of HCl = 0.335 ml

pH=?

Number of moles of NaOH and HCl is calculated by using equation 1 and converting volume in litres

moles of NaOH = 0.0131

moles of HCl= 0.025 moles

The ratio of moles is 1:1 . To find the unreacted moles of acid and base which does not participated in neutralization so the difference of number of moles of acid minus number of moles of base is taken.

difference of moles = 0.0119 moles ( NaOH moles is more)

Molarity can be calculated by using equation 1 in (25 +75 ml) litre of solution

molarity = $\frac{0.0119}{0.1}$

= 0.11 M (pOH Concentration)

14 = pH + pOH

pH = 14 - 0.11

pH = 13.89

3 0

3 years ago