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3 years ago

10

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

2 answers:

soldi70 [24.7K]3 years ago

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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Explain the reason potassium was visible when using the cobalt glass. Describe what occured.

Korolek [52]

Explanation:

If potassium is burnt the ions go into a high state of energy. Once they cool, it gives off energy in the form of a visible spectrum which has a characteristic color Now, The cobalt glass blocks out yellow light, and potassium ion which is purple in color is visible.

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3 years ago

Help!! 29 pointss

yaroslaw [1]

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4 years ago

When a solution is considered to have a high concentration of hydrogen ions (H+) is also considered to be?

charle [14.2K]

Answer: option A. acidic

Explanation: acidic solution is characterized by the presence of Hydrogen ion.

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3 years ago

50.0 mL of 0.200 M HNO2 is titrated to its equivalence point with 1.00 M NaOH. What is the pH at the equivalence point?

yKpoI14uk [10]

Answer:

8.279

Explanation:

The pH can be determined by hydrolysis of a conjugate base of weak acid at the equivalence point.

At the equivalence point, we have

$n_{NaOH}=n_{HNO_2}$

= 25.00 x 0.200

= 5.00 m-mol

= 0.005 mol

Volume of the base that is added to reach the equivalence point is

$\frac{0.005}{1.00} \times 1000= 5.00 \ mL$

Number of moles of $NO^-_{2}=n_{HNO_2}$

= 0.005 mol

Volume at the equivalence point is 25 + 5 = 30.00 mL

Therefore, concentration of $NO^-_{2}= \frac{5}{30}$

= 0.167 M

Now the ICE table :

$NO^-_2 + H_2O \rightarrow HNO_3 + OH^-$

I (M) 0.167 0 0

C (M) -x +x +x

E (M) 0.167-x x x

Now, the value of the base dissociation constant is ,

$K_w=K_a \times K_b$ $(K_w \text{ is the ionic product of water })$

$K_b =\frac{K_w}{K_a}$

$K_b =\frac{1 \times 10^{-14}}{4.6 \times 10^{-4}}$

= $2.174 \times 10^{-11}$

Base ionization constant, $K_b = \frac{\left[HNO_2\right] \left[OH^- \right]}{\left[NO^-_2 \right]}$

$2.174 \times 10^{-11}=\frac{x^2}{0.167 -x}$

$x= 1.9054 \times 10^{-6}$

So, $[OH^-]=1.9054 \times 10^{-6 } \ M$

pOH =- $\log[OH^-]$

= $- \log(1.9054 \times 10^{-6} \ M)$

=5.72

Now, since pH + pOH = 14

pH = 14.00 - 5.72

= 8.279

Therefore the ph is 8.279 at the end of the titration.

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3 years ago

Which element(s) is being reduced in the redox reaction below? 6 nh4clo4 (s) + 10 al(s) 5 al2o3 (g) + 6 hcl (g) + 3 n2 (g) + 9 h

vitfil [10]

The answer is Cl because its oxidation number goes from a +7 down to a -1.

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3 years ago