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VARVARA [1.3K]
2 years ago
15

Suppose you carry out a titration involving 2.50 molar NaOH and an unknown concentration of HBr. To bring the reaction to its en

d point, you add 49.2 milliliters of NaOH to 35.0 milliliters of HBr.
What is the concentration of the HBr solution?

A. 0.711 M
B. 1.41 M
C. 1.78 M
D. 3.51 M
Chemistry
1 answer:
givi [52]2 years ago
7 0

Answer:

B

Explanation:

Brainly Patrol ima stop you here

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Fatty acids are carboxylic acids with a long, unbranched hydrocarbon chain. There are three main classes of fatty acids. Classif
LuckyWell [14K]

Answer:

Coconut oil, Olive oil and Sunflower oil

Explanation:

Fatty acids are carboxylic acids with a long unbranched chain of carbon and hydrogen atoms.

There are three main classes of fatty acids which are explained as under:

1. Saturated Fatty acids: These fatty acids have long carbon chain with two hydrogen atoms bonded to each carbon atom. This saturation of fatty acids make the fatty acids more stable towards high temperature. These fatty acids becomes solid at room temperature. Coconut oil and butter are the examples of saturated fatty acids.

2. Monounsaturated Fatty Acids: In a long carbon chain, if there is a carbon atom which is double bonded with another carbon atom and rest is saturated with hydrogen atoms, because of this single double-bond, the fatty acid is termed as monounsaturated fatty acids. These fatty acids are liquid at room temperature but solidify in refrigerator. Olive oil is an example of such fatty acids.

3. Polyunsaturated Fatty Acids: In a long carbon chain, if there are two or more than two carbon atoms which are double bonded with each other and rest is saturated with hydrogen atoms, because of multiple double bonds, such fatty acids are termed as polyunsaturated fatty acids. Because of higher unsaturation, these fatty acids are liquid in both normal room temperature and in refrigerator. Such unsaturation also make them unfit for cooking purposes. Sunflower oil, Soyabean oil and Flaxseed oil are examples of polyunsaturated fatty acids.

8 0
3 years ago
A rock that weighs 12.2 g is added to a graduated cylinder with 30.0 mL of water in it. The meniscus rises to 54.4 mL. What is t
GuDViN [60]

Answer:

              0.50 g/mL

Explanation:

Formula of Density,

                          D = mass / volume

Mass = 12.2 g

Volume = 54.4 mL - 30.0 mL = 24.4 mL

Putting values,

                          D = 12.2 g / 24.4 mL

                          D = 0.50 g/mL

4 0
2 years ago
A 44.8 g rhodium sample contains how many rhodium atoms
Tresset [83]

Answer:

2.63*10^23

Explanation:

1 mol rhodium = 102.91

44.8g/1 mol * 1 mol/ 102.91mol * 6.022*10^23/1 mol =

2.63*10^23

6 0
3 years ago
A target is shielded from a radioactive sample by a sheet of metal. What kind of radiation can pass through the sheet of metal a
weeeeeb [17]

A. o. I know gamma can I Hope this helps


A now i remember yeah gamma only

4 0
3 years ago
Read 2 more answers
The Balmer series, named after Johann Balmer, is a portion of the hydrogen emission spectrum produced from the transitions betwe
horrorfan [7]

Explanation:

The wavelength of the balmer series is calculated using the following steps;

- Find the Principle Quantum Number for the Transition

- Calculate the Term in Brackets

- Multiply by the Rydberg Constant

- Find the Wavelength

The Balmer series in a hydrogen atom relates the possible electron transitions down to the n = 2 position to the wavelength of the emission that scientists observe.

The λ symbol represents the wavelength, and RH is the Rydberg constant for hydrogen, with RH = 1.0968 × 107 m−1

n=7 to n=2

- The principal quantum numbers are 2 and 7.

-  (1/2²) − (1 / n²₂)

For n₂ = 7, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 7²)

= (1/4) − (1/49)

= 0.2230

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.2230

= 2445864 m−1

- λ = 1 / 2445864 m−1

= 4.08 × 10−7 m

= 408 nanometers

≈ 410nm

n=6 to n=2

- The principal quantum numbers are 2 and 6.

-  (1/2²) − (1 / n²₂)

For n₂ = 6, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 6²)

= (1/4) − (1/36)

=  0.2222

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 3/16

= 2437090 m−1

- λ = 1 / 2437090 m−1

= 4.10 × 10−7 m

= 410 nanometers

n=5 to n=2

- The principal quantum numbers are 2 and 5.

-  (1/2²) − (1 / n²₂)

For n₂ = 5, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 5²)

= (1/4) − (1/25)

= 0.21

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.21

= 2303280 m−1

- λ = 1 / 2303280 m−1

= 4.34 × 10−7 m

= 434 nanometers

n=4 to n=2

- The principal quantum numbers are 2 and 4.

-  (1/2²) − (1 / n²₂)

For n₂ = 4, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 4²)

= (1/4) − (1/16)

= 0.1875

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.1875

= 2056500 m−1

- λ = 1 / 2056500 m−1

= 4.86 × 10−7 m

= 486 nanometers

n=3 to n=2

- The principal quantum numbers are 2 and 3.

-  (1/2²) − (1 / n²₂)

For n₂ = 3, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 3²)

= (1/4) − (1/9)

= 0.13889

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.13889

= 1523345 m−1

- λ = 1 / 1523345 m−1

= 6.56 × 10−7 m

= 656 nanometers

7 0
3 years ago
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