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Karo-lina-s [1.5K]
3 years ago
13

What is the oxidation number of C in NaHCO3?

Chemistry
1 answer:
Phoenix [80]3 years ago
4 0
Na = +1
H = +1
O = -2
Total charge is 0:
1 + 1 + C + 3 x -2 = 0
C = 4
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Acceleration is defined as the rate of change for which characteristic?
weeeeeb [17]

Answer:

Acceleration is defined as the rate of change for velocity.

4 0
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limestone breaks down when heated to form quicklime and carbon dioxide. what type of reaction is this?
Trava [24]
No chemical because, its changeing the stuff 
7 0
3 years ago
An unknown compound displays singlets at δ 2.1 ppm and 2.56 ppm in the ratio of 3:2. what is the structure of the compound?
olasank [31]

1) As can be seen from any 1H NMR chemical shift ppm tables, hydrogens which have δ values from 2ppm to 2.3ppm are hydrogens from carbon which is bonded to a carbonyl group. From this, we can conclude that our hydrogens belong to the type, but from 2 different alkyl groups because of 2 different signals.

 

2) So, one alkyl group is CH3 and second one can be CH or CH2.

 

3) If we know that ratio between two types of hydrogens is 3:2, it can be concluded that second alkyl group is CH2. 


4) Finally, we don't have any other signals and it indicates that part of the compound which continues on CH2 is exactly the same as the first part.

The ratio remains the same, 3:2 ie 6:4

7 0
3 years ago
A researcher studying the nutritional value of a new candy places a 4.90 g sample of the candy inside a bomb calorimeter and com
snow_lady [41]

Answer:

449730.879 cal/g

Explanation:

Given data:

Mass of sample = 4.9 g

Change in temperature  = 2.08 °C  (275.23 k)

Heat capacity of calorimeter = 33.50 KJ . K⁻¹

Solution:

C(candy) = Q/m

Q = C (calorimeter) × ΔT

C(candy) = C (calorimeter) × ΔT / m

C(candy) =  33.50 KJ . K⁻¹ × 275.23 K / 4.90 g

C(candy) = 9220.205 KJ / 4.90 g

C(candy) =  1881.674 KJ / g

It is known that,

1 KJ /g = 239.006 cal/g

1881.674 × 239.006 = 449730.879 cal/g

8 0
3 years ago
A certain gas has a volume of 500.0 mL at 77.0°C and 600KPa. Calculate the temperature if the volume decreased to 400.0 mL and t
Jlenok [28]

Answer:

354.67K

Explanation:

Applying

P₁V₁/T₁ = P₂V₂/T₂................. Equation 1

Where Where P₁ = initial pressure, T₁ = Initial temperature, V₁ = Initial Volume, P₂ = Final pressure, V₂ = Final Volume, T₂ = Final Temperature.

From the question, we are ask to look for the final temperature,

Therefore we make T₂ the subject of the equation

T₂ = P₂V₂T₁/P₁V₁............. Equation 2

Given: P₁ = 600 kPa, V₁ = 500 mL, T₁ = 77 °C = (273+77) = 350 K, P₂ = 760 kPa, V₂ = 400.0 mL

Substitute these values into equation 2

T₂ = (760×400×350)/(600×500)

T₂ = 354.67 K

3 0
3 years ago
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