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Butoxors [25]
1 year ago
3

380 kelvin to celsius

Chemistry
1 answer:
Vanyuwa [196]1 year ago
7 0

Answer

106.85 Celsius

Explanation

Given

380 Kelvin

What to find:

To convert 380 K to Celsius.

Solution:

The formula to convert Kelvin to Celsius is given by:

^0C=K-273.15

Put K = 380 into the formula:

^0C=380-273.15=106.85

380 Kelvin to celsius is 106.85 Celsius

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A meal containing a burger, fries, and a milkshake contains 53.0 grams of fat, 38.0 grams of protein, and 152 grams of carbohydr
RSB [31]

Answer:

286 minutes

Explanation:

Given the content of fat in a meal= 53g

the content of protien in a meal= 38g

the content of carbohydrate in a meal= 152g

Also given that fuel value of protien = 17 KJ/g

the fuel value of fat = 38 KJ/g

the fuel value of carbohydrate = 17 KJ/g

Now to burn all the molecules the energy recquired is content of molecule times the fuel value.

the energy recquired to burn fat = 53\times 38=2014KJ

the energy recquired to burn protien = 38\times 17=646KJ

the energy recquired to burn carbohydrate =152 \times 17=2584KJ

The total energy=2014+646+2584=5244KJ

The energy spent on swimming is 1100KJ/hour

for 1 hour 1100 KJ are spent

to spend 5244 KJ of energy the time taken is = \frac{5244}{1100}=4.76hours

We know 1 hour=60 minutes

The total time = 4.76\times 60=286minutes

6 0
4 years ago
What is the pCu of the resulting solution if 20.00 mL of 0.08 M EDTA (H4Y) is added to 15.00 mL of 0.10 M CuSO4 and buffered at
katrin [286]

Answer:

The answer is "5.4".

Explanation:

BoH + HCL =BCL +H_2o \\\\At eq \\\\N_1V_1=N_2V_2 \\\\v_2=20 \ ml\\\\[BCL]=\frac{20 \times 0.08}{20+20}=0.04\\\\pH = \frac{1}{2} [pkw - pk_b - \log e]\\\\pk_b = 2 pH - Pkw + \Log C\\\\pK_b=5.4

3 0
3 years ago
A certain liquid X has a normal boiling point of 111.20 celsius and a boiling point elevation constant Kb=0.95. calculate the bo
Goshia [24]

Answer: the boiling point is = 137.325°C

Explanation:

From the formula: ∆Tb= Kb*m

From the question, Kb= 0.95, m= 27.5, T1= 111.2°C

Substitute into ∆Tb= Kb*m

∆Tb= 0.95*27.5= 26.125

∆Tb= T2-T1

Hence

T2- 111.2=26.125

T2= 26.125+ 111.2= 137.325°C

8 0
3 years ago
A chemical reaction that gives off heat as it proceeds is said to be:.
Dafna1 [17]
Exothermic.

To remember this, ‘exo’ means outside, and ‘thermic’ means heat. It gives the outside (exo), heat (thermic).
7 0
3 years ago
How are a proton and a neutron alike?
baherus [9]

A. Each is found outside the nucleus

5 0
3 years ago
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