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Ludmilka [50]
3 years ago
6

What is something cold scientist study?

Chemistry
1 answer:
netineya [11]3 years ago
6 0
Something that scientists study that is cold could be liquid nitrogen!
You might be interested in
One gallon of gasoline in an automobiles engine produces on average 9.50 kg of carbon dioxide, which is a greenhouse gas; that i
vagabundo [1.1K]

Answer:

The value is U  =1.3908 *10^{11} \  kg

Explanation:

From the question we are told that

mass of carbon dioxide produced by one gallon of gasoline is m  =  9.50 \  kg

The number of cars is N =  40 \  million =  40 *10^{6} \  cars

The distance covered by each car is d =  7930 \ mi

The rate is R  =  23.6 \  mi/ gallon

Generally the amount of gasoline used by one car is mathematically represented as

G  =  \frac{d}{R}

=> G  =  \frac{7930}{23.6}

=> G  =   366 \  gallons

Generally the amount of gasoline used by N cars is

H  =  N  *  G

=> H  =  40*10^{6}  *  366

=> H  = 1.464*10^{10} \  gallons

Generally the annual production of carbon dioxide is mathematically represented as

U  =  m *  H

=> U  =9.50 *   1.464*10^{10}

=> U  =1.3908 *10^{11} \  kg

8 0
4 years ago
A sample of methane gas collected at a pressure of 0.884 atm and a temperature of 6.00 C is found to occupy a volume of 26.5 L.
Genrish500 [490]

Answer:

1.03mole

Explanation:

Given parameters:

Pressure = 0.884 atm

Temperature  = 6°C  = 273 + 6  = 279k

Volume  = 26.5L

Unknown:

Number of moles  = ?

Solution:

To solve this problem, we use the ideal gas equation:

          PV  = nRT

P is the pressure

V is the volume

n is the number of moles

R is the gas constant  = 0.082atmdm³mol⁻¹K⁻¹  

T is the temperature

        n  = \frac{PV}{RT}    = \frac{0.884 x 26.5}{0.082 x 279}    = 1.03mole

4 0
3 years ago
How close a measured value is to the value is to the accepted value
sveta [45]
You could use percent error. Absolute | | of accepted value minus experimental value \ divided by accepted value multiply by 100 and that is your answer


8 0
3 years ago
What are the possible values of ml for an electron in a d orbital?
spin [16.1K]
The correct answer for the question that is being presented above is this one: "-2, -1, 0, 1, 2." T<span>he possible values of ml for an electron in a d orbital are -2, -1, 0, 1, 2. </span>Since the allowed values for mℓ range from −ℓ to +ℓ, once you know the value for ℓ you know the values for mℓ."
3 0
3 years ago
P O R F A V O R
ale4655 [162]

Answer:

a. 45×10³ kg

b. 1.25 kg

c. 5443200 s

d. 2.69×10⁻⁴ m/s²

e. 2.57×10⁻⁶N

f. 7.48×10⁻³ m /s²

g. 2.45 Pa

h. 10 m/s

Explanation:

The SI units are: kg, m, s, N, K, A, Pa, J and cd

a. 1 g is the mass for 1 cm³. We convert the m³ to cm³

45 m³. 1×10⁶ cm³ / 1 m³ = 45×10⁶ cm³

By the way, 45×10⁶ cm³ = 45×10⁶ g

We convert the g to kg →  45×10⁶ g . 1 kg / 1000 g = 45×10³ kg

b. As 1 g = 1 cm³,  we convert the cm³ to g and then, the g to kg

1250 cm³ = 1250 g → 1250 g . 1kg / 1000 g = 1.25 kg

c. 1 day has 24 hours; 1 hour has 60 minutes; 1 minute has 60 seconds

1 hour has 3600 s. Then 24 h . 3600 s / 1 h = 86400 s

86400 s/d. 63 d = 5443200 s

d. 1 min² = 3600 s²

97 cm / 3600 s² = 0.0269 cm/s²  / 100 = 2.69×10⁻⁴ m/s²

e. 927 g.cm / min² / 3600 s² = 0.2575 g.cm/s² → dyn

We need to convert dyn to N

1 dyn = 10⁻⁵N → 0.2575  dyn . 10⁻⁵N / 1dyn = 2.57×10⁻⁶N

f. 1 m/s² = 12960 km/h²

12960 km/h² . 1 m/s² / 97 km/h² = 7.48×10⁻³ m /s²

g. 2500 g/cm² . 1kg / 1000 g = 2.5 kg/cm²

1 Pa = 1.02kg/cm²

2.5 kg/cm² . 1 Pa / 1.02 kg/cm² = 2.45 Pa          

h. 1 h = 3600 s

36 km / 3600 s = 0.01 km /s → 0.01 km . 1000 m / 1 km = 10

= 10 m/s

5 0
3 years ago
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