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rodikova [14]
3 years ago
10

Light rays enter a translucent material what happening to the light rays

Chemistry
1 answer:
artcher [175]3 years ago
6 0

When this happens, light changes speed and the light ray bends, either toward or away from what we call the normal line, an imaginary straight line that runs perpendicular to the surface of the object.

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What the common uses for Bohrium?
lora16 [44]

Answer:

Explanation:

Bohrium's most stable isotope, bohrium-270, has a half-life of about 1 minute. It decays into dubnium-266 through alpha decay. Since only a few atoms of bohrium have ever been made, there are currently no uses for bohrium outside of basic scientific research.

5 0
3 years ago
Read 2 more answers
What is the mass of 20.0 L of sulfur dioxide (SO2) at STP?
larisa86 [58]

Answer:

Mass = 57.05 g

Explanation:

Given data:

Volume of SO₂ = 20.0 L

Temperature = standard = 273 K

Pressure = standard = 1 atm

Mass of SO₂ = ?

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

n = PV/RT

n = 1 atm ×  20.0 L / 0.0821 atm.L/ mol.K× 273 k

n =  20.0  / 22.41/mol

n = 0.89 mol

Mass of SO₂:

Mass = number of moles × molar mass

Mass = 0.89 mol × 64.1 g/mol

Mass = 57.05 g

6 0
3 years ago
Need help plzz
Sveta_85 [38]

_____ are types of active transport.

(A)Diffusion and osmosis

<u>(B)Engulfing and transport proteins </u>

(C)Osmosis and engulfing

(D)Transport proteins and diffusion

5 0
3 years ago
Chemistryyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy
mrs_skeptik [129]
Water is your answer!!!!!!!!!
5 0
3 years ago
A 1000g Ni rod, heated to 150. °C was placed in 1.00 kg of water at 25.0 °C. The final temperature of the water was 26.3 °C. Wha
Shtirlitz [24]

Answer : The molar specific heat of Ni is, 2.576J/mole^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of Ni = ?

c_1 = specific heat of water = 4.18J/g^oC

m_1 = mass of Ni = 1000 g

m_2 = mass of water = 1 kg = 1000 g

T_f = final temperature of water = 26.3^oC

T_1 = initial temperature of Ni = 150^oC

T_2 = initial temperature of water = 25^oC

Now put all the given values in the above formula, we get

1000g\times c_1\times (26.3-150)^oC=-1000g\times 4.18J/g^oC\times (26.3-25)^oC

c_1=0.0439J/g^oC

Now we have to calculate the molar specific heat of Ni.

\text{Molar specific heat of Ni}=\text{Specific heat of Ni}\times \text{Molar mass of Ni}=(0.0439J/g^oC)\times (58.69g/mole)=2.576J/mole^oC

Therefore, the molar specific heat of Ni is, 2.576J/mole^oC

4 0
3 years ago
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