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astraxan [27]
3 years ago
13

The very common mineral shown in this photograph is commonly a pink- to cream-colored mineral with wavy, light-colored lines. it

does not effervesce. what mineral is it?
Chemistry
1 answer:
musickatia [10]3 years ago
8 0
The very common mineral shown in the figure that is referred in this problem that is commonly a pink- to cream-colored mineral with wavy, light-colored lines and does not effervesce would be feldspar. It make up about 41 percent weight of the Earth's crust. It is a group of rocks that contains tectosilicate compounds.
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The first law of thermodynamics is observed when
krek1111 [17]

Answer:

Explanation:

The first law of Thermodynamics is known as Conservation because it explains that energy is always maintained within a closed system and cannot be created or destroyed. Therefore, this is observed when there is no longer change in temperature in a system. Mainly because the energy is not being transferred to and from another system. Without this transfer of energy, the energy itself gets conserved within the system and the temperature no longer fluctuates.

8 0
2 years ago
Which of the following answers is true for the following statement?
Vsevolod [243]

it is either "aweak acid or a lousy (or very weak) acid"

6 0
3 years ago
How many elements are there​
Levart [38]

Answer:

there r 118 elements in periodic table..

hope it helps.

6 0
3 years ago
Read 2 more answers
Calculate the Molarity when a 6.11 mL solution of 0.1 H2SO4 is diluted with 105.12 mL of water
barxatty [35]

Molarity after dilution : 0.0058 M

<h3>Further explanation </h3>

The number of moles before and after dilution is the same  

The dilution formula

 M₁V₁=M₂V₂

M₁ = Molarity of the solution before dilution  

V₁ = volume of the solution before dilution  

M₂ = Molarity of the solution after dilution  

V₂ = Molarity volume of the solution after dilution

M₁=0.1 M

V₁=6.11

V₂=105.12

\tt M_2=\dfrac{M_1.V_1}{V_2}=\dfrac{0.1\times 6.11}{105.12}=0.0058~M

5 0
3 years ago
A sample of nitrogen gas had a volume of 500. ml, a pressure in its closed container of 740 torr and a temperature 25 degrees c
Semenov [28]
Question:
              <span>A sample of nitrogen gas had a volume of 500mL, a pressure in its closed container of 740 torr and a temperature of 25°c. what was the volume of gas when the temperature was changed to 50°c and the new pressure was 760 torr?

Answer:

Data Given:
                   V</span>₁  =  500 mL

                   P₁  =  740 torr

                   T₁  =  25 °C + 273  =  298 K

                   V₂  =  ?

                   P₂  =  760 torr

                   T₂  =  50 °C + 273  =  323 K

Solution:
             Let suppose the gas is acting Ideally, then According to Ideal Gas Equation,

                      P₁ V₁ / T₁  =  P₂ V₂ / T₂

Solving for V₂,

                      V₂  =  (P₁ V₁ T₂) ÷ (T₁ P₂)

Putting Values,
  
                      V₂  =  (740 torr × 500 mL × 323 K) ÷ (298 K × 760 torr)

                      V₂  =  527.68 mL
7 0
3 years ago
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