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Olegator [25]
1 year ago
10

What is an example of taking action or doing service in the cement industry

Chemistry
1 answer:
yarga [219]1 year ago
6 0

Answer:

mining of clay limestone and then heated to a certain temperature of 1450⁰ in a cement kiln

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What is another common name for the lanthanoids on the periodic table?
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Lanthanide Lanthanoid, also called Lanthanide
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Which atom has the largest atomic radius
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francium , in the Periodic table the atomic radius increases from top to bottom in a group, and decreases from left to right across a period. making helium is the smallest element, and francium the largest.
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3 years ago
When the final velocity is less than the initial velocity, this is speed time acceleration deceleration
WITCHER [35]

Hello!

When the final velocity is less than the initial velocity, this is deceleration

<h2>Why?</h2>

Acceleration is defined as the physical magnitude that measures the change in velocity with time. The units to express acceleration are speed over time.

The equation for acceleration is: a=\frac{v-v_o}{t-t_o}

Where: a=acceleration, v=final velocity, vo=initial velocity, t=final time, to=initial time.

If the final velocity is less than the initial velocity, then the acceleration is negative, and that is called deceleration. An example of this is when a car brakes.

Have a nice day!



6 0
2 years ago
Read 2 more answers
When an equation is balanced what is the same on both side?
sammy [17]
"equal to"

1=1
1 is equal to 1

Is this what you mean?
3 0
2 years ago
Read 2 more answers
An engine operates on a Carnot cycle that uses 1mole of an ideal gas as the
sattari [20]

Answer:

Step 1;

q = w = -0.52571 kJ, ΔS = 0.876 J/K

Step 2

q = 0, w = ΔU = -7.5 kJ, ΔH = -5.00574 kJ

Explanation:

The given parameters are;

P_i = 100 N·m

T_i = 327 K

P_f = 90 N·m

Step 1

For isothermal expansion, we have;

ΔU = ΔH = 0

w = n·R·T·ln(P_f/P_i) = 1 × 8.314 × 600.15 × ln(90/100) = -525.71

w ≈<em> -0.52571</em> kJ

At state 1, q = w = -0.52571 kJ

ΔS = -n·R·ln(P_f/P_i) = -1 × 8.314 × ln(90/100) ≈ 0.876

ΔS ≈ 0.876 J/K

Step 2

q = 0 for adiabatic process

ΔU = 25×(27 - 327) = -7,500

w = ΔU = <em>-7.5 kJ</em>

ΔH = ΔU + n·R·ΔT

ΔH = -7,500 + 8.3142 × 300 = -5,005.74

ΔH = ΔU = <em>-5.00574 kJ</em>

6 0
2 years ago
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