Answer:
The mass of tin is 164 grams
Explanation:
Step 1: Data given
Specific heat heat of tin = 0.222 J/g°C
The initial temeprature of tin = 80.0 °C
Mass of water = 100.0 grams
The specific heat of water = 4.184 J/g°C
Initial temperature = 30.0 °C
The final temperature = 34.0 °C
Step 2: Calculate the mass of tin
Heat lost = heat gained
Qlost = -Qgained
Qtin = -Qwater
Q = m*c*ΔT
m(tin)*c(tin)*ΔT(tin) = -m(water)*c(water)*ΔT(water)
⇒with m(tin) = the mass of tin = TO BE DETERMINED
⇒with c(tin) = the specific heat of tin = 0.222J/g°C
⇒with ΔT(tin) = the change of temperature of tin = T2 - T1 = 34.0°C - 80.0°C = -46.0°C
⇒with m(water) = the mass of water = 100.0 grams
⇒with c(water) = the specific heat of water = 4.184 J/g°C
⇒with ΔT(water) = the change of temperature of water = T2 - T1 = 34.0° C - 30.0 °C = 4.0 °C
m(tin) * 0.222 J/g°C * -46.0 °C = -100.0g* 4.184 J/g°C * 4.0 °C
m(tin) = 163.9 grams ≈ 164 grams
The mass of tin is 164 grams
Answer:
Compounds are represented by chemical formulas.
Explanation:
Answer:
N, O, R
Explanation:
Thanks for answering bro, you saved me
Answer: 27.09 ppm and 0.003 %.
First, <u>for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. </u>Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.
Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.
So, according to the <em>law of ideal gases,</em>
PV = nRT
where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)
The moles of CO will be,
n = 35 mg x
x
→ n = 0.00125 mol
We clear V from the equation and substitute P = 0.92 atm and
T = -30 ° C + 273.15 K = 243.15 K
V = 
→ V = 0.0271 L
As 1000 cm³ = 1 L then,
V = 0.0271 L x
= 27.09 cm³
<u>Then the acceptable concentration </u><u>c</u><u> of CO in ppm is,</u>
c = 27 cm³ / m³ = 27 ppm
<u>To express this concentration in percent by volume </u>we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:
c = 27.09
x
x 100%
c = 0.003 %
So, <u>the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is </u>27.09 ppm and 0.003 %.
This is an incomplete question, here is a complete question.
Manganese commonly occurs in nature as a mineral. The extraction of manganese from the carbonite mineral rhodochrosite, involves a two-step process. In the first step, manganese (II) carbonate and oxygen react to form manganese (IV) oxide and carbon dioxide:

In the second step, manganese (IV) oxide and aluminum react to form manganese and aluminum oxide:

Write the net chemical equation for the production of manganese from manganese (II) carbonate, oxygen and aluminum. Be sure your equation is balanced.
Answer : The net chemical equation for the production of manganese is:

Explanation :
The given two chemical reactions are:
(1) 
(2) 
First we are multiplying reaction 1 by 3, and reaction 2 by 2, we get:
(1)
(2) 
Now we are adding both the reactions, we get the overall chemical reaction.

The
is common on both side, by cancelling it, we get:
The net chemical equation for the production of manganese is:
