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STALIN [3.7K]
3 years ago
9

PLS HELP

Chemistry
2 answers:
alexandr1967 [171]3 years ago
5 0

Answer:

Explanation:

A 12.48 g sample of an unknown metal, heated to 99.0 °C was then plunged into 50.0 mL of 25.0 °C water. The temperature of the water rose to 28.1 Go to calculating final temperature when mixing two samples of water ... Problem #1: A 610. g piece of copper tubing is heated to 95.3 °C and placed in an ... The two rings are heated to 65.4 °C and dropped into 12.4 mL of water at 22.3 °C. ... Problem #4: A 5.00 g sample of aluminum (specific heat capacity = 0.89 J g¯1

Simora [160]3 years ago
5 0
Someone put the answer btw
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3 years ago
Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.7 g of ethane is m
Bond [772]

Answer:

m_{H_2O}=4.86gH_2O

Explanation:

Hello,

In this case, the described chemical reaction is:

C_2H_6+\frac{7}{2} O_2\rightarrow 2CO_2+3H_2O

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

n_{C_2H_6}=2.7gC_2H_6*\frac{1molC_2H_6}{30gC_2H_6} =0.09molC_2H_6

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

n_{C_2H_6}^{consumed\ by \ O_2}=13.0gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6}{\frac{7}{2} molO_2}=0.116molC_2H_6

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

m_{H_2O}=0.09molC_2H_6*\frac{3molH_2O}{1molC_2H_6} *\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=4.86gH_2O

Best regards.

3 0
3 years ago
How many significant figures are in 1.00 L
likoan [24]

Answer:

3 significant numbers

6 0
3 years ago
Arrange the following oxides in order of increasing acidity.
zmey [24]

Answer:

Based on the Modern Periodic table, there is an increase in the electropositivity of the atom down the group as well as increases across a period. On comparing the electropositivities of the mentioned oxides central atom, it is seen that Ca is most electropositive followed by Al, Si, C, P, and S is the least electropositive.  

With the decrease in the electropositivity, there is an increase in the acidity of the oxides. Thus, the increasing order of the oxides from the least acidic to the most acidic is:  

CaO > Al2O3 > SiO2 > CO2 > P2O5 > SO3. Hence, CaO is the least acidic and SO3 is the most acidic.  

5 0
3 years ago
What is the final pressure of a system (atm) that has the volume increased from 0.75 l to 1.1 l with an initial pressure of 1.25
lana66690 [7]
Boyle's law states that pressure is inversely proportional to volume of gas at constant temperature 
PV = k
where P - pressure , V - volume and k - constant 
P1V1 = P2V2
where parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation 
substituting these values in the equation 
1.25 atm x 0.75 L = P x 1.1 L
P = 0.85 atm 
final pressure is B) 0.85 atm 
3 0
3 years ago
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