Answer:
answer is option a..............
 
        
                    
             
        
        
        
In Na2O, what is the oxidation state of oxygen? In Na2O oxidation state of Na is 1+
        
                    
             
        
        
        
Answer:
THE SPECIFIC HEAT OF THE ALLOY IS 0.9765 J/g K
Explanation:
Mass of alloy = 33 g
Initial temperature of alloy = 93°C
Mass of water = 50 g
Initail temp. of water = 22 °C 
Heat capacity of calorimeter = 9.20 J/K
Final temp. = 31.10 °C 
specific heat of alloy = unknown
specific heat capacity of water = 4.2 J/g K
Heat = mass * specific heat * change in temperature = m c ΔT
Heat = heat capcity * chage in temperature = Δ H * ΔT
In calorimetry;
Heat lost by the alloy = Heat gained by water + Heat of the calorimeter
                      mc ΔT = mcΔT + Heat capacity * ΔT
33 * C * ( 93 - 31.10) = 50 * 4.2 * ( 31.10 -22) + 9.20 * ( 31.10 -22)
33 * C * 61.9 = 50 * 4.2 * 9.1 + 9.20 * 9.1
2042.7 C = 1911 + 83,72
C = 1911 + 83.72 / 2042.7
C = 1994.72 /2042.7
C =0.9765 J/g K
The specific heat of the alloy is 0.9765 J/ g K
 
        
             
        
        
        
Answer:
sodium bromide (NaBr) potassium hydroxide (KOH) magnesium chloride (MgCl2) silicon dioxide (SiO2) sodium oxide (Na2O)
Explanation:
 
        
             
        
        
        
I could only find 7! 
- independent variable 
- dependent variable 
- control group
- experimental group 
- constant 
- observation 
- inference