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Burka [1]
2 years ago
10

Sample A: 300 mL of 1M sodium chloride

Chemistry
1 answer:
____ [38]2 years ago
7 0

Answer:

Sample A contains the smaller volume

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Using the following bond energies:
uranmaximum [27]

Answer:

The answer is "Option b".

Explanation:

Given equation and value:

\Rightarrow C_2H_2 \ (g)+ \frac{5}{2}O_2 \ (g) \longrightarrow 2CO_2\ (g)+H_2O\ (g)

\left\begin{array}{ccc}C \equiv  C&839\\C-H &413\\O=O &495\\ C=O &799\\ O-H &467 \end{array}\right \\\\

calculating equation value:

\Rightarrow 839 + 2(413) +\frac{5}{2} (495) \longrightarrow 2(2)799+2 \times 467 +E

\Rightarrow 839 + 826 +1237.5 = 3196+934 +E\\\\\Rightarrow 2902.5 =4130 +E\\\\\Rightarrow 2902 = 4130 +E\\\\\Rightarrow 2902 - 4130 +E\\\\\Rightarrow E = -1228 \ KJ \\\\

4 0
3 years ago
A 250-ml sample of na2so4 is reacted with an excess of bacl2. if 5.28 g baso4 is precipitated, what is the molarity of the na2so
Oksanka [162]
To get the molarity you need to follow this equation
                       moles of solute
Molarity (M = -----------------------
                        Liters of solution

But before you apply that equation you need to find the moles of solute and the liters of solution. Follow this equation

Na2SO4 + BaCl2 = BaSO4 + 2 NaCl

Solution

Moles of BaSO4 = 5.28 g 
                               ---------------
                                233.43 g / mol
                             =  0.0226  moles
Moles of NaSO4 = 0.0226 
                  0.0226 mole
Molarity = -----------------
                  0.250 L
              =  0.0905 mol / L

So the answer is 0.0905 mol / L
3 0
3 years ago
Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 1.30 MJ o
Sunny_sXe [5.5K]

Answer:

97 000 g Na

Explanation:

The absortion (or liberation) of energy in form of heat is expressed by:

q=m*Cp*ΔT

The information we have:

q=1.30MJ= 1.30*10^6 J

ΔT = 10.0°C = 10.0 K (ΔT is the same in °C than in K)

Cp=30.8 J/(K mol Na)

If you notice, the Cp in the question is in relation with mol of Na. Before using the q equation, we can find the Cp in relation to the grams of Na.

To do so, we use the molar mass of Na= 22.99g/mol

Cp= \frac{30.8J}{K*mol Na}*\frac{1 mol Na}{22.99 g Na}=1.34\frac{J}{K*g Na}

Now, we are able to solve for m:

m=\frac{1.30*10^6 J}{1.34\frac{J}{K*g Na} *10.0K}=\frac{1.30*10^6J}{13.4\frac{J}{g Na} }  = 9.70*10^4 g Na=97 Kg Na=97 000 g Na

7 0
3 years ago
Read 2 more answers
A. What is a period on the periodic table?
malfutka [58]

Answer:

In chemical bonding: Arrangement of the elements. The horizontal rows of the periodic table are called periods. Each period corresponds to the successive occupation of the orbitals in a valence shell of the atom, with the long periods corresponding to the occupation of the orbitals of a d subshell.

Explanation:

7 0
3 years ago
Read 2 more answers
PLEASE HELP!! 40 POINTS
PSYCHO15rus [73]

Answer:

1) 6.524779402×10^(-17)  

2)521.1g

3)113

Explanation:

5 0
3 years ago
Read 2 more answers
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