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natulia [17]
4 years ago
8

3x+5y=-16 -2x+6y=-36

Chemistry
1 answer:
saul85 [17]4 years ago
7 0
I believe the answer is 3,-5
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What is the mass of 0.500 moles of chlorine gas?
emmainna [20.7K]

Answer: m = 17.5 g Cl

Explanation: To find the mass of Cl we will need to convert the moles of Cl to mass using the relationship of 1 mole = molar mass of Cl.

0.500 mole Cl x 35 g Cl / 1 mole Cl

= 17. 5 g Cl

8 0
3 years ago
This is the chemical formula for talc Mg3(Si2O5)2(OH)2(the main ingredient in talcum powder):
Elena-2011 [213]

Answer:

0.022 mol O

Explanation:

Mg3(Si2O5)2(OH)2

We can see that 1 mol of this substance has 3 mol of Mg.

Oxygen altogether  is 5*2 (from (Si2O5)2) + 2(from(OH)2) = 10 +2 = 12

So, 1 mol of this substance has 12 mol oxygen.

So,  1 mol of this substance contains 3 mol Mg and  12 mol O, or

ratio Mg : O = 3 : 12 = 1 : 4

1 mol Mg ----- 4 mol O

0.055 mol Mg ---x mol O

x = 0.055*4/1 = 0.220 mol O

8 0
4 years ago
Material safety data sheets are used to give the information about a solvent. The excerpt below is part of an MSDS. Which inform
Kitty [74]

Answer:

the answer is A

Explanation:

hope this helps

7 0
3 years ago
Read 2 more answers
PLEASE HELPPPP! HOW DO I DO THIS???
kow [346]

Answer:

1552.83J Released

Explanation:

1. mass/m=225

Initial temp:86C, final:32.5C

Changed Temp: 32.5-86= -53.5C

s=0.129 J/gC

Formula: q= m times s times changed Temp.

q=(225)(0.129)(-53.5)

q= -1552.83 J

q=1552.83 J Released

8 0
3 years ago
Integrated rate law for second order unimolecular irreversible
kirill115 [55]

Answer:

The rate law for second order unimolecular irreversible reaction is

\frac{1}{[A]} = k.t + \frac{1}{[A]_{0} }

Explanation:

A second order unimolecular irreversible reaction is

2A → B

Thus the rate of the reaction is

v = -\frac{1}{2}.\frac{d[A]}{dt} = k.[A]^{2}

rearranging the ecuation

-\frac{1}{2}.\frac{k}{dt} = \frac{[A]^{2}}{d[A]}

Integrating between times 0 to <em>t </em>and between the concentrations of [A]_{0} to <em>[A].</em>

\int\limits^0_t -\frac{1}{2}.\frac{k}{dt} =\int\limits^A_{0} _A\frac{[A]^{2}}{d[A]}

Solving the integral

\frac{1}{[A]} = k.t + \frac{1}{[A]_{0} }

5 0
3 years ago
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