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nadezda [96]
3 years ago
15

How does government ensure a safe environment for inverters? ​

Chemistry
1 answer:
Anestetic [448]3 years ago
7 0

hope it helps please brainliest

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At a certain temperature the rate of this reaction is first order in SO₃ with a rate constant of 0.208 s⁻¹:
NNADVOKAT [17]

Answer:

0.30 M

Explanation:

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t  = ?

[A_0] is the initial concentration  = 1.36 M

k is the rate constant = 0.208 s⁻¹

t = 7.30 seconds

So,  

[A_t]=1.36\times e^{-0.208\times 7.30}\ M=0.30\ M

8 0
3 years ago
Calculate the empirical formula for DCBN, given that the percent composition is 48.9%C, 1.76%H, 41.2%Cl, and 8.15%N.
sweet [91]

Answer:

empirical formula is C7H3NCl2

Explanation:

too much work too explain and im lazy

5 0
3 years ago
Gaseous methane (CH4) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO) and gaseous water (H2O) . Suppo
Leto [7]

Answer:

0 g.

Explanation:

Hello,

In this case, since the reaction between methane and oxygen is:

CH_4+2O_2\rightarrow CO_2+2H_2O

If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:

n_{CH_4}=0.963gCH_4*\frac{1molCH_4}{16gCH_4}=0.0602molCH_4\\ \\n_{CH_4}^{consumed}=7.5gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2} =0.117molCH_4

Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.

left\ over=0g

Regards.

7 0
3 years ago
The following reaction is a _______ reaction.
jasenka [17]
Compound reaction :)
7 0
3 years ago
Does anyone know how to do this? the solubility of barium carbonate, BaCO3, is 0.0100 g/L. Its molar mass is 197.3 g/mol. What i
g100num [7]
It's simple, just follow my steps.

1º - in 1 L we have 0.0100~g of BaCO_3

2º - let's find the number of moles.

\eta=\frac{m}{MM}

\eta=\frac{0.0100}{197.3}

\boxed{\boxed{\eta=5.07\times10^{-5}~mol}}

3º - The concentration will be

C=5.07\times10^{-5}~mol/L

But we have this reaction

BaCO_3\rightleftharpoons Ba^{2+}+CO_3^{2-}

This concentration will be the concentration of Ba^{2+}~~and~~CO_3^{2-}

K_{sp}=\frac{[Ba^{2+}][CO_3^{2-}]}{[BaCO_3]}

considering [BaCO_3]=1~mol/L

K_{sp}=[Ba^{2+}][CO_3^{2-}]

and

[Ba^{2+}]=[CO_3^{2-}]=5.07\times10^{-5}~mol/L

We can replace it

K_{sp}=(5.07\times10^{-5})*(5.07\times10^{-5})

K_{sp}\approx25.70\times10^{-10}

Therefore the K_{sp} is:

\boxed{\boxed{\boxed{K_{sp}\approx2.57\times10^{-11}}}}
4 0
3 years ago
Read 2 more answers
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