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3 years ago

An -ate or -ite at the end of a compound name ususally indiates that the compound contains...

Chemistry

1 answer:

Zarrin [17]3 years ago

4 0

Answer: Oxygen.

Explanation: The -ate is used for the ion that has the largest number of Oxygen atoms. The -ite would be used for the ion with the smaller amount of oxygen atoms.

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The rainforest contains what percentage of all of the species in the world?

Anuta_ua [19.1K]

Answer:

I think 50%

Explanation:

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3 years ago

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The maximum amount of nickel(II) cyanide that will dissolve in a 0.220 M nickel(II) nitrate solution is...?

sweet [91]

Answer : The maximum amount of nickel(II) cyanide is $5.84\times 10^{-12}M$

Explanation :

The solubility equilibrium reaction will be:

$Ni(CN)_2\rightleftharpoons Ni^{2+}+2CN^-$

Initial conc. 0.220 0

At eqm. (0.220+s) 2s

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ni^{2+}][CN^-]^2$

Now put all the given values in this expression, we get:

$3.0\times 10^{-23}=(0.220+s)\times (2s)^2$

$s=5.84\times 10^{-12}M$

Therefore, the maximum amount of nickel(II) cyanide is $5.84\times 10^{-12}M$

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3 years ago

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Troyanec [42]

Answer: Hope This Helps

Explanation:

1: The nephrons contain a filter called the glomerulus. The fluid that is filtered from the blood then travels down through a tubule. The tubule adjusts the level of wastes that will leave the body in the urine.

(I dont know the second part so hope this half answer helps a little bit. Sorry.)

8 0

3 years ago

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Which type of ions do metals form?

alexandr1967 [171]

Answer:

metal atoms lose electrons to form positively charged ions. non-metal atoms gain electrons to form negatively charged ions

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3 years ago

g How many moles of NaOH are present in a sample if it is titrated to its equivalence point with 44.02 mL of 0.0885 M H2SO4? 2Na

Lisa [10]

Answer:

$n_{base}=3.90x10^{-3}molNaOH$

Explanation:

Hello!

In this case, since the sulfuric acid and sodium hydroxide react in a 1:2 mole ratio, given the reaction, we realize they have the following mole ratio at the equivalence point:

$2*n_{acid}=n_{base}$

Which in terms of concentrations and volumes is:

$2*M_{acid}V_{acid}=n_{base}$

Thus, we can plug in the volume and concentration of acid to find the moles of base:

$n_{base}=0.04402L*0.0885\frac{mol}{L} \\\\n_{base}=3.90x10^{-3}molNaOH$

Best regards!

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3 years ago