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2 years ago

6

Where am I missing an electron?

1 answer:

shtirl [24]2 years ago

4 0

There should be three electrons.
Four neutrons and four protons.

let me know if i helped or if I was wrong

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Please please please please help

deff fn [24]

I’ll give u my number and I’ll send u a pic of the answers

8 0

2 years ago

What is the mass of 2.64 moles of water

drek231 [11]

The mass of 2.64 moles of water is 47.56 grams

4 0

2 years ago

Solid NaI is slowly added to a solution that is 0.0079 M Cu+ and 0.0087 M Ag+.Which compound will begin to precipitate first?NaI

NeTakaya

Answer :

AgI should precipitate first.

The concentration of $Ag^+$ when CuI just begins to precipitate is, $6.64\times 10^{-7}M$

Percent of $Ag^+$ remains is, 0.0076 %

Explanation :

$K_{sp}$ for CuI is $1\times 10^{-12}$

$K_{sp}$ for AgI is $8.3\times 10^{-17}$

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgI has a smaller than CuI then AgI should precipitate first.

Now we have to calculate the concentration of iodide ion.

The solubility equilibrium reaction will be:

$CuI\rightleftharpoons Cu^++I^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Cu^+][I^-]$

$1\times 10^{-12}=0.0079\times [I^-]$

$[I^-]=1.25\times 10^{-10}M$

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

$AgI\rightleftharpoons Ag^++I^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ag^+][I^-]$

$8.3\times 10^{-17}=[Ag^+]\times 1.25\times 10^{-10}M$

$[Ag^+]=6.64\times 10^{-7}M$

Now we have to calculate the percent of $Ag^+$ remains in solution at this point.

Percent of $Ag^+$ remains = $\frac{6.64\times 10^{-7}}{0.0087}\times 100$

Percent of $Ag^+$ remains = 0.0076 %

8 0

2 years ago

What has to happen to join two monosaccharide units into a disaccharide?

omeli [17]

A condensation reaction forming a glycosidic bond. so in other words a monosaccharide joining together to form a disaccharide.

5 0

2 years ago

A mileage test is conducted for a new car model. Thirty randomly selected cars are driven for a month and the mileage is measure

Leni [432]

Answer: $(27.81,\ 29.39)$

Explanation:

Given : Sample size : n= 30 , it means it is a large sample (n≥ 30), so we use z-test .

Significance level : $\alpha: 1-0.95=0.05$

Critical value: $z_{\alpha/2}=1.96$

Sample mean : $\overline{x}=28.6$

Standard deviation : $\sigma=2.2$

The formula to find the confidence interval is given by :-

$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

i.e. $28.6\pm (1.96)\dfrac{2.2}{\sqrt{30}}$

i.e. $28.6\pm 0.787259889321$

$\approx28.6\pm 0.79=(28.6-0.79,28.6+0.79)=(27.81,\ 29.39)$

Hence, the 95% confidence interval for the mean mpg in the entire population of that car model = $(27.81,\ 29.39)$

4 0

2 years ago