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2 years ago

PLEASE HELP, dana wants...

Chemistry

2 answers:

tensa zangetsu [6.8K]2 years ago

8 0

Answer:

Explanation:

umka21 [38]2 years ago

6 0

Answer:

I think A, not sure

Explanation:

her weight depends on the force of gravity

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Which element has the least density at STP?

vladimir2022 [97]

Answer:

Helium, this noble gas has the least density at STP

Explanation:

8 0

2 years ago

What is the percent composition of nitrogen in a 2.57 g sample of Al(NO3)3?

Lisa [10]

Answer:

19.8% of Nitrogen

Explanation:

In the Al(NO₃)₃ there are:

1 atom of Al

3 atoms of N

And 9 atoms of O

The molar mass of Al(NO₃)₃ is:

1 Al * (26.98g/mol) = 26.98g/mol

3 N * (14g/mol) = 42g/mol

9 O * (16g/mol) = 144g/mol

26.98 + 42 + 144 = 212.98g/mol

We can do a conversion using these molar masses to find the mass of nitrogen is the sample, that is:

2.57g * (42g/mol / 212.98g/mol) =

0.51g N

Percent composition of nitrogen is:

0.51g N / 2.57g * 100

= 19.8% of Nitrogen

6 0

2 years ago

Calculate the volume of a sample of mercury with a density of 14.6 g/mL and a mass of 1.00 g. The answer is assumed to be in mL.

Darya [45]

Answer:

0.0685 mL

Explanation:

To find the volume of the sample, divide the mass by the density.

(1.00 g)/(14.6 g/mL) = 0.0685 mL

3 0

3 years ago

TRIPLE POINTS!!! I NEED HELP!!! PLEASE EXPLAIN TOO!!! What is the energy of a quantum of light with a frequency of 4.31 x 1014 1

Lelechka [254]

Answer:

The answer is

$2.86 \times {10}^{ - 19} \: J$

Explanation:

The energy of a quantum of light can be found by using the formula

where

E is the energy

f is the frequency

h is the Planck's constant which is

6.626 × 10-³⁴ Js

From the question

f = 4.31 × 10¹⁴ Hz

We have

E = 4.31 × 10¹⁴ × 6.626 × 10-³⁴

We have the final answer as

$2.86 \times {10}^{ - 19} \: J$

Hope this helps you

3 0

2 years ago

Solid NaBr is slowly added to a solution that is 0.073 M in Cu+ and 0.073 M in Ag+.Which compound will begin to precipitate firs

saul85 [17]

Answer :

AgBr should precipitate first.

The concentration of $Ag^+$ when CuBr just begins to precipitate is, $1.34\times 10^{-6}M$

Percent of $Ag^+$ remains is, 0.0018 %

Explanation :

$K_{sp}$ for CuBr is $4.2\times 10^{-8}$

$K_{sp}$ for AgBr is $7.7\times 10^{-13}$

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

Now we have to calculate the concentration of bromide ion.

The solubility equilibrium reaction will be:

$CuBr\rightleftharpoons Cu^++Br^-$

The expression for solubility constant for this reaction will be,

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

$4.2\times 10^{-8}=0.073\times [Br^-]$

$[Br^-]=5.75\times 10^{-7}M$

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

$AgBr\rightleftharpoons Ag^++Br^-$

The expression for solubility constant for this reaction will be,

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

$7.7\times 10^{-13}=[Ag^+]\times 5.75\times 10^{-7}M$

$[Ag^+]=1.34\times 10^{-6}M$

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

Percent of $Ag^+$ remains = $\frac{1.34\times 10^{-6}}{0.073}\times 100$

Percent of $Ag^+$ remains = 0.0018 %

3 0

2 years ago

PLEASE HELP, dana wants...​

PLEASE HELP, dana wants...