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weeeeeb [17]
3 years ago
13

I am stuck... Chemistry is the worst and anything is appreciated!

Chemistry
2 answers:
Soloha48 [4]3 years ago
8 0
I think it is 110 mL but I’m not totally sure.
const2013 [10]3 years ago
7 0

Answer:

2.78 mL of solution will be needed to make 4.5 M solution with 0.50 grams of NaOH.

Explanation:

Moles of sodium hydroxide = n

n=\frac{0.50 g}{40 g/mol}=0.0125 mol

Moles (n)=Molarity(M)\times Volume (L)

Volume of sodium hydroxide solution =V

Molarity of the sodium nitrate solution = 4.5 M

V=\frac{0.0125 mol}{4.5 M}=0.00278 L

V  = 0.00278 L = 2.78 mL

2.78 mL of solution will be needed to make 4.5 M solution with 0.50 grams of NaOH.

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Which of the following statements is true? Oceanic crust is thicker than continental crust. Both oceanic and continental crust i
zhuklara [117]
The crust is the thickest layer of the Earth.

3 0
3 years ago
national defense, currency, post office, foreign affairs, & interstate commerce. b. charter local governments, education, pu
katen-ka-za [31]

Answer And Explanation:

Option C is correct.

Lend and borrow money, taxation, law enforcement, charter banks and transportation.

Some of the powers that were mentioned in the other options that weren't concurrent powers (that is, they belong to either the state government alone or the federal government alone) & disqualified them from being the answer include:

National defence (federal), Currency (federal), foreign affairs (federal), intrastate commerce (state) etc.

3 0
3 years ago
Which type of bond is present in hydrogen sulfide (H2S)? The table of electronegativities is given.
Paraphin [41]

nonpolar covalent bond (with LD force)

6 0
3 years ago
The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151
vitfil [10]

Answer:-  The natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for ^1^5^1_E_u and ^1^5^3_E_u as 150.919860 and 152.921243 amu, respectively.  Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of ^1^5^1_E_u as n then the abundance of ^1^5^3_E_u would be 1-n .

Let's plug in the values in the formula:

151.964=150.919860(n)+152.921243(1-n)

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

151.964-152.921243=150.919860n-152.921243n

-0.957243=-2.001383n

negative sign is on both sides so it is canceled:

0.957243=2.001383n

n=\frac{0.957243}{2.001383}

n=0.478

The abundance of ^1^5^1_E_u is 0.478 which is 47.8%.  

The abundance of ^1^5^3_E_u is = 1-0.478

= 0.522 which is 52.2%

Hence, the natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .


3 0
3 years ago
11. What is the concentration (in M) of an ammonia (NH3) solution if 12.23 grams of ammonia are
photoshop1234 [79]

Answer:  1.284M NH3

Explanation:  (12.23 grams)/(17.0 gramms/mole) = 0.7191 moles

Dissolved in 560.0 ml (=0.5600L)

(0.7191 moles)/(0.560L) = 1.284M  (4 sig figs)

5 0
3 years ago
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