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NISA [10]
3 years ago
7

Can you please help me with this question

Chemistry
1 answer:
myrzilka [38]3 years ago
4 0

Answer:

First blank: spiral arms

Second blank: center

Explanation:

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Dahasolnce [82]

Answer:0,25 g/cm3 creo

Explanation:densidad es masa entre volumen

5 0
3 years ago
Read 2 more answers
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
Calculate the maximum volume (in mL) of 0.143 M HCl that each of the following antacid formulations would be expected to neutral
Nuetrik [128]

Answer:

a. The maximum volume of 0.143 M HCl required is 154.4 mL.

b. The maximum volume of 0.143 M HCl required is 135.7 mL.

Explanation:

a.

Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O

Mass of aluminum hydroxide = 350 mg =  0.350 g ( 1mg = 0.001 g)

Moles of aluminum hydroxide = \frac{0.350 g}{78 g/mol}=0.004487 mol

According to reaction ,3 moles of HCl neutralize 1 mole of aluminum hydroxide.Then 0.004487 mole of aluminum hydroxide will be neutralize by :

\frac{3}{1}\times 0.004487 mol=0.01346 mol of HCl.

Mg(OH)_2+2HCl\rightarrow MgCL_2+2H_2O

Mass of magnesium hydroxide = 250 mg =  0.250 g ( 1mg = 0.001 g)

Moles of magnesium hydroxide = \frac{0.250 g}{58 g/mol}=0.004310 mol

According to reaction ,2 moles of HCl neutralize 1 mole of magnesium hydroxide.Then 0.004310  mole of magnesium hydroxide will be neutralize by :

\frac{2}{1}\times 0.004310 mol=0.008621 mol of HCl.

Total moles of HCl required to neutralize both :

0.01346 mol + 0.008621 mol = 0.02208 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

Molarity=\frac{\text{Total moles of HCl}{\text{Volume in Liter}}

V=\frac{0.02208 mol}{0.143 M}=0.1544 L

1 L = 1000 mL

0.1544 L = 154.4 mL

The maximum volume of 0.143 M HCl required is 154.4 mL.

b.

CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2

Mass of calcium carbonate = 970mg =  0.970 g ( 1mg = 0.001 g)

Moles of calcium carbonate = \frac{0.970 g}{100 g/mol}=0.00970 mol

According to reaction ,2 moles of HCl neutralize 1 mole of calcium carbonate.Then 0.00970 mole of calcium carbonate will be neutralize by :

\frac{2}{1}\times 0.00970 mol=0.0194 mol of HCl.

Total moles of HCl required to neutralize calcium carbonate : 0.0194 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

Molarity=\frac{\text{Total moles of HCl}}{\text{Volume in Liter}}

V=\frac{0.0194 mol}{0.143 M}=0.1357 L

1 L = 1000 mL

0.1357 L = 135.7 mL

The maximum volume of 0.143 M HCl required is 135.7 mL.

4 0
3 years ago
.
Darina [25.2K]

Answer:

C

Explanation:

4× CH2 = C4H8

7 0
2 years ago
What is the chemical symbol of an element
11Alexandr11 [23.1K]

A chemical symbol is a shorthand method of representing an element. Instead of writing out the name of an element, we represent an element name with one or two letters. As you know, the periodic table is a chemist's easy reference guide. ... Each element is represented by a chemical symbol consisting of letters

3 0
3 years ago
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