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irina [24]
3 years ago
11

Describe the changes to the coastline pictured above. Look particularly at North and South Monomy Island.

Chemistry
1 answer:
s344n2d4d5 [400]3 years ago
3 0

             Where is the picture?                                                                         

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At 741 torr and 44°C, 7.10 g of a gas occupy a volume of 5.40 L. What is the molar mass
amm1812

The molar mass of the gas is 35. glmol.

the molar mass of a chemical compound is described as the mass of a sample of that compound separated by the amount of substance in that sample, measured in moles.

The molar mass is a bulk, not molecular, effects of a substance.

<h3>What is molar mass and how is it calculated?</h3>

The molar mass is the mass of one mole of a sampling. To find the molar mass, count the atomic masses (atomic weights) of all of the atoms in the molecule. Find the atomic mass for each element by using the mass shown in the Periodic Table or table of atomic weights.

<h3>What is molar ?</h3>

Molar refers to the unit of concentration molarity, which is equivalent to the number of moles per liter of a solution. In chemistry, the word most often guides to molar concentration of a solute in a solution. Molar attention has the units mol/L or M.

To learn more about the molar mass, refer

brainly.com/question/15476873

#SPJ9

4 0
2 years ago
Ionic formula for sodium oxide
Bas_tet [7]
The ionic formula of sodium oxide would be Na20
5 0
3 years ago
What is the melting point of a solution in which 2.5 grams of sodium chloride is added to 230 mL of water?
Zina [86]
0.976 should be the answer. 
5 0
3 years ago
Read 2 more answers
If 30.0 mL of Ca(OH)2 with an unknown concentration is neutralized by 40.0 mL of 0.175 M HCl, what is the concentration of the C
Over [174]

Answer:

Concentration of Ca(OH)₂:

0.117 M.

Explanation:

How many moles of HCl is consumed?

Note the unit of concentration: moles per liter solution.

c(\text{HCl}) = 0.175\;\text{M} = 0.175\;\text{mol}\cdot\textbf{L}^{-1}.

Convert milliliters to liters.

V(\text{HCl})=40.0\;\text{mL} = 0.0400\;\text{L}.

n(\text{HCl}) = c(\text{HCl})\cdot V(\text{HCl})= 0.175\;\text{mol}\cdot\text{L}^{-1} \times 0.0400\;\text{L}= 7.00\times 10^{-3}\;\text{mol}.

How many moles of NaOH in the solution?

Refer to the equation. The coefficient in front of Ca(OH)₂ is 1. The coefficient in front of HCl is 2. In other words, it takes two moles of HCl to neutralize one mole of Ca(OH)₂. That 7.00\times 10^{-3}\;\text{mol} of HCl will neutralize only half that much Ca(OH)₂.

\displaystyle n(\text{Ca}(\text{OH})_2)=\frac{1}{2}\;n(\text{HCl}) = 3.50\times 10^{-3}\;\text{mol}.

What's the concentration of the Ca(OH)₂ solution?

Concentration is the number of moles of solute per unit volume.

\displaystyle c(\text{Ca}(\text{OH})_2) = \frac{n(\text{Ca}(\text{OH})_2)}{V(\text{Ca}(\text{OH})_2)} = \frac{3.50\times 10^{-3}\;\text{mol}}{0.0300\;\text{L}}=0.117\;\text{mol}\cdot\text{L}^{-1}.

3 0
3 years ago
A student has a mixture of salt (NaCl) and sugar (C12H22O11). To determine the percentage, the student measures out 5.84 grams o
julsineya [31]

Answer:

The volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml

Explanation:

Here we have the reaction of AgNO₃ and NaCl as follows;

AgNO₃(aq) + NaCl(aq)→ AgCl(s) + NaNO₃(aq)

Therefore, one mole of silver nitrate, AgNO₃, reacts with one mole of sodium chloride, NaCl, to produce one mole of silver choride, AgCl, and one mole of sodium nitrate, NaNO₃,

Therefore, since 5.84 grams of NaCl which is 58.44 g/mol, contains

Number \, of \, moles, n  = \frac{Mass}{Molar \ mass} =  \frac{5.84}{58.44} = 0.09993 \ moles \ of \ NaCl

0.09993 moles of NaCl will react with 0.09993 moles of AgNO₃

Also, as 1.0 M solution of AgNO₃ contains 1 mole per 1 liter or 1000 mL, therefore, the volume of AgNO₃ that will contain 0.09993 moles is given as follows;

0.09993 × 1 Liter/mole= 0.09993 L = 99.93 mL

Therefore, the volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml.

3 0
4 years ago
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