NaCl+AgC2H3O2--> NaC2H3O2+AgCl what is this balanced

30g of naoh is dissolved in 1.5 liter solution the active mass of naoh is?

Pepsi [2]

Answer:

0.5mol/L

Explanation:

First, let us calculate the number of mole NaOH = 23 + 16 + 1 = 40g/mol

Mass of NaOH from the question = 30g

Number of mole = Mass /Molar Mass

Number of mole = 30/40 = 0.75mol

Volume = 1.5L

Active mass = mole/Volume

Active mass = 0.75mol/1.5L

Active mass = 0.5mol/L

7 0

2 years ago

What is the electron configuration for a calcium atom? 01s². 2s 2.2p5.3s 2.3p 5.4s¹ 02s², 2s ², 2p5.3s 2.3p 6,4s1 01s 2. 2s 2. 2

Wewaii [24]

Answer:

Ca = 1s²2s²2p⁶3s²3p⁶4s²

Explanation:

Calcium is located in the second column and the 4th row of the periodic table. Thus, there are 2 valence electrons and the highest quantum number is 4. This is represented by 4s². However, all of the previous orbitals and electrons must be listed beforehand, making the entire electron configuration: 1s²2s²2p⁶3s²3p⁶4s².

7 0

1 year ago

A mixture of caco3 and (nh4)2co3 is 60.7 % co3 by mass. part a find the mass percent of caco3 in the mixture.

blagie [28]

The mixture contains:

CaCO3 + (NH4)2CO3 in which the amount of carbonate CO3 = 60.7% by mass

Let, the total mass = 100 grams

Mass of CaCO3 = x grams

Mass of (NH4)2CO3 = y grams

Thus, x + y = 100 ------------(1)

Mass of CO3 = 60.7% = 60.7 g

Molar mass of CO3 = 60 g/mol

Total # moles of CO3 = 60.7 g/60 g.mol-1 = 1.012 moles

The total moles of CO3 comes from CaCO3 and (NH4)2CO3. Therefore,

moles CaCO3 + moles (NH4)2CO3 = 1.012

mass CaCO3/molar mass CaCO3 + mass (NH4)2 CO3/molar mass = 1.012

x/100 + y/96 = 1.012---------(2)

based on equation 1 we can write: y = 100-x

x/100 + (100-x)/96 = 1.012

x = 71.2 g

Mass of CaCO3 = 71.2 g

7 0

2 years ago

During the following chemical reaction, 46.3 grams of C3H6O react with 73.2 grams of O2

ra1l [238]

Answer:

a) O2 is the limiting reactant

b) 75.70 grams CO2 (theoretical yield)

c) There remains 12.81 grams of C3H6O

d) The actual yield CO2 is 34.29 grams

Explanation:

Step 1: Data given

Mass of C3H6O = 46.3 grams

Mass of O2 = 73.2 grams

Molar mass of C3H6O = 58.08 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

C3H6O + 4O2 → 3 CO2 + 3H2O

Step 3: Calculate moles C3H6O

Moles C3H6O = mass C3H6O / molar mass C3H6O

Moles C3H6O = 46.3 grams / 58.08 g/mol

Moles C3H6O = 0.793 moles

Step 4: Calculate moles O2

Moles O2 = 73.2 grams / 32 g/mol

Moles O2 = 2.29 moles

Step 5: Calculate limiting reactant

For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O

O2 is the limiting reactant. It will completely be consumed. (2.29 moles).

C3H6O is in excess. There will react 2.29/4 = 0.5725 moles C3H6O

There will remain 0.793 - 0.5725 = 0.2205 moles C3H6O

This is 0.2205 moles * 58.08 g/mol = 12.81 grams

Step 6: Calculate moles of CO2

For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O

For 2.29 moles O2 we need 3/4 * 2.29 = 1.72 moles CO2

This is 1,72 moles * 44.01 g/mol = 75.70 grams CO2

Step 7: Calculate actual yield

% yield = 45.3 % = 0.453 = (actual yield / theoretical yield)

actual yield = 0.453 * 75.70 = 34.29 grams

3 0

2 years ago

PLS HELP ASAP I WILL GIVE BRAINLIEST FOR THE ANSWER THAT IS RIGHT!Based on the diagram below, in what period is the atom located

katen-ka-za [31]

Answer:

6

Explanation:

im guessing because there is no diagram

3 0

1 year ago