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Fudgin [204]
3 years ago
5

How do i know if a reaction is exo/endothermic?

Chemistry
1 answer:
oee [108]3 years ago
8 0

Answer:

You would if a reaction is exo/endothermic because an endothermic reaction soaks up heat. An exothermic reaction releases heat.

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A powdered drink mix is stirred into a glass of water to form a solution. If any more is added it will not dissolve
Brut [27]
SaturTED, hope this helps, m8
6 0
3 years ago
A chemist makes up a solution by dissolving 42.0 g of Mg(NO3)2 in enough water to produce a final solution volume of 259 mL. To
Igoryamba

Answer:

The molar mass of Mg(NO₃)₂, 148.3 g/mol.

Explanation:

Step 1: Given data

  • Mass of Mg(NO₃)₂ (solute): 42.0 g
  • Volume of solution: 259 mL = 0.259 L

Step 2: Calculate the moles of solute

To calculate the moles of solute, we need to know the molar mass of Mg(NO₃)₂, 148.3 g/mol.

42.0 g × 1 mol/148.3 g = 0.283 mol

Step 3: Calculate the molarity of the solution

M = moles of solute / liters of solution

M = 0.283 mol / 0.259 L

M = 1.09 M

7 0
2 years ago
Day and night are caused by what
Aleks04 [339]

Answer:

THE GOD ALMIGHTY!

Explanation:

6 0
3 years ago
What is the percent yield of CuS for the following reaction given that you start with 15.5 g of Na2S and 12.1 g CuSO4? The actua
notka56 [123]

Answer:

(B) 42.1%

Explanation:

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Given: For Na_2S

Given mass = 15.5 g

Molar mass of Na_2S = 78.0452 g/mol

<u>Moles of Na_2S = 15.5 g / 78.0452 g/mol = 0.1986 moles</u>

Given: For CuSO_4

Given mass = 12.1 g

Molar mass of CuSO_4 = 159.609 g/mol

<u>Moles of CuSO_4 = 12.1 g / 159.609 g/mol = 0.0758 moles</u>

According to the given reaction:

Na_2S+CuSO_4\rightarrow Na_2SO_4+CuS

1 mole of Na_2S react with 1 mole of CuSO_4

0.1986 mole of Na_2S react with 0.1986 mole of CuSO_4

Available moles of CuSO_4 = 0.0758 moles

Limiting reagent is the one which is present in small amount. Thus, CuSO_4 is limiting reagent. (0.0758 < 0.1986)

The formation of the product is governed by the limiting reagent. So,

1 mole of CuSO_4 gives 1 mole of CuS

0.0758 mole of CuSO_4 gives 0.0758 mole of CuS

Molar mass of CuS = 95.611 g/mol

Mass of CuS = Moles × Molar mass = 0.0758 × 95.611 g = 7.2473 g

Theoretical yield = 7.2473 g

Given experimental yield = 3.05 g

% yield = (Experimental yield / Theoretical yield) × 100 = (3.05/7.2473) × 100 = 42.1 %

<u>Option B is correct.</u>

6 0
3 years ago
7.26 of a hydrate, Cu(NO3)2.xH2O, formed 2.4 g copper(II) oxide.
goldfiish [28.3K]

Number of moles= mass/ molar mass

Or n=m/MM

n = number of moles

m = mass

MM = molar mass

1) n CuO = 2.4g / 79.54g/mol = 0.03 mol CuO

2) n Cu(NO3)2.xH2O = 7.26 g / 205.6 = 0.035 moles of Cu(NO3)2.xH2O

3) 205.6 g

Cu = 63.5 g

N = 14g

O =16g

H= 1 g

63.5+ (14+(16*3))*2+1*2+16 =205.6 g

4) yes is 188g

5) I don’t know, I assume was 1

6 0
3 years ago
Read 2 more answers
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