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Reptile [31]
4 years ago
6

What is the mass of silver metal produced from 6.35 g of copper when copper reacts with excess silver nitrate in a single replac

ement reaction? ____ g?
Chemistry
1 answer:
frutty [35]4 years ago
8 0
<span>21.6 grams The balanced equation for the reaction is Cu + 2AgNO3 ==> 2Ag + Cu(NO3)2 So you can easily determine that for each mole of copper used, 2 moles of silver are produced. So let's lookup the atomic weights of copper and silver. Atomic weight copper = 63.546 Atomic weight silver = 107.8682 Moles copper = 6.35 g / 63.546 g/mol = 0.099927611 mol Since we'll produce 2 moles of silver per mole of copper used, we get 0.099927611 mol * 2 = 0.199855223 mol And to get the mass, just multiply by the atomic weight of silver. So 0.199855223 * 107.8682 = 21.55802316 g Finally, round to 3 significant figures, giving 21.6 g.</span>
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You dig a small hole in the soil and the next day it rains. The hole fills with water but does not
MakcuM [25]
The soil is mostly made of clay
3 0
3 years ago
Calculate the Molar Mass in g/mol of Ammonium Sulfate, (NH4)2(SO4)
saw5 [17]

Answer:

132g/mol

Explanation:

The problem here is to find the molar mass of the compound in g/mol

To solve this problem, we simply sum the atomic masses of the atoms in the compound:

   (NH₄)₂ SO₄:

  Atomic mass of N  = 14g

   Atomic mass of H  = 1g

  Atomic mass of S  = 32g

  Atomic mass of O = 16g

So;

   Molar mass = 2[(14 + 4(1))] + 32 + 4(16)  = 132g/mol

6 0
3 years ago
Name three examples of how the atom has changed since Rutherford's discovery of the nucleus .
NikAS [45]

Answer:

Rutherford overturned Thomson's model in 1911 with his well-known gold foil experiment, in which he demonstrated that the atom has a tiny, high- mass nucleus. In his experiment, Rutherford observed that many alpha particles were deflected at small angles while others were reflected back to the alpha source.

8 0
3 years ago
Cual es el calor específico de una sustancia que tiene una masa de 15 g y una capacidad térmica de 0.550 calC
mr Goodwill [35]

Answer:

Explanation:

Q = Ce . m .ΔT

Q : calor  

Ce : calor especifico

m: masa  

ΔT : variación de temperatura

capacidad térmica : 0,550 cal / °C

lectura :

por 1 °C se tiene 0,550 cal

por lo tanto  tenemos datos de la temperatura y del calor

pero no olvidar las unidades en el sistema internacional :

Ce : J / kg . K

J: joules

kg: kilogramo

K: kelvin  

pasar de gramos a kilogramos

pasar de calorías a joules

pasar de grado celsius a kelvin

1000g equivale a 1kg

15g     equivale a 0,015 kg

K= °C + 273 ⇒ formula para pasar de grado celsius a kelvin

K=  1 + 273

K= 274

1 caloría equivale a  4,184 joules

0,550 caloría  equivale a 2,3012 joules

ahora como todos los datos ya están en el S.I remplazamos en la formula

Q = Ce . m .ΔT

2,3012 = Ce . 0,015.274

Ce=0,5599 J / kg. K

6 0
3 years ago
How many moles are dissolved in 1.5 L of a 5.7 M solution?
Scrat [10]

Answer:

8.55 moles

Explanation:

The formula for calculating molarity is moles of solute divided by liters of solution. In this case:

5.7=\frac{x}{1.5}

x=5.7\cdot 1.5=8.55 moles. Hope this helps!

4 0
3 years ago
Read 2 more answers
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