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Papessa [141]
3 years ago
9

por favor es urgente!!!!!!!!.... 1 -¿cuántos átomos- gramos de cloro hay en 88,75 g de dicho elemento? 2-¿calcular los moles de

níquel en 18,6g de este metal y los átomos que contiene? 3-¿ determinar el peso de 3,5 moles de HNO? 4-¿hallar ¿cuántas moléculas habrá en 20g de H2 SO4? 5-¿a cuántas moléculas equivale 8 moles de cloruro de sodio?
Chemistry
1 answer:
Mamont248 [21]3 years ago
6 0

Answer:

don't understand your language

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What is the <br> formula for gunpowder
mrs_skeptik [129]

Here is the chemical equation for gun powder, in it’s simple form:

<span>2 KNO3 + S + 3 C → K2S + N2 + 3 CO2.</span>

This is the same simplified formula, only balanced:

<span>10 KNO3 + 3 S + 8 C → 2 K2CO3 + 3 K2SO4 + 6 CO2 + 5 N2.</span>

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What would happen to the pressure if the volume were reduced to 0.5 l and the temperature increased to 260 ∘c?
Vsevolod [243]
<span>The ideal gas law. PV=nRT pressure x volume = moles x Faraday's constant x Temp Kelvin (C+273) Original data Pressure 1 atmosphere Volume 1 liter Temp 25C = 298K New data Volume 0.5 liter pressure X Temp 260C = 533K P1v1T1 = P2v2T2 plug and chug. (1)(1)(293) = (x)(0.5)(533) Solve for X, which is the new pressure. </span>
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What is the electron arrangement of bromine
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4 years ago
The total bonding energy for the products of a reaction is 2535 kJ/mol and the bonding energy of the reactants is 1375 kJ/mol. C
Dmitry [639]

Answer:

-1160kj/mol

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7 0
3 years ago
n unknown metal is either aluminum, iron or lead. If 150. g of this metal at 150.0 °C was placed in a calorimeter that contains
Nitella [24]

Answer : The metal used was iron (the specific heat capacity is 0.449J/g^oC).

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of unknown metal = ?

c_2 = specific heat of water = 4.184J/g^oC

m_1 = mass of unknown metal = 150 g

m_2 = mass of water = 200 g

T_f = final temperature of water = 34.3^oC

T_1 = initial temperature of unknown metal = 150.0^oC

T_2 = initial temperature of water = 25.0^oC

Now put all the given values in the above formula, we get

150g\times c_1\times (34.3-150.0)^oC=-200g\times 4.184J/g^oC\times (34.3-25.0)^oC

c_1=0.449J/g^oC

Form the value of specific heat of unknown metal, we conclude that the metal used in this was iron (Fe).

Therefore, the metal used was iron (the specific heat capacity is 0.449J/g^oC).

6 0
4 years ago
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