An unknown metal is very reactive; it reacts violently with water and must be stored in a moisture-free environment. It is solid at room temperature. Which group does the metal likely belong to?

Answers

Answer 1

Answer:

This Metal belongs to Group I (Alkali Metals).

Explanation:

Alkali Metals present in Group I are considered as the most reactive metals in periodic table. There reactivity increases tremendously down the group. The reactivity is mainly due to less ionization energies. Therefore, going from top to bottom along the group the ionization energies decreases hence, increasing there reactivity respectively.

Alkali metals when reacted with water undergoes an exothermic reaction resulting the formation of corresponding metal hydroxide and hydrogen gas i.e.

2 M + 2 H₂O → 2 MOH + H₂

Also, these metals are in solid state at room temperature (i.e. 25 °C) and their boiling points are as follow,

Lithium = 180.5 °C

Sodium = 97.79 °C

Potassium = 63.5 °C

Rubidium = 39.48 °C

Cesium = 28.44 °C

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Given: Z = 0.43Y + 12 ; What is Y when Z = 28 ?
Can someone explain this one?

Answers

Z=0.43y+12=28=0.43y+12=28-12=0.43y=16=0.43y,then u divide 0.43y by itself and by 16,you will get y=37.209

Which statement must be true for any chemical reaction at equilibrium?(1) The concentration of the products is greater than the concentration of the reactants.
(2) The concentration of the products is less than the concentration of the reactants.
(3) The concentration of the products and the concentration of the reactants are equal.
(4) The concentration of the products and the concentration of the reactants are constant.

Answers

The correct answer is 3. The concentration of the product and the concentration of the reactants are equal. That is because matter cannot be created or destroyed and the product cannot randomly become something else in the reaction, nor can a reactant change into something different. Simply put, you cannot turn things like water into gold because there will always be an equilibrium.

Answer:

Answer is 4 The concentration of the products and the concentration of the reactants are constant.

Explanation:

What ion must be present for a compound to be considered a base?

Answers

it is OH- (hydroxide ion)

A helium-ﬁlled balloon contains 130 mL of gas at a pressure of 0.97 atm. What volume will the gas occupy at standard pressure?

Answers

Boyle's Law
G: V1= 130 mL
P1= 0.97 atm P2= 1 atm (standard pressure)

U: V2=?
F: V2 = V1 P1/P2
S: V2 = (130 mL) (0.97 atm) / 1 atm
Answer : V2 = 126.1 mL volume of the gas occupy @ STP

Which sample of co2 has a definite shape and definite volume

Answers

The answer is Solid.

This is on account of the substances that develop a solid are packed in a settled, firmly pressed geometric plan.

in a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

Answers

The partial pressure exerted by a gas in a mixture, depends on the mole

fraction of the gas and the pressure exerted by the mixture.

The partial pressure of H₂O is 20 atm.

Reasons:

Given parameters are;

Explosion equation is 4C₃H₅N₃O₉ → 12CO₂(g) + O₂(g) + 6N₂(g) + 10H₂O(g)

Amount of nitroglycerine = 227 g

Molar mass of nitroglycerine = 227 g/mol

Required:

Partial pressure of the water vapor

Solution:

Number of moles of nitroglycerine in the reaction = 1 mole

Therefore;

Number of moles of CO₂ = 12/4 = 3 moles

Number of moles of O₂ = 0.25 moles

Number of nitrogen, N = 1.5 moles

Number of moles of H₂O = 2.5 moles

$Mole \ fraction \ of \ H_2O, \ X_(H_2O) = (2.5)/(2.5 + 1.5 + 0.25 + 3) = (10)/(29)$

According to Raoults law, we have;

The partial pressure of H₂O = $X_(H_2O) * P_$

Therefore, partial pressure of H₂O = $(10)/(29) * 58$ = 20 atm.

Learn more here:

brainly.com/question/10165688

This is an incomplete question, here is a complete question.

Nitroglycerine (C₃H₅N₃O₉) explodes with tremendous force due to the numerous gaseous products. The equation for the explosion of Nitroglycerine is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

A scientist conducts an experiment to characterize a bomb containing nitroglycerine. She uses a steel, ridge container for the test.

Volume of rigid steel container: 1.00 L

Molar mass of Nitroglycerine: 227 g/mol

Temperature: 300 K

Amount of Nitroglycerine tested: 227 g

Value for ideal gas constant, R: 0.0821 L.atm/mol.K

In a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

Answer : The partial pressure of the water vapor is, 20.01 atm

Explanation :

First we have to calculate the moles of $C_3H_5N_3O_9$

$\text{Moles of }C_3H_5N_3O_9=\frac{\text{Given mass }C_3H_5N_3O_9}{\text{Molar mass }C_3H_5N_3O_9}=(227g)/(227g/mol)=1mol$

Now we have to calculate the moles of $CO_2,O_2,N_2\text{ and }H_2O$

The balanced chemical reaction is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

From the balanced chemical reaction we conclude that,

As, 4 moles of $C_3H_5N_3O_9$ react to give 12 moles of $CO_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $(12)/(4)=3$ moles of $CO_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 1 moles of $O_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $(1)/(4)=0.25$ moles of $O_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 6 moles of $N_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $(6)/(4)=1.5$ moles of $N_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 10 moles of $H_2O$

So, 1 moles of $C_3H_5N_3O_9$ react to give $(10)/(4)=2.5$ moles of $H_2O$

Now we have to calculate the mole fraction of water.

$\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CO_2+\text{Moles of }O_2+\text{Moles of }N_2}$

$\text{Mole fraction of }H_2O=(2.5)/(2.5+3+0.25+1.5)=0.345$

Now we have to calculate the partial pressure of the water vapor.

According to the Raoult's law,

$p_(H_2O)=X_(H_2O)* p_T$

where,

$p_(H_2O)$ = partial pressure of water vapor gas = ?

$p_T$ = total pressure of gas = 58 atm

$X_(H_2O)$ = mole fraction of water vapor gas = 0.345

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

$p_(H_2O)=X_(H_2O)* p_T$

$p_(H_2O)=0.345* 58atm=20.01atm$

Therefore, the partial pressure of the water vapor is, 20.01 atm

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