What element is solid at room temperature shiny malleable and a good conductor of heat

How much heat is required to convert 0.3 kilograms of ice at 0°C to water at the same temperature?

grin007 [14]

It is C honestly I just looked it up lollllll but I know it's right

6 0

3 years ago

HURRY!! 30 points!!

AnnZ [28]

Answer:

The amount of each element that begins photosynthesis equals the amount of each element that is produced

Explanation:

3 0

3 years ago

Neutrons have neither charge nor mass.<br> TRUE<br> FALSE

iren2701 [21]

Answer:

FALSE

nuetrons do indeed have no charge, however they are nearly 2000 more times massive than electrons.

7 0

3 years ago

Balance each of the following redox reactions occurring in basic solution.MnO−4(aq)+Br−(aq)→MnO2(s)+BrO−3(aq)Express your answer

Ahat [919]

Answer : The balanced chemical equation is,

$2MnO_4^-(aq)+Br^-(aq)+H_2O(l)\rightarrow 2MnO_2(s)+BrO_3^-(aq)+2OH^-(aq)$

Explanation :

Rules for the balanced chemical equation in basic solution are :

First we have to write into the two half-reactions.
Now balance the main atoms in the reaction.
Now balance the hydrogen and oxygen atoms on both the sides of the reaction.
If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the more number of oxygen are present.
If the hydrogen atoms are not balanced on both the sides then adding hydroxide ion $(OH^-)$ at that side where the less number of hydrogen are present.
Now balance the charge.

The half reactions in the basic solution are :

Reduction : $MnO_4^-(aq)+2H_2O(l)+3e^-\rightarrow MnO_2(s)+4OH^-(aq)$ ......(1)

Oxidation : $Br^-(aq)+6OH^-(aq)\rightarrow BrO_3^-(aq)+3H_2O(l)+6e^-$ .......(2)

Now multiply the equation (1) by 2 and then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

$2MnO_4^-(aq)+Br^-(aq)+H_2O(l)\rightarrow 2MnO_2(s)+BrO_3^-(aq)+2OH^-(aq)$

8 0

3 years ago

A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0- L vessel at 300 K . The following equilibr

umka2103 [35]

Answer:

The concentration of N2 at the equilibrium will be 0.019 M

Explanation:

Step 1: Data given

Number of moles of NO = 0.10 mol

Number of moles of H2 = 0.050 mol

Number of moles of H2O = 0.10 mol

Volume = 1.0 L

Temperature = 300K

At equilibrium [NO]=0.062M

Step 2: The balanced equation

2NO(g) + 2H2(g) → N2(g) + 2H2O(g)

Step 3: Calculate the initial concentration

Concentration = Moles / volume

[NO] = 0.10 mol / 1L = 0.10 M

[H2] = 0.050 mol / 1L = 0.050 M

[H2O] = 0.10 mol / 1L = 0.10 M

[N2] = 0 M

Step 4: Calculate the concentration at the equilibrium

[NO] at the equilibrium is 0.062 M

This means there reacted 0.038 mol (0.038M) of NO

For 2 moles NO we need 2 moles of H2 to produce 1 mol N2 and 2 moles of H2O

This means there will also react 0.038 mol of H2

The concentration at the equilibrium is 0.050 - 0.038 = 0.012 M

There will be porduced 0.038 moles of H2O, this means the final concentration pf H2O at the equilibrium is 0.100 + 0.038 = 0.138 M

There will be produced 0.038/2 = 0.019 moles of N2

The concentration of N2 at the equilibrium will be 0.019 M

5 0

4 years ago