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Paha777 [63]
1 year ago
15

Robert places a piece of magnesium metal (Mg) into a test tube of hydrochloric acid (HCl). The magnesium begins to bubble and fi

zz, releasing hydrogen gas (H2) and the test tube gets warm. Which equation represents the reaction Robert observed?
Chemistry
2 answers:
olchik [2.2K]1 year ago
8 0

Answer:

See below.

Explanation:

The balanced equation is

Mg(s) + 2HCl ---->   MgCl2 + H2(g)

ohaa [14]1 year ago
5 0

Answer:

Mg_{(s)} + 2HCl_{(aq)} ---> MgCl_2_{(aq)} + H_2_{(g)}

Explanation:

The reaction is a simple replacement reaction in which hydrogen gas is liberated when it gets replaced by Magnesium in aqueous solution of hydrochloric acid.

<em>1 mole of Mg reacts with 2 moles of HCl to give 1 mole MgCl_2 and 1 mole H_2 gas. The balanced equation for the reaction Robert observed is as follows:</em>

 Mg_{(s)} + 2HCl_{(aq)} ---> MgCl_2_{(aq)} + H_2_{(g)}

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AVprozaik [17]
Light acts as a wave so when you burn a certain element it generates a specific wavelength which represents a specific color light. ^-^
7 0
1 year ago
Cu + 2AgNO3 es002-1.jpg 2Ag + Cu(NO3)2 How many moles of copper must react to form 0.854 mol Ag?
marin [14]
Balance Chemical Equation is as follow,

<span>                        Cu + 2 AgNO</span>₃     →    2 Ag + Cu(NO₃)₂

According to Balance Equation,

                   2 Moles of Ag is produced by reacting  =  1 Mole of Cu
So, 
     0.854 Moles of Ag will be produced by reacting  =  X Moles of Cu

Solving for X,
                             X  =  (0.854 mol × 1 mol) ÷ 2 mol

                             X  =  0.427 Moles of Cu
Result:
            0.854 Moles of Ag 
are produced by reacting 0.427 Moles of Cu.
4 0
1 year ago
The solubility of N2 in blood can be a serious problem (the "bends") for divers breathing compressed air (78% N2 by volume) at d
OLga [1]

Answer:

The volume is 19.7 mL

Explanation:

<u>Step 1</u>: Given data

Pressure at sea level = 1.00 atm

Pressure at 50 ft = 2.47535 atm

kH for N2 in water at 25°C is 7.0 × 10−4 mol/L·atm

Molarity (M) = kH x P

<u>Step 2</u>: Calculate molarity

M at sea level:

M = 7.0*10^-4 * (1.00atm * 0.78) = 5.46*10^-4 mol/L

M at 50ft:

M = 7.0*10^-4 * (2.47535atm * 0.78) = 13.5*10^-4 mol/L

We should find the volume of N2. To find the volume whe have to find the number of moles first. This we calculate by calculating the difference between M at 50 ft and M at sea level.

13.5*10^-4 mol/L - 5.46*10^-4 mol/L = 8.04*10^-4 mol/L

Step 3: Calculate volume

P*V=nRT

with P = 1.00 atm

with V = TO BE DETERMINED

with n =  8.04*10^-4 mol/L  *1L = 8.04*10^-4

with R= 0.0821 atm * L/ mol *K

with T = 25 °C = 273+25 = 298 Kelvin

To find the volume, we re-organize the formula to: V=nRT/P

V= (8.04*10^-4 mol * 0.0821 (atm*L)/(mol*K)* 298K ) / 1.00atm = 0.0197L = 19.7ml

The volume is 19.7 mL

5 0
2 years ago
What is the percent composition by mass of nitrogen in the compound N2H4 (gram-formula mass = 32 g/mol)?
Alex

Answer:

\large \boxed{93\, \% }

Explanation:

1. Calculate the molar mass of N₂H₄

2N = 2 × 14 = 28

4H = 2 ×  1  = <u>  4</u>

           Tot. = 32

2. Calculate the mass percent of N

\text{\% of element} = \dfrac{\text{mass of element}}{\text{mass of compound}} \times 100 \, \% = \dfrac{\text{28}}{\text{32}} \times 100 \, \% =\mathbf{88 \, \%}\\\\\text{The percentage of N in N$_{2}$H$_{4}$ is $\large \boxed{\mathbf{88\, \% }}$}}

3 0
2 years ago
Read 2 more answers
When 1.365 g of anthracene, C14H10, is combusted in a bomb calorimeter that has a water jacket containing 500.0 g of water, the
luda_lava [24]
<h3>The enthalpy of combustion per mole of anthracene : 7064 kj/mol(- sign=exothermic)</h3><h3>Further explanation  </h3>

The law of conservation of energy can be applied to heat changes, i.e. the heat received/absorbed is the same as the heat released  

Q in = Q out  

Heat can be calculated using the formula:  

Q = mc∆T  

Heat released by anthracene= Heat absorbed by water

Heat absorbed by water =

\tt Q=500\times 4.18\times 25.89=54110.1~J

mol of  anthracene (MW=178,23 g/mol)

\tt \dfrac{1.365}{178.23}=0.00766

The enthalpy of combustion per mole of anthracene :

\tt \Delta H=-\dfrac{Q}{n}=\dfrac{54110.1}{0.00766}=-7063981.7~J/mol\approx -7064~kJ/mol

8 0
1 year ago
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