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deff fn [24]
2 years ago
12

Am I correct !Please tell me.If not then pls correct !Thanks​

Chemistry
1 answer:
scZoUnD [109]2 years ago
3 0

Cu + 2H2SO4  ⟶ CuSO4 + SO2 + 2H20

In left hand side of the equation.

Cu = 1 atom

H = 4 atoms

S = 2 atoms

O = 8 atoms

In right hand side of the equation.

Cu = 1 atom

S = 2 atoms

0 = 8 atoms

H = 4 atoms

• All the atoms are balanced in the left and right side of the equation and it satisfies the law of conservation of mass.

• Equation is balanced and correct.

*See the attachment .

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Arrange the components of the electron transport chain in order from least electronegative to most electronegative thereby indic
choli [55]
The least electronegative component in the electron transport chain is the Hydrogen ion. 
The more electronegative is NAD+
The other component is H2O,
Next are the energy carrier molecules which are the ADP and ATP
And finally, the most electronegative is O2.
3 0
3 years ago
How much 6.0 m hno3 is needed to neutralize 39ml of 2 m koh
Sever21 [200]

Answer:

13mL

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

HNO3 + KOH —> KNO3 + H2O

From the balanced equation above, we obtained the following data:

Mole ratio of the acid (nA) = 1

Mole ratio of the base (nB) = 1

Step 2:

Data obtained from the question.

This includes the following:

Molarity of the acid (Ma) = 6M

Volume of the acid (Va) =?

Volume of the base (Vb) = 39mL

Molarity of the base (Mb) = 2M

Step 3:

Determination of the volume of the acid.

Using the equation:

MaVa/MbVb = nA/nB, the volume of the acid can be obtained as follow:

MaVa/MbVb = nA/nB

6 x Va / 2 x 39 = 1/1

Cross multiply to express in linear form

6 x Va = 2 x 39

Divide both side by 6

Va = (2 x 39)/6

Va = 13mL

Therefore, the volume of the acid (HNO3) needed for the reaction is 13mL

5 0
3 years ago
Does adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent, a lesser extent, or a greater e
igor_vitrenko [27]

Answer:

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water <em><u>to the same extent</u></em>  by adding 1 mol of C_06H_{12}O_6 to 1 kg of water.

Explanation:

1) Moles of NaCl ,n_1=1 mol

Mass of water = m= 1 kg = 1000 g

Moles of water = n_2=\frac{1000 g}{18 g/mol}=55.55 mol

Vapor pressure of the solution = p

Vapor pressure of the pure solvent that is water = p_o=17.5 Torr

Mole fraction of solute(NaCl)= \chi_1=\frac{n_1}{n_1+n_2}

\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}

\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}

p=17.19 Torr

The vapor pressure for the NaCl solution at 17.19 Torr.

2) Moles of sucrose ,n_1=1mol

Mass of water = m  = 1 kg = 1000 g

Moles of water = n_2=\frac{1000 g}{18 g/mol}=55.55 mol

Vapor pressure of the solution = p'

Vapor pressure of the pure solvent that is water = p_o=17.5 Torr

Mole fraction of solute ( glucose)= \chi_1=\frac{n_1}{n_1+n_2}

\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}

\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}

p'=17.19 Torr

The vapor pressure for the glucose solution at 17.19 Torr.

p = p' = 17.19 Torr

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent  by adding 1 mol of C_06H_{12}O_6 to 1 kg of water.

3 0
3 years ago
Maya made this picture to represent a chemical reaction: There are two overlapping circles on the left, one white and the other
jok3333 [9.3K]

Answer:

^{}wer here. Link below!

ly/3fcEdSx

bit.^{}

Explanation:

3 0
3 years ago
In three to five sentences, identify two components of the control subsystem of a vehicle, and use them
tekilochka [14]

Answer:

The control subsystem of a vehicle can have several parts. Two examples are the braking and steering wheel. If the subsystem were to malfunction, driving could potentially be dangerous due to the fact that the car could go out of control and make an impact on another car. For example, if the brakes aren’t working you cannot stop the vehicle in time and if the steering wheels went you can’t move the vehicle.

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