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maxonik [38]
3 years ago
6

I need help in chemistry 10th grade (sophomore)

Chemistry
2 answers:
LuckyWell [14K]3 years ago
6 0

Answer:

Here is the answers I am pretty sure sorry if I am not 100%

Explanation:

1.  2

2. 4

3. 5

4. 2

5. 2

6. 1

7. 3

8. 5

9. 4

10. 4

11. 3

12. 5

13. 6

14. 2

15. 2

16. 1

17. 3

And sorry I am not sure on the rest I did all I could. You have a wonderful day! Please mark brainliest ! Much love to all.

Goryan [66]3 years ago
3 0

Answer:

1) 2

2)4

3)5

4)2

5)2

6)1

7)3

8)5

9)4

10)4

11)3

12)5

13)6

14)2

15)2

16)1

17)2

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How are communication signals on the ground sent to spacecraft?
gogolik [260]

Answer:

d) the signals must be changed to another form at ground stations to travel through space.

Explanation:

Spacecraft use transmitters to communicate with people on Earth. This are sent in the form of signals.

3 0
2 years ago
Which of the following statements correctly describe(s) the driving forces for diffusion of Na+ and K+ ions through their respec
melomori [17]

Question:

Which of the following statements correctly describe(s) the driving forces for diffusion of Na+ and K+ ions through their respective channels? Select all that apply.

A)The diffusion of Na+ ions into the cell is facilitated by the Na+ concentration gradient across the plasma membrane.

B)The diffusion of Na+ ions into the cell is impeded by the electrical gradient across the plasma membrane.

C)The diffusion of K+ ions out of the cell is impeded by the K+ concentration gradient across the plasma membrane.

D)The diffusion of K+ ions out of the cell is impeded by the electrical gradient across the plasma membrane. The electrochemical gradient is larger for Na+ than for K+.

Answer:

"The concentration gradient and the electro-chemical gradient" describes  the driving forces for diffusion of Na+ and K+ ions through their respective channels

Explanation:

The Na ions diffusion inside the cell is facilitated by the concentration gradient of the Na ions which is present across the plasma membrane. Hence, the diffusion of the K ions which is present outside the cell and will be impeded due to the electrical gradient which is present near the plasma membrane. Thus, the electro-chemical gradient is greater as compared to the Na ion than that of the K ion.

6 0
3 years ago
What is the cell potential of an electrochemical cell that has the half -reactions shown below? Ni^ 2+ +2e Ni A| A|^ 3+ +3e^ -
Nadya [2.5K]
<h3>Answer:</h3>

A. 1.4 V

<h3>Explanation:</h3>

We are given the half reactions;

Ni²⁺(aq) + 2e → Ni(s)

Al(s) → Al³⁺(aq) + 3e

We are required to determine the cell potential of an electrochemical cell with the above half-reactions.

E°cell = E(red) - E(ox)

From the above reaction;

Ni²⁺ underwent reduction(gain of electrons) to form Ni

Al on the other hand underwent oxidation (loss of electrons) to form Al³⁺

The E.m.f of Ni/Ni²⁺ is -0.25 V and that of Al/Al³⁺ is -1.66 V

Therefore;

E°cell = -0.25 V - (-1.66 V)

         = -0.250V + 1.66 V

         = + 1.41 V

         = + 1.4 V

Therefore, the cell potential will be +1.4 V

3 0
3 years ago
Base your answer on the information below. in liquid water, an equilibrium exists between h2o() molecules, h+(aq) ions, and oh–(
kicyunya [14]

Answer:

OH^-

Explanation:

Any substance that is able to neutralize acidity in the stomach is generally known as an antacid. There are various kinds of antacids that are in common use. It should be noted that the stomach is usually slightly acidic.

Milk of magnesia is the substance magnesium hydroxide with chemical formula Mg(OH)2. A solution of milk of magnesia contains Mg^2+ and OH^-.

Hence the negative ion contained in milk of magnesia is the hydroxide ion  OH^-.

8 0
3 years ago
a concentration solution of H2so4 is 59.4% by mass (m/m) and has a density of 1.83 g/mL. How many mL of the solution would be re
Blababa [14]

Answer: 41.5 mL

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n}{V_s}

where,

n = moles of solute

V_s = volume of solution in L

Given : 59.4 g of H_2SO_4 in 100 g of solution  

moles of H_2SO_4=\frac{\text {given mass}}{\text {molar mass}}=\frac{59.4g}{98g/mol}=0.61

Volume of solution =\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.83g/ml}=54.6ml

Now put all the given values in the formula of molality, we get

Molality=\frac{0.61\times 1000}{54.6ml}=11.2M

To calculate the volume of acid, we use the equation given by neutralisation reaction:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of stock acid which is H_2SO_4

M_2\text{ and }V_2 are the molarity and volume of dilute acid which is H_2SO_4

We are given:

M_1=11.2M\\V_1=mL\\M_2=0.30M\\V_2=1550mL

Putting values in above equation, we get:

11.2\times V_1=0.30\times 1550\\\\V_1=41.5mL

Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid

4 0
3 years ago
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