Which explanation correctly describes why it is important for a cell’s surface-area-to-volume ratio to not be too small?

Which element has the most protons:<br>Krypton<br>Argon<br>Nitrogen<br>Chlorine

stealth61 [152]

The answer would be A) Kyrpton

---

Krypton has 36 protons.

Argon has 18 protons.

Nitrogen has 7 protons.

Chlorine has 17 protons.

After looking at all of these numbers we can see that the element with the highest number of protons is krypton, at 36 protons.

~Hope I could help! Stay safe!~

7 0

3 years ago

Read 2 more answers

The picture shows an aerialist walking on a tightrope and holding a balancing bar. An action-reaction pair of forces exist betwe

noname [10]

Answer:

C

Explanation:

3 0

2 years ago

A strip of copper 370 μm thick and 4.5 mm wide is placed in a uniform magnetic field B with arrow of magnitude 0.30 T, with B wi

777dan777 [17]

Explanation:

The given data is as follows.

Magnetic field (B) = 0.30 T

Current (I) = 24 A

n = $370 \times 10^{-6} m$

charge on electron (e) = $1.602 \times 10^{-19}$ C

Formula to calculate the potential difference is as follows.

V = $\frac{B \times I/d}{n \times e
}$

= $\frac{0.30 T \times 24 amps}{8.47 \times 10^{28} electrons/ m^{3} \times 370 \times 10^{-6} m \times 1.602 \times 10^{-19} C)}$

= $1.38 \times 10^{-6}$ V

Therefore, we can conclude that value of potential difference is $1.38 \times 10^{-6}$ V.

8 0

3 years ago

One lap of an engineered hamster track measures 255 cm. To run 10. meters, how many laps should the hamster run? Report your ans

Maru [420]

Answer:

3.9 laps

Explanation:

Step 1: Given data

Length of the hamster track (L): 255 cm

Total distance to be run (D): 10 m

Step 2: Convert "D" to centimeters

We will use the relationship 1 m = 100 cm.

10 m × (100 cm/1 m) = 1.0 × 10³ cm

Step 3: Calculate the number of laps (n) that the hamster should run

We will use the following expression.

n = D/L

n = 1.0 × 10³ cm/255 cm

n = 3.9

4 0

3 years ago

What volume of 0.100 m sodium chloride must be added to 75.0 ml of 0.200 m lead(ii) nitrate to precipitate all of the lead ions?

Svetradugi [14.3K]

Reaction for precipitation is
Pb(NO3)2(aq) + 2Nacl(aq)→PbCL2(s)+2NaNO3(aq)
The given reaction is
The mol of Nacl required is =2×mol of Pb(NO3)2 which is present.
∴M(Nacl)×V(Nacl) =2×M(Pb(NO3)2 ×V(Pb(NO3)2
o.100M×V(Nacl)=2×0.200 MX75.0ML
V(Nacl = 300mL.

5 0

3 years ago