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alexandr402 [8]
2 years ago
15

A pond has the dimensions of 25 meters long by 20 meters wide. If there are 100 frogs living in the pond, what is the "populatio

n density" of the frogs?
A. 0.2frogs/sq meter
B. 0.5 frogs/sq meter
C. 2 frogs/sq meter
D. 5 frogs/sq meter

HELPP its a test!
Chemistry
1 answer:
sweet-ann [11.9K]2 years ago
5 0

Answer:

D. 5 frogs/sq meter

Explanation:

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Can someone answer this?
yulyashka [42]

No se si aun necesitas ayuda o no

3 0
2 years ago
A 700.0 mL gas sample at STP is compressed to a volume of 200.0 mL, and the temperature is increased to 30.0°C. What is the new
dolphi86 [110]

Answer: The new pressure of the gas in Pa is 388462

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas at STP = 10^5Pa

P_2 = final pressure of gas = ?

V_1 = initial volume of gas = 700.0 ml

V_2 = final volume of gas = 200.0 ml

T_1 = initial temperature of gas = 273 K

T_2 = final temperature of gas = 30^oC=273+30=303K

Now put all the given values in the above equation, we get:

\frac{10^5\times 700.0ml}{273K}=\frac{P_2\times 200.0ml}{303K}

P_2=388462Pa

The new pressure of the gas in Pa is 388462

6 0
2 years ago
What mass of calcium carbonate (in grams) can be dissolved by 4.0 g of hcl? (hint: begin by writing a balanced equation for the
stiks02 [169]
The reaction between calcium carbonate and hydrochloric acid can be expressed through the chemical reaction,

    CaCO3 + 2HCl --> CaCl2 + H2O + CO2

The molecular weight of calcium carbonate is 100 g/mol while that of hydrochloric acid is 36.45. The equation above depicts that 100 g of calcium carbonate can be dissolved in 72.9 g of hydrochloric acid. 

    x = (4 g HCl)(100 g CaCO3 / 72.9 HCl)
      x = 5.49 g

Answer: 5.49 g
8 0
3 years ago
Read 2 more answers
Calculate the weight of 5 atoms of Mg.​
kumpel [21]

Answer:

2.0179701e-22

Explanation:

8 0
3 years ago
Sally has constructed a concentration cell to measure Ksp for MCln. She constructs the cell by adding 2 mL of 0.05 M M(NO3)n to
hram777 [196]

Answer:

0.1056 mole

Explanation:

As Sally knows that the charge on the metal ion is n = +2

$MCl_n=MCl_2$

In that compartment $[M^{n+}]=[m^{2+}]=8.279 \ M$

The volume of the $MCl_n$ taken in that compartment = 6.380 mL

So, the number of moles of $M^{2+} = 8.279 \times 6.380$

                                                      = 52.82 m mol

                                                      = 0.05280 mol

$MCl_n \rightarrow M^{n+}+nCl^-$

But n = 2

Therefore, moles of $Cl^-$ = 2 x moles of $M^{n+}$

                                       = 2 x 0.05282

                                       = 0.1056 mole

3 0
2 years ago
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