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3 years ago

Which describes the difference between frogs and bacteria?

Chemistry

2 answers:

Zielflug [23.3K]3 years ago

8 0

Answer:the last one

Explanation:it not that hard

likoan [24]3 years ago

6 0

Answer:the first one

Explanation:Bacteria living on the skin of frogs could save them from a deadly virus, ... With Differences in the Composition of the Skin Microbiome of a Wild

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roundup an herbicide manufactured by monsanto has the formula c3h8no5p. how many moles of molecules are there in a 500g sample o

uranmaximum [27]

N= m/Mr Find the Mr of C3H8NO5P and then divide the mass by the Mr. Sorry I don't have periodic table with me so I can't calculate the Mr for you

3 0

3 years ago

SiCl4(l) + H2O(l) → SiO2(s) + HCl(aq)

Sveta_85 [38]

I assume you’re looking for a balanced equation.
SiCl4 + 2H2O = SiO2 + 4HCl

5 0

3 years ago

A hot air balloon is filled with 15 moles helium gas and 5 moles nitrogen gas. What is the volume of the balloon at 1.01 atm and

kvasek [131]

Given:

Moles of He = 15

Moles of N2 = 5

Pressure (P) = 1.01 atm

Temperature (T) = 300 K

To determine:

The volume (V) of the balloon

Explanation:

From the ideal gas law:

PV = nRT

where P = pressure of the gas

V = volume

n = number of moles of the gas

T = temperature

R = gas constant = 0.0821 L-atm/mol-K

In this case we have:-

n(total) = 15 + 5 = 20 moles

P = 1.01 atm and T = 300K

V = nRT/P = 20 moles * 0.0821 L-atm/mol-K * 300 K/1.01 atm = 487.7 L

Ans: Volume of the balloon is around 488 L

3 0

3 years ago

2. What is the electron configuration of an electrically neutral atom of magnesium?

alekssr [168]

Are u talking about electron sublevel config or where the electrons show in the "rings" of the atom

7 0

3 years ago

A tank contains 90 kg of salt and 2000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain

sashaice [31]

Answer:

a) 90 kg

b) 68.4 kg

c) 0 kg/L

Explanation:

Mass balance:

$-w=\frac{dm}{dt}$

w is the mass flow

m is the mass of salt

$-v*C=\frac{dm}{dt}$

v is the volume flow

C is the concentration

$C=\frac{m}{V+(6-3)*L/min*t}$

$-v*\frac{m}{V+(6-3)*L/min*t}=\frac{dm}{dt}$

$-3*L/min*\frac{m}{2000L+(3)*L/min*t}=\frac{dm}{dt}$

$-3*L/min*\frac{dt}{2000L+(3)*L/min*t}=\frac{dm}{m}$

$-3*L/min*\int_{0}^{t}\frac{dt}{2000L+(3)*L/min*t}=\int_{90kg}^{m}\frac{dm}{m}$

$-[ln(2000L+3*L/min*t)-ln(2000L)]=ln(m)-ln(90kg)$

$-ln[(2000L+3*L/min*t)/2000L]=ln(m/90kg)$

$m=90kg*[2000L/(2000L+3*L/min*t)]$

a) Initially: t=0

$m=90kg*[2000L/(2000L+3*L/min*0)]=90kg$

b) t=210 min (3.5 hr)

$m=90kg*[2000L/(2000L+3*L/min*210min)]=68.4kg$

c) If time trends to infinity the division trends to 0 and, therefore, m trends to 0. So, the concentration at infinit time is 0 kg/L.

6 0

3 years ago