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3 years ago

How do cells respond to contact with other cells?

1 answer:

6 0

It depends on the function and activity of both cells. Most normal cells crease division if they came into contact with other cells however the immune system interact directly with pathogens or foreign organisms to destroy them.

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If the reactants are 10 kg what is the mass of the products

nlexa [21]

If the mass of both the reactants is 10kg then the mass of the products also equals 10kg.

It is due to the law of conservation of mass.

Mass can neither be created nor be destroyed.

7 0

3 years ago

Hello guys can you help me on this.

RideAnS [48]

Answer:

1.Sulfur dioxide and nitrogen oxide

2.a)it forms carbonic acid

3.the community can use renewable energy like solar and wind power cause they produce less pollution

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3 years ago

Which sediment type requires the least amount of energy to be eroded

castortr0y [4]

Answer:

SEDIMENTARY ROCK

Explanation:

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3 years ago

5.(06.04B MC)

NARA [144]

Answer:

Explanation:

1. The reaction will proceed backward, shifting the equilibrium position to the left.

2. The reaction will proceed forward, shifting the equilibrium position to the right.

3. Either add more of the products ( H2O or Cl2) or remove the reactant (HCl or O2)

7 0

2 years ago

What is the density of air at 75 F and latm pressure? Express in lbm/ft3 and kg/m3

ser-zykov [4K]

Answer : The density of air in $lbm/ft^3$ and $kg/m^3$ is, $0.0743lbm/ft^3$ and $1.19kg/m^3$ respectively.

Explanation :

$PV=nRT\\\\PV=\frac{m}{M}RT\\\\P=\frac{m}{V}\frac{RT}{M}\\\\P=\rho \frac{RT}{M}\\\\\rho=\frac{PM}{RT}$

where,

P = pressure of air = 1 atm

V = volume of air

T = temperature of air = 297 K

The conversion used for the temperature from Fahrenheit to degree Celsius is:

$^oC=(^oF-32)\times \frac{5}{9}$

$^oC=(75-32)\times \frac{5}{9}=24^oC$

The conversion used for the temperature from degree Celsius to Kelvin is:

$K=273+^oC$

$K=273+24=297K$

n = number of moles

m = mass of air

M = average molar mass of air = 28.97 g/mole

$\rho$ = density of air = ?

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

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

$\rho=\frac{PM}{RT}$

$\rho=\frac{(1atm)\times (28.97g/mole)}{(0.0821L.atm/mol.K)\times (297K)}$

$\rho=1.19g/L$

Now we have to calculate density in $lbm/ft^3$ .

Conversion used :

$1g/L=0.0624lbm/ft^3$

So,

$1.19g/L=\frac{1.19g/L}{1g/L}\times 0.0624lbm/ft^3=0.0743lbm/ft^3$

The density of air in $lbm/ft^3$ is, $0.0743lbm/ft^3$

Now we have to calculate density in $kg/m^3$ .

Conversion used :

$1g/L=1kg/m^3$

So,

$1.19g/L=1.19kg/m^3$

The density of air in $kg/m^3$ is, $1.19kg/m^3$

8 0

3 years ago