What does M stand for

Given the following equation: Cu + 2 AgNO3 ---> Cu(NO3)2 + 2

Temka [501]

Given the following equation; Cu + 2AgNO3 = Cu(NO3)2 + 2Ag, 48.97 grams of Cu are needed to react with 262g of AgNO3.

<h3>How to calculate mass of substances?</h3>

The mass of a substance can be calculated using the following steps:

Cu + 2AgNO3 = Cu(NO3)2 + 2Ag

1 mole of Cu react with 2 moles of AgNO3

Molar mass of AgNO3 = 169.87 g/mol
Molar mass of Cu = 63.5g/mol

moles of AgNO3 = 262g/169.87g/mol = 1.54mol

1.54 moles of AgNO3 will react with 0.77 moles of Cu.

mass of Cu = 0.77 × 63.5 = 48.97g

Therefore, given the following equation; Cu + 2AgNO3 = Cu(NO3)2 + 2Ag, 48.97 grams of Cu are needed to react with 262g of AgNO3.

Learn more about mass at: brainly.com/question/6876669

8 0

2 years ago

What would the world be like without elements? please

Lisa [10]

Answer:

It would be nothing. Quite literally nothing. No Oxygen, no dirt, no anything.

Explanation:

3 0

2 years ago

Read 2 more answers

The nuclear binding energy of one nitrogen-14 atom is (BLANK) x 10A J. Round to 3

d1i1m1o1n [39]

mass defect = mass of constituents - mass of atom

N has 7p and 9n

proton mass ~ 1.00728 amu

neutron mass ~ 1.00866 amu

electron mass ~ 0.000549 amu

Nitrogen mass ~ 14.003074 amu

mass defect = (7*1.00728)-(7*1.00866)-(7*0.000549) - 14.003074

= 0.11235amu

convert to energy, the binding energy = 1.68x10^-11 J

5 0

3 years ago

4 . suppose you are setting up a reaction that requires an iodide salt and are planning to use sodium iodide. however, at the la

xxTIMURxx [149]

You will need to weight out less

6 0

3 years ago

Three kilograms of steam is contained in a horizontal, frictionless piston and the cylinder is heated at a constant pressure of

lakkis [162]

Answer:

Final temperature: 659.8ºC

Expansion work: 3*75=225 kJ

Internal energy change: 275 kJ

Explanation:

First, considering both initial and final states, write the energy balance:

$U_{2}-U_{1}=Q-W$

Q is the only variable known. To determine the work, it is possible to consider the reversible process; the work done on a expansion reversible process may be calculated as:

$dw=Pdv$

The pressure is constant, so: $w=P(v_{2}-v_{1} )=0.5*100*1.5=75\frac{kJ}{kg}$ (There is a multiplication by 100 due to the conversion of bar to kPa)

So, the internal energy change may be calculated from the energy balance (don't forget to multiply by the mass):

$U_{2}-U_{1}=500-(3*75)=275kJ$

On the other hand, due to the low pressure the ideal gas law may be appropriate. The ideal gas law is written for both states:

$P_{1}V_{1}=nRT_{1}$

$P_{2}V_{2}=nRT_{2}\\V_{2}=2.5V_{1}\\P_{2}=P_{1}\\2.5P_{1}V_{1}=nRT_{2}$

Subtracting the first from the second:

$1.5P_{1}V_{1}=nR(T_{2}-T_{1})$

Isolating $T_{2}$ :

$T_{2}=T_{1}+\frac{1.5P_{1}V_{1}}{nR}$

Assuming that it is water steam, n=0.1666 kmol

$V_{1}=\frac{nRT_{1}}{P_{1}}=\frac{8.314*0.1666*373.15}{500} =1.034m^{3}$

$T_{2}=100+\frac{1.5*500*1.034}{0.1666*8.314}=659.76$ ºC

7 0

2 years ago