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3 years ago

12

In the following equation , which compounds are the reactants? NaCL+AgNo>NaNo+AgCl

1 answer:

NeX [460]3 years ago

3 0

The first two are always the reactants the products come after so they are last

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The answer is A.) Gravity

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Calculate the reading on voltmeter v²

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The reading of the voltmeter can be determined by finding the potential difference across the 2Ω resistance by using the value of current in the circuit. V=IR, here V is the potential difference across a resistance R through which a current I is flowing.

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What occurs when an object travels in a curved path

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A 230.-mL sample of a 0.240 M solution is left on a hot plate overnight; the following morning, the solution is 1.75 M. What vol

Gwar [14]

Answer:

The volume of water evaporated is 199mL

Explanation:

Concentration is calculated with the following formula

$C=\frac{n}{V}$

where n is the number of moles of solute and V is the volume of the solution (in this case is the same as the solvent volume) in liters.

So we isolate the variable n to know the amount of moles, using the volume given in liters

$230mL=0.23L$

$n=C*V=0.240 M*0.23L=0.055 mol$

Now, we isolate the variable V to know the new volume with the new concentration given.

$V=\frac{n}{C} =0.055mol/1.75M=0.031L=31mL$

Finally, the volume of water evaporated is the difference between initial and final volume.

$V_{ev}= V_{i} -V_{f} =230mL-31mL=199mL$

4 0

3 years ago

An engine draws energy from a hot reservoir with a temperature of 1250 K and exhausts energy into a cold reservoir with a temper

dimulka [17.4K]

Answer:

The power output of this engine is $P = 17.5 W$

The the maximum (Carnot) efficiency is $\eta_c = 0.7424$

The actual efficiency of this engine is $\eta _a = 0.46$

Explanation:

From the question we are told that

The temperature of the hot reservoir is $T_h = 1250 \ K$

The temperature of the cold reservoir is $T_c = 322 \ K$

The energy absorbed from the hot reservoir is $E_h = 1.37 *10^{5} \ J$

The energy exhausts into cold reservoir is $E_c = 7.4 *10^{4} J$

The power output is mathematically represented as

$P = \frac{W}{t}$

Where t is the time taken which we will assume to be 1 hour = 3600 s

W is the workdone which is mathematically represented as

$W = E_h -E_c$

substituting values

$W = 63000 J$

So

$P = \frac{63000}{3600}$

$P = 17.5 W$

The Carnot efficiency is mathematically represented as

$\eta_c = 1 - \frac{T_c}{T_h}$

$\eta_c = 1 - \frac{322}{1250}$

$\eta_c = 0.7424$

The actual efficiency is mathematically represented as

$\eta _a = \frac{W}{E_h}$

substituting values

$\eta _a = \frac{63000}{1.37*10^{5}}$

$\eta _a = 0.46$

7 0

4 years ago