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2 years ago

How many moles of C are needed to react with 0.460 mole SO2 ? Equation: 5C(s)+2SO2(g)→CS2(l)+4CO(g)

Chemistry

1 answer:

stepan [7]2 years ago

3 0

Answer:

12345678909ygtufyghkkgjfjvhbkjn,bhmgncfxcgvhbm

Explanation:

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At a cost of $1600/oz, how much would you have to pay for a solid cubic foot of gold?

Alex777 [14]

We'll begin by calculating the mass in ounce (oz) of a cube foot (ft³) of gold. This can be obtained as follow:

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Density of gold = 19298 oz/ft³

Volume of gold = 1 ft³

Density = mass /volume

19298 = mass / 1

Finally, we shall determine the cost of 19298 oz of gold. This can be obtained as follow:

1 oz = $ 1600

Therefore,

19298 oz = 19298 × 1600

19298 oz = $ 30876800

Therefore, a solid cube foot of gold (i.e 19298 oz) will cost $ 30876800

Learn more: brainly.com/question/15407624

4 0

2 years ago

The complete orbital notation diagram of an atom is shown.

Leno4ka [110]

Answer:

atoms are little things things thinbs timmy amrfiv

Explanation:

5 0

1 year ago

2.00 L of 0.800 M NaNO3 must be prepared from a solution known to be 2.50 M in concentration.How many mL are required? Plus don'

Morgarella [4.7K]

Answer:

1.36 × 10³ mL of water.

Explanation:

We can utilize the dilution equation. Recall that:

$\displaystyle M_1V_1= M_2V_2$

Where <em>M</em> represents molarity and <em>V</em> represents volume.

Let the initial concentration and unknown volume be <em>M</em>₁ and <em>V</em>₁, respectively. Let the final concentration and required volume be <em>M</em>₂ and <em>V</em>₂, respectively. Solve for <em>V</em>₁:

$\displaystyle \begin{aligned} (2.50\text{ M})V_1 &= (0.800\text{ M})(2.00\text{ L}) \\ \\ V_1 & = 0.640\text{ L} \end{aligned}$

Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:
$\displaystyle 2.00\text{ L} - 0.640\text{ L} = 1.36\text{ L}$

Convert this value to mL:
$\displaystyle 1.36\text{ L} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 1.36\times 10^3\text{ mL}$

Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.

8 0

2 years ago

A helium-filled balloon has a volume of 2.48 L and a pressure of 150 kPa. The volume of the balloon increases to 2.98 L after yo

Tema [17]

Answer:

THE NEW PRESSURE OF THE HELIUM GAS AT 2.98 L VOLUME IS 124.8 kPa.

AT AN INCREASE ALTITUDE, THERE IS A LOWER PRESSURE ENVIRONMENT AND THE HELIUM GAS PRESSURE DECREASES AND HENCE AN INCREASE IN VOLUME.

Explanation:

The question above follows Boyle's law of the gas law as the temperature is kept constant.

Boyle's law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.

Mathematically, P1 V1 = P2 V2

P1 = 150 kPa = 150 *10^3 Pa

V1 = 2.48 L

V2 = 2.98 L

P2 = ?

Rearranging the equation, we obtain;

P2 = P1 V1 / V2

P2 = 150 kPa * 2.48 / 2.98

P2 = 372 *10 ^3 / 2.98

P2 = 124.8 kPa.

The new pressure of the gas when at a height which increases the volume of the helium gas to 2.98 L is 124.8 kPa.

6 0

2 years ago

Which phase is heat energy being released?

scZoUnD [109]

The answer is B
Vaporization

6 0

2 years ago