All categories

3 years ago

Use the following equation.

Mathematics

1 answer:

ludmilkaskok [199]3 years ago

3 0

Answer:

1 mole of Oxygen and two moles of Hydrogen

Step-by-step explanation:

The mass of oxygen equal to one mole of oxygen is 15.998 grams and the mass of one mole of hydrogen is 1.008 g.

You might be interested in

Avery has 6 yards of ribbon. He needs 1/3 yard to make 1 bow.<br>How many bows can Avery make

nikitadnepr [17]

18 18 18 18 18 18 18 18 18 18

5 0

3 years ago

Find the mode for the set of data. 70, 80, 50, 30, 40, 50, 90, 80, 50

m_a_m_a [10]

The mode is 50 because it occurs the most frequently (three times) out of any of the values.

5 0

3 years ago

Find the side indicated by the variable. Round to the nearest tenth.

zepelin [54]

Answer:

b ≈ 25.6

Step-by-step explanation:

From the figure attached,

By applying tangent rule in the given triangle,

tan(32°) = $\frac{\text{Opposite side}}{\text{Adjacent side}}$

tan(32°) = $\frac{16}{b}$

b = $\frac{16}{\text{tan}(32)}$

b = $\frac{16}{0.62487}$

b = 25.605

b ≈ 25.6

5 0

3 years ago

A spherical balloon currently has a radius of 19cm. If the radius is still growing at a rate of 5cm or minute, at what rate is a

miv72 [106K]

Answer:

22670.8 cm³/min

Step-by-step explanation:

Given:

Radius of the balloon at a certain time (r) = 19 cm

Rate of growth of radius is, $\frac{dr}{dt}=5\ cm/min$

The rate at which the air is pumped in the balloon can be calculated by finding the rate of increase in the volume of the balloon.

So, first we find the volume of the sphere in terms of 'r'. As the balloon is spherical in shape, the volume of the balloon is equal to the volume of a sphere. Therefore,

Volume of balloon is given as:

$V=\frac{4}{3}\pi r^3$

Now, rate of increase of volume is obtained by differentiating both sides of the equation with respect to time 't'.

Differentiating both sides with respect to time 't', we get:

$\frac{dV}{dt}=\frac{d}{dt}(\frac{4}{3}\pi r^3)\\\\\frac{dV}{dt}=\frac{4\pi}{3}(3r^2)(\frac{dr}{dt})\\\\\frac{dV}{dt}=4\pi r^2(\frac{dr}{dt})$

Now, plug in 19 cm for 'r', 5 cm per minute for $\frac{dr}{dt}$ and solve for $\frac{dV}{dt}$ . This gives,

$\frac{dV}{dt}=4\pi (19 cm)^2(5\ cm/min)\\\\\frac{dV}{dt}=4\times 3.14\times 361\times 5\ cm^3/min\\\\\frac{dV}{dt}=22670.8\ cm^3/min$

Therefore, the rate at which the air is being pumped into the balloon is 22670.8 cm³/min.

4 0

4 years ago

Please answer don’t need to show work

bulgar [2K]

Answer:

1,042

Step-by-step explanation:

3 0

4 years ago

Read 2 more answers