All categories

2 years ago

The m equivalent of 758,422,344 cm equals (to 3 sf)

Chemistry

1 answer:

denis23 [38]2 years ago

8 0

1) convert cm to m

$= 758,422,344\ cm\ x\ \frac{1\ m}{100\ cm} \\= 7,584,223.44\ m$

2) sig figs

When making a bigger number smaller, the exponent will be a positive number.

The answer will be $7.58$ × $10^{6}$ meters after rounding

You might be interested in

1. Name the compound present in a sample of dry air.

aev [14]

Answer:

not air

Explanation:

I think so

hahaha

7 0

2 years ago

What is the volume, in liters, of 6.8 mol of Kr gas at STP?

Virty [35]

Answer:

$\boxed {\boxed {\sf D. \ 152 \ L}}$

Explanation:

Any gas at standard temperature and pressure (STP) has a volume of 22.4 liters per mole or 22.4 L/mol. We can create a proportion with this value.

$\frac {22.4 \ L \ Kr}{1 \ mol \ Kr}$

Multiply both sides of the equation by 6.8 moles of krypton.

$6.8 \ mol \ Kr *\frac {22.4 \ L \ Kr}{1 \ mol \ Kr}$

The units of moles of krypton will cancel.

$6.8 *\frac {22.4 \ L \ Kr}{1 }$

The denominator of 1 can be ignored, so this becomes a simple multiplication problem.

$68 * 22.4 \ L \ Kr$

$152.32 \ L \ Kr$

If we round to the nearest whole number, the 3 in the tenths place tells us to leave the 2 in the ones place.

$152 \ L \ Kr$

6.8 moles of krypton gas at standard temperature and pressure is equal to <u>152 liters</u>.

6 0

3 years ago

A chemist wants to extract copper metal from copper chloride solution. The chemist places 0.25 grams of aluminum foil in a solut

netineya [11]

Answer: About 0.20 grams of copper (II) is formed, and some aluminum is left in the reaction mixture

7 0

2 years ago

Describe nuclear atomic model described by Thomson

Ivenika [448]

It’s so hard but try
Nicole’s Tesla’s

4 0

2 years ago

How many grams of Co2 are produced from 12 g of CH4 and 133 g of O2

Andru [333]

Answer:

$m_{CO_2}=33gCO_2$

Explanation:

Hello,

In this case the undergoing chemical reaction is:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

Next we identify the limiting reactant with the given amounts of methane and oxygen by computing the available moles of methane and the moles of methane consumed by the 133 g of oxygen:

$n_{CH_4}^{available}=12g*\frac{1mol}{16g} =0.75molCH_4\\n_{CH_4}^{consumed\ by\ O_2}=133gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2}=2.08molCH_4$

In such a way, the limiting reactant is methane with 0.75 moles which produce the following mass of carbon dioxide:

$m_{CO_2}=0.75molCH_4*\frac{1molCO_2}{1molCH_4} *\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=33gCO_2$

Best regards.

7 0

2 years ago

Read 2 more answers