All categories

3 years ago

How many significant digits are in 1.2x10^0

1 answer:

8 0

1.2 x 10^3 is equivalent to 1200. 1 & 2 are signficant because they are real numbers, the 0's do not play a role in significant figures in this setup particularly because no decimal point is given. However, if a decimal point was present it would have 4 signficiant digits and you equation would look something more like, 120.0 x 10^1.

Hope this explains it a bit more!

You might be interested in

A chemist needs 1.36 mol of strontium bromide, SrBr2, for a reaction. What mass of strontium chloride should the chemist use?

yaroslaw [1]

Mass SrBr2 = 247.42 g/mol x 1.36 mol
= 336 g

5 0

4 years ago

Read 2 more answers

for a neutralization reaction, would you expect the magnitude of q to increase, decrease, or stay the same if the concentration

ikadub [295]

For a neutralization reaction, the value of q(heat of neutralization) is doubled when the concentration of only the acid is doubled.

A neutralization reaction is a reaction in which an acid reacts with a base to yield salt and water. Ionically, a neutralization reaction goes as follows; H^+(aq) + OH^-(aq) ------> H20(l).

The heat of neutralization (Q) of the system depends on the concentration of the solutions. Since Q is dependent on concentration, if the concentration of any of the reactants is doubled, more heat is evolved hence Q is doubled.

Learn more: brainly.com/question/10323185

4 0

3 years ago

What is the mole ratio of NO2 to O2

Mazyrski [523]

The mole ratio is 4 NO2 to 3 O2; 4:3

8 0

3 years ago

I need help please!!! Name the parts making up the female part of the flower and state what each part does. Please use your own

Greeley [361]

<h2>Answer:-</h2>

The female part is the pistil. The pistil usually is located in the center of the flower and is made up of three parts: the stigma, style, and ovary. The stigma is the sticky knob at the top of the pistil. It is attached to the long, tubelike structure called the style.

6 0

3 years ago

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0

4 years ago