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3 years ago

9

Write 657000 in scientific notation.

1 answer:

Anika [276]3 years ago

4 0

Answer:

600000+50000+7000

Explanation:

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How much sucrose (g) do you need to weight in order to prepare 19.16 g of a 13.1 % (weight percent) solution?

Nonamiya [84]

<span>2.51 grams
You want to prepare 19.16 g of some solution which will have 13.1% of it's mass being sucrose. So we just need to perform some simple multiplication: 19.16g * 0.131 = 2.50996g
Rounding to 3 significant figures gives 2.51 g.</span>

3 0

3 years ago

Given 3.4 grams of x compound with a molar mass of 85 g and 4.2 grams of y compound with a molar mass of 48 g How much of compou

Ulleksa [173]

Answer:

$4.36~g~XY$

Explanation:

In this case, we can start with the reaction:

$2X + Y_2~->~2XY$

If we check the reaction, we will have 2 X and Y atoms on both sides. So, <u>the reaction is balanced</u>. Now, the problem give to us two amounts of reagents. Therefore, we have to find the <u>limiting reagent</u>. The first step then is to find the moles of each compound using the <u>molar mass</u>:

$3.4~g~X\frac{1~mol~X}{85~g~X}=0.04~mol~X$

$4.2~g~Y_2\frac{1~mol~Y_2}{48~g~Y_2}=0.0875~mol~Y_2$

Now, we can <u>divide by the coefficient</u> of each compound (given by the balanced reaction):

$\frac{0.04~mol~X}{1}=~0.04$

$\frac{0.0875~mol~Y_2}{2}=0.04375$

The smallest value is for "X", therefore this is our <u>limiting reagent</u>. Now, if we use the <u>molar ratio</u> between "X" and "XY" we can calculate the moles of XY, so:

$0.04~mol~X\frac{2~mol~XY}{2~mol~X}=0.04~mol~XY$

Finally, with the molar mass of "XY" we can calculate the grams. Now, we know that 1 mol X = 85 g X and 1 mol $Y_2$ = 48 g $Y_2$ (therefore 1 mol Y = 24 g Y). With this in mind the <u>molar mass of XY</u> would be 85+24 = 109 g/mol. With this in mind:

$0.04~mol~XY\frac{109~g~XY}{1~mol~XY}=4.36~g~XY$

I hope it helps!

6 0

4 years ago

Which of the following are likely to form an ionic bond

sattari [20]

Answer:

No question is posted.

But Ionic bond is mostly between metals live Na, K, Ca, Mg and strong non metals like Cl, O, F

Explanation:

3 0

3 years ago

Every spring has an equilibrium position. Which statements describe a spring at its equilibrium position ? Check all that apply.

Naddika [18.5K]

Answer:

the answer is The elastic potential energy is zero

Explanation:

The elastic potential energy is zero

3 0

4 years ago

Read 2 more answers

In a gas grill, 29 lbs propane C3H8 are

dimulka [17.4K]

Answer : The mass of combustion products formed are 134 lbs.

Explanation :

The balanced chemical reaction will be:

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

Given :

Mass of $C_3H_8$ = 29 lbs = 13154.2 g

conversion used : 1 lbs = 453.592 g

Molar mass of $C_3H_8$ = 44 g/mole

First we have to calculate the moles of $C_3H_8$ .

$\text{ Moles of }C_3H_8=\frac{\text{ Mass of }C_3H_8}{\text{ Molar mass of }C_3H_8}=\frac{13154.2g}{44g/mole}=298.9moles$

Now we have to calculate the moles of $CO_2$ and $H_2O$ .

From the balanced chemical reaction we conclude that,

As, 1 mole of $C_3H_8$ react to give 3 moles of $CO_2$

So, 298.9 mole of $C_3H_8$ react to give $298.9\times 3=896.7$ moles of $CO_2$

and,

As, 1 mole of $C_3H_8$ react to give 4 moles of $H_2O$

So, 298.9 mole of $C_3H_8$ react to give $298.9\times 4=1195.6$ moles of $H_2O$

Now we have to calculate the mass of $CO_2$ and $H_2O$ .

Molar mass of $CO_2$ = 44 g/mole

Molar mass of $H_2O$ = 18 g/mole

$\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass }CO_2$

$\text{Mass of }CO_2=896.7mole\times 44g/mole=39454.8g=86.98lbs$

and,

$\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass }H_2O$

$\text{Mass of }H_2O=1195.6mole\times 18g/mole=21520.8g=47.44lbs$

The total mass of products = Mass of $CO_2$ + Mass of $H_2O$

The total mass of products = 86.98 + 47.44 = 134.42 ≈ 134 lbs

Therefore, the mass of combustion products formed are 134 lbs.

6 0

3 years ago