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2 years ago

7

What volume of product is produced when 20.0 Mg of liquid hydrogen reacts with excess liquid oxygen in a rocket propellant react

ion? Gas is formed 101.3 to 5 kPa and 40.0 °C

1 answer:

Zinaida [17]2 years ago

4 0

Good luck g hope you get it right

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What kinds of rocks would you expect to see on Mars? What evidence do you have

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like what are they made of? if so iron, due to iron oxide appearing as red hence why mars is red

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3 years ago

What are the 8 properties of matter

SIZIF [17.4K]

color (intensive)

density (intensive)

volume (extensive)

mass (extensive)

boiling point (intensive): the temperature at which a substance boils.

melting point (intensive): the temperature at which a substance melts.

5 0

3 years ago

A 254.5 g sample of a white solid is known to be a mixture of KNO3, BaCl2, and NaCl. When 116.5 g of this mixture is dissolved i

Margaret [11]

Answer:

Mass of KNO3 in the original mix is 146.954 g

Explanation:

mass of $KNO_3$ in original 254.5 mixture.

moles of $BaSO_4 = \frac{mass}{Molecular\ Weight}$

moles of $BaSO_4 = \frac{68.3}{233.38}$

= 0.2926 mol of BaSO4

Therefore,

0.2926 mol of BaCl2,

mass of $BaCl_2 = mol\times molecular weight$

$= 0.2926\times 208.23$

= 60.92 g

the AgCl moles $= \frac{mass}{Molecular\ Weight}$

$= \frac{199.1}{143.32}$

= 1.3891 mol of AgCl

note that, the Cl- derive from both, $BACl_2 and NaCl$

so

mole of Cl- f NaCl $= (1.3891) - (0.2926\times 2) = 0.8039$ mol of Cl-

mol of NaCl = 0.8039 moles

$mass = mol\times Molecular\ Weight = 0.8039 \times 58 = 46.626\ g \ of \ NaCl$

then

KNO3 mass = 254.5 - 60.92-46.626 = 146.954 g of KNO_3

Mass of KNO3 in the original mix is 146.954 g

8 0

2 years ago

An unknown diprotic acid (H2A) requires 44.391 mL of 0.111 M NaOH to completely neutralize a 0.58 g sample. Calculate the approx

Anna [14]

Answer:

$M=235.42g/mol$

Explanation:

Hello!

In this case, given this is an acid-base neutralization and we are considering a diprotic acid, we can write the following mole-mole relationship:

$2n_{acid}=n_{base}$

It means that the moles of acid can be computed given the volume and concentration of NaOH:

$n_{acid}=\frac{M_{base}V_{base}}{2} =\frac{0.044391L*0.111mol/L}{2} \\\\n_{acid}=2.46x10^{-4}mol$

It means that the approximate molar mass of the acid is:

$M=\frac{m_{acid}}{n_{acid}} \\\\M=\frac{m_{acid}}{n_{acid}} =\frac{0.58g}{2.46x10^{-3}mol}\\\\M=235.42g/mol$

Best regards!

5 0

2 years ago

HELP ME THIS IS 1 MONTH OVER DUE PLS HURRY

AVprozaik [17]

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2false
3true
4true

4 0

2 years ago

Read 2 more answers