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stich3 [128]
3 years ago
14

Convert 5 pounds and 10 ounces into ounces

Chemistry
1 answer:
igomit [66]3 years ago
7 0
90 oz because 5lbs is equal to 80 oz (5*16=80) plus 10oz = 90
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Asexual is growth and repair , 4 daughter cells and cell divides once dna replicates and the rest is sexual reproduction besides divides one
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4 years ago
Please help me im going out of my way and giving you extra points I really need some assistance
Hitman42 [59]

Answer:

x=2.8moles

Explanation:

first step balance the chemical equation

C2H6+O2-->CO2+2H2O

use more ratio to find the moles of water

1mole of C2H6= 2 moles of H2O

1.4mole of C2H6=?x

cross multiply

x=2.8 moles of H2O

4 0
3 years ago
When a solid with a mass of 53.78 g is added to 50.0 ml of water in a graduated cylinder, the volume increases to 56.4 ml. Calcu
Bad White [126]

Answer:

<h3>The answer is 8.40 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}  \\

volume = final volume of water - initial volume of water

volume = 56.4 - 50 = 6.4 mL

We have

density =  \frac{53.78}{6.4}  \\  = 8.403125

We have the final answer as

<h3>8.40 g/mL</h3>

Hope this helps you

3 0
3 years ago
Given the following, determine ΔG°f at 298 K for SnO. Sn(s) + SnO2(s) → 2SnO(s) ; ΔG° = 12.0 kJ at 298K
Virty [35]

Answer:

The value of change in Gibbs free energy for tin(II) oxide solid is -251.9 kJ/mol.

Explanation:

Sn(s) + SnO_2(s)\rightarrow 2SnO(s), \Delta G_{f}^{o} = 12.0 kJ

\Delta G_{f,SnO_2}^{o}= -515.8kJ/mol

\Delta G_{f,Sn}^{o}= 0 kJ/mol

\Delta G_{f,SnO}^{o}=?

\Delta G_{f}^{o}=\sum ((\Delta G_{f}^{o})_ {products})-\sum ((\Delta G_{f}^{o})_ {reactants})

12.0 kJ=(2 mol\times \Delta G_{f,SnO}^{o})-(1mol\times 0 kJ/mol+1 mol\times -515.8 kJ/mol)

\Delta G_{f,SnO}^{o}=-251.9 kJ/mol

The value of change in Gibbs free energy for tin(II) oxide solid is -251.9 kJ/mol.

3 0
3 years ago
One gallon milk is equal to how many milliliters of milk?
Black_prince [1.1K]

Answer:

3785.411784 mL

Explanation:

6 0
3 years ago
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