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3 years ago

When you slide a box across the floor what force must your push be stronger than?

Chemistry

1 answer:

weeeeeb [17]3 years ago

6 0

If I remember correctly, the answer is Friction force.

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A scoop of potassium oxide was placed in

kotykmax [81]

Answer:

I just need points sorry

8 0

2 years ago

Which body changes occur during adolescence?

cupoosta [38]

The answer is C,growth spurts,puberty,& sexual maturity

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2 years ago

Read 2 more answers

C12H7Cl3FNaO2 what are them elements in that chemical formula

LuckyWell [14K]

Assuming you are asking for the names of the elements in that formula , the answer is

carbon
hydrogen
chlorine
fluorine
sodium
oxygen

8 0

3 years ago

For the reaction Na2CO3+Ca(NO3)2⟶CaCO3+2NaNO3 how many grams of calcium carbonate, CaCO3, are produced from 79.3 g of sodium car

Alexus [3.1K]

Answer:

74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.

Explanation:

The balanced reaction is:

Na₂CO₃ + Ca(NO₃)₂ ⟶ CaCO₃ + 2 NaNO₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

Na₂CO₃: 1 mole
Ca(NO₃)₂: 1 mole
CaCO₃: 1 mole
NaNO₃: 2 mole

Being the molar mass of the compounds:

Na₂CO₃: 106 g/mole
Ca(NO₃)₂: 164 g/mole
CaCO₃: 100 g/mole
NaNO₃: 85 g/mole

then by stoichiometry the following quantities of mass participate in the reaction:

Na₂CO₃: 1 mole* 106 g/mole= 106 g
Ca(NO₃)₂: 1 mole* 164 g/mole= 164 g
CaCO₃: 1 mole* 100 g/mole= 100 g
NaNO₃: 2 mole* 85 g/mole= 170 g

You can apply the following rule of three: if by stoichiometry 106 grams of Na₂CO₃ produce 100 grams of CaCO₃, 79.3 grams of Na₂CO₃ produce how much mass of CaCO₃?

$mass of CaCO_{3} =\frac{79.3 grams of Na_{2} CO_{3} *100 grams of of CaCO_{3}}{106 grams of Na_{2} CO_{3}}$

mass of CaCO₃= 74.81 grams

74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.

6 0

3 years ago

calcualte pressure at STP in 10.0 L vessel after reaction of 1.0 L hydrochloride acid (concentration 35% and density 1.28 g/cub.

sattari [20]

Answer:

The pressure in the vessel is 13,3 atm.

Explanation:

The reaction that occurs in vessel (where limestone is 96% of CaCO₃) is:

2 HCl (aq)+ CaCO₃ (s) → CaCl₂(aq)+ H₂O(l)+ CO₂(g)

The increase in the pressure of the vessel after the reaction is by formation of a gas (CO₂). So we have to find the produced moles of this gas and apply the gas ideal law to find the pressure.

We have to find the limit reactant, to do so, we have to calculate the moles of each reactant in the reaction, the one that have the less moles will be the limit reactant:

HCl:

1,0L × (35/100) × (1000 cm³/1L) × (1,28 g/ 1cm³) × (1mol HCl/ 36,46 g) ÷ 2mol

(Concentration) (L to cm³) (cm³ to g) (g to mol) (moles of reaction)

moles of HCl= 6,14 mol

CaCo₃:

1,0 kg × (96/100) × (1000 g/1kg) × (1 mol/100,09g)

(Limestone) (CaCo₃ in limestone) (kg to g) (g to mol)

moles of CaCo₃= 9,59 mol

So, reactant limit is HCl

This reaction have a yield of 97%. So, the CO₂ moles are:

6,14 mol × 97÷ = 5,96 mol CO₂

The ideal gas formula to obtain pressure is:

P = nRT/V

Where: n = 5,96mol; R= 0,082 atm×L/mol×K; T = 273,15 (until STP conditions) and V= 10,0 L

Replacing this values in the equation the pressure is

P = 13,3 atm

I hope it helps!

7 0

3 years ago