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belka [17]
2 years ago
14

Place the

Chemistry
1 answer:
9966 [12]2 years ago
3 0
Potassium, aluminum, silicon, magnesium neon
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When light hits the Earth at a greater angle, what does this cause?​
crimeas [40]
Angle of Solar Radiation and Temperature
5 0
2 years ago
What type of energy does a person experience when jumping on a trampoline?
Fofino [41]

Answer:

Knetic

Explanation:

When you jump on a trampoline, your body has kinetic energy that changes over time. As you jump up and down, your kinetic energy increases and decreases with your velocity. Your kinetic energy is greatest, just before you hit the trampoline on the way down and when you leave the trampoline surface on the way up.

7 0
2 years ago
An actacide tablet containing Mg(OH)2 (MM = 58.3g / (mol)) is titrated with a 0.100 M solution of HNO3. The end point is determi
PtichkaEL [24]

Answer:

0.0583g

Explanation:

The equation of the reaction is;

2HNO3(aq) + Mg(OH)2(aq) -------> Mg(NO3)2(aq) + 2H2O(l)

From the question, number of moles of HNO3 reacted= concentration × volume

Concentration of HNO3= 0.100 M

Volume of HNO3 = 20.00mL

Number of moles of HNO3= 0.100 × 20/1000

Number of moles of HNO3 = 2×10^-3 moles

From the reaction equation;

2 moles of HNO3 reacts with 1 mole of Mg(OH)2

2×10^-3 moles reacts with 2×10^-3 moles ×1/2 = 1 ×10^-3 moles of Mg(OH)2

But

n= m/M

Where;

n= number of moles of Mg(OH)2

m= mass of Mg(OH)2

M= molar mass of Mg(OH)2

m= n×M

m= 1×10^-3 moles × 58.3 gmol-1

m = 0.0583g

6 0
2 years ago
COCI2 has an effusion rate of 0.00172 m/sec. Which of the gases below would have an effusion rate of 0.00323 m/sec?
geniusboy [140]
CO por qué si y punto, chao
8 0
2 years ago
what mass of carbon dioxide gas would be produced if 10g of calcium carbonate reacted with an excess of hydrochloric acid?
LuckyWell [14K]
Answer is: 4,4 grams <span>of carbon dioxide gas would be produced.
</span>Chemical reaction: CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O.
m(CaCO₃) = 10 g.
n(CaCO₃) = 10 g ÷ 100 g/mol.
n(CaCO₃) = 0,1 mol.
From chemical reaction: n(CaCO₃) : n(CO₂) = 1 : 1.
n(CO₂) = 0,1 mol.
m(CO₂) = n(CO₂) · M(CO₂).
m(CO₂) = 0,1 mol· 44 g/mol.
m(CO₂) = 4,4 g.
7 0
3 years ago
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