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3 years ago

What is the weight of a 30.0 kg object? 3.0 N 300 N 6.8 N 132 N

Chemistry

2 answers:

andreyandreev [35.5K]3 years ago

6 0

Explanation:

$w \: = mg \\ = 30\:kg \times 10\:ms^{-2}\\=300\:kg\:ms^{-2} \\ = 300\:N$

Ivan3 years ago

6 0

Answer:

300N

Explanation:

Weight and mass are related by the following equation

Weight = Mass x Acceleration due to gravity

From the question given,

Mass = 30kg

Acceleration due to gravity = 10m/s^2

Weight = 30 x 10

Weight = 300N

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How many grams of oxygen are produced when 1.05 moles of hydrogen gas is<br> produced?

vivado [14]

Answer:

1.058337 grams of hydrogen and 2H2 + O2 ==> 2H2O hydrogen peroxide

mols H2 = grams/molar mass

Using the coefficients in the balanced equation, convert mols H2 to mols O2.

Now convert mols O2 to grams. That's grams mols O2 x molar mass O2.

and it could also produce H2O water but no air but it could make other things

8 0

2 years ago

1. Write the balanced chemical equation for the synthesis of magnesium phosphide from its elements, including the word “energy”

Evgen [1.6K]

Answer:

3Mg(s) +2P(s) -------> Mg3P2(s) + energy

Keq= [Mg3P2]/[Mg]^3 [P]^2

Explanation:

The equation for the formation of magnesium phosphide from its elements is;

3Mg(s) +2P(s) -------> Mg3P2(s) + energy

Hence we can see that three moles of magnesium atoms combines with two moles of phosphorus atoms to yield one mole of magnesium phosphide. The equation written above is the balanced chemical reaction equation for the formation of the magnesium phosphide.

The equilibrium expression for the reaction K(eq) will be given by;

Keq= [Mg3P2]/[Mg]^3 [P]^2

8 0

3 years ago

Lawrencium-262 has a half-life of 4 hr. How much of a 40 mg sample remains after 12 hours?

CaHeK987 [17]

Answer:

5 mg

Explanation:

If one half life is 4 hours, then 3 half lives is 12 hours.

This means that the sample will decay to 1/8 of its original amount.

So, the answer is 40(1/8) = 5 mg.

3 0

1 year ago

Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho

Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

$2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\$

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

$n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2$

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

$m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3$

c) The leftover is computed as follows:

$m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\$

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0

2 years ago

What were Lamarck's ideas about evolution and why were those ideas incorrect

kow [346]

What were Lamarck's ideas about evolution and why were those ideas incorrect

4 0

2 years ago

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