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Advocard [28]
2 years ago
5

Using g= 10 m/s?, find the weight of a 3 kg mass.

Chemistry
1 answer:
Wewaii [24]2 years ago
3 0

Answer:

30 N

Explanation:

weight = mass * g

w = mg

w 3*10

w = 30

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Which could be the catalyst in a chemical reaction? A. reactants B. product C. enzyme D. activated complex
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C. Enzyme because they catalyze biochemical reactions
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3 years ago
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If a gas has a volume of 1000 ML at a temperature of 23°C and a pressure of 100 mmhg, what is it’s volume under standard conditi
Colt1911 [192]

Answer:

119.7 mL.

Explanation:

  • From the general law of ideal gases:

<em>PV = nRT.</em>

where, P is the pressure of the gas.

V is the volume of the container.

n is the no. of moles of the gas.

R is the general gas constant.

T is the temperature of the gas (K).

  • For the same no. of moles of the gas at two different (P, V, and T):

<em>P₁V₁/T₁ = P₂V₂/T₂.</em>

  • P₁ = 100.0 mmHg, V₁ = 1000.0 mL, T₁ = 23°C + 273 = 296 K.
  • P₂ = 1.0 atm = 760.0 mmHg (standard P), V₂ = ??? mL, T₂ = 0.0°C + 273 = 273.0 K (standard T).

<em>∴ V₂ = (P₁V₁T₂)/(T₁P₂) </em>= (100.0 mmHg)(1000.0 mL)(273.0 K)/(296 K)(760.0 mmHg) =  121.4 <em>mL.</em>

8 0
3 years ago
Compared to orange light, violet light
natulia [17]

Answer:

i think it was lll,lV ㄸㄸㅇㅊㅎㅍㅋ

5 0
3 years ago
Which two types of information are written in an element's box in the periodic table?
GenaCL600 [577]

Answer:

Yes it is B,D.

Explanation:

Each box represents an element and contains its atomic number, symbol, average atomic mass, and (sometimes) name. The elements are arranged in seven horizontal rows, called periods or series, and 18 vertical columns, called groups.

8 0
2 years ago
When 125 mL of 0.150 M Pb(NO3)2 is mixed with 145 mL of 0.200 M KBr, 4.92 g of PbBr2 is collected. Calculate the percent yield.
Semenov [28]

Answer:

Y = 92.5 %

Explanation:

Hello there!

In this case, since the reaction between lead (II) nitrate and potassium bromide is:

Pb(NO_3)_2+2KBr\rightarrow PbBr_2+2KNO_3

Exhibits a 1:2 mole ratio of the former to the later, we can calculate the moles of lead (II) bromide product to figure out the limiting reactant:

0.125L*0.150\frac{molPb(NO_3)_2}{L} *\frac{1molPbBr_2}{1molPb(NO_3)_2} =0.01875molPbBr_2\\\\0.145L*0.200\frac{molKBr}{L} *\frac{1molPbBr_2}{2molKBr} =0.0145molPbBr_2

Thus, the limiting reactant is the KBr as it yields the fewest moles of PbBr2 product. Afterwards, we calculate the mass of product by using its molar mass:

0.0145molPbBr_2*\frac{367.01gPbBr_2}{1molPbBr_2} =5.32gPbBr_2

And the resulting percent yield:

Y=\frac{4.92g}{5.32g} *100\%\\\\Y=92.5\%

Regards!

4 0
3 years ago
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