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madreJ [45]
3 years ago
13

How much work does the climber do on his pack if his pack weighs 90 N and he climbs to a height of 30 m?

Chemistry
1 answer:
Vladimir79 [104]3 years ago
8 0

Answer:

2,700j

Explanation:

yupppppppp

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A basketball weighs approximately 1.35 kg. What is this mass in grams (g)?<br><br> step by step
choli [55]

Answer:

1350 g

Explanation: just add a 0

8 0
3 years ago
A gas occupies a volume of 40.0 milliliters at 20 c if the volume is increased too 80.0 milliliters at constant pressure, the re
Troyanec [42]
Charles law gives the relationship between temperature and volume of gases. It  states that the volume of gas is directly proportional to temperature at constant pressure. 
V / T = k 
where V - volume and T - temperature in Kelvin and k - constant 
\frac{V1}{T1} =  \frac{V2}{T2}
where parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation 
T1 - 20 °C + 273 = 293 K
substituting these values in the equation 
\frac{40.0 mL}{293 K}  =  \frac{80.0 mL}{T}
T = 586 K
temperature in celsius = 586 K - 273 = 313 °C
new temperature is 313 °C
8 0
3 years ago
What is the acronym of DNA?? ​
g100num [7]
Abbreviation. DNA, which stands for deoxyribonucleic acid, is defined as a nucleic acid that contains the genetic code.
3 0
3 years ago
P 4 + 5O 2 P 4 O 10 , 1.5 moles of product was made in 30 seconds. What is the rate of reaction? 0.011 g/min 210 g/min 380 g/min
Airida [17]

Answer: 850.0 g/min.

Explanation:

  • The rate of the reaction = (ΔC/Δt) where,

ΔC is the change in concentration of reactants or products.

Δt is the change in time of the reaction proceeding.

  • The rate is needed to be calculated in (g/min).
  • We need to calculate the amount of the product in (g) via using the relation (n = mass / molar mass).
  • mass (g) = n x molar mass,
  • n = 1.5 moles and molar mass of P₄O₁₀ = 283.88 g/mol.
  • m = 1.5 x 283.88 = 425.82 g.
  • ΔC = 425.82 g and Δt = 30 s / 60 = 0.5 min.
  • The rate of the reaction = ΔC / Δt = (425.82 g / 0.5 min) = 851.64 g/min.

<em>can be approximated to 850.0 g/min.</em>

6 0
3 years ago
Read 2 more answers
What volume of 0.235 M H2SO4 is needed to titrate 40.0 mL of 0.0500 M Na2CO M NaCO3 solution?
Andreyy89

Answer: The volume of 0.235 M H_2SO_4 needed to titrate 40.0 mL of 0.0500 M Na_2CO_3 is 8.51 ml

Explanation:

To calculate the volume of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is Na_2CO_3

We are given:

n_1=2\\M_1=0.235M\\V_1=?mL\\n_2=2\\M_2=0.0500M\\V_2=40.0mL

Putting values in above equation, we get:

2\times 0.235\times V_1=2\times 0.0500\times 40.0\\\\V_1=8.51mL

Thus volume of 0.235 M H_2SO_4 needed to titrate 40.0 mL of 0.0500 M Na_2CO_3 is 8.51 ml

5 0
3 years ago
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