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Maksim231197 [3]
2 years ago
9

ХА

Chemistry
1 answer:
kvasek [131]2 years ago
8 0

Answer:

B. Biosphere.

Explanation:

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The weight of the body decrease inside water why?​
Lady_Fox [76]

Penurunan atau kehilangan massa otot bisa menimbulkan penurunan berat badan yang tidak direncanakan

3 0
3 years ago
Which type of radiation particle, emitted from a nuclear reaction, is most similar to a helium nucleus?
olasank [31]
Alpha particles because two protons and two neutrons 
6 0
3 years ago
Read 2 more answers
A gas at constant temperature has a pressure of 404.6 kPa with a volume of 12 ml. If the volume changes to 43ml, what is the new
blagie [28]

Answer:

The answer is

<h2>112.912 kPa</h2>

Explanation:

The new pressure can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the new pressure

P_2 =  \frac{P_1V_1}{V_2}  \\

404.6 kPa = 404600 Pa

From the question we have

P_2 =  \frac{404600 \times 12}{43}  =  \frac{4855200}{43}  \\  = 112911.6279... \\  = 112912

We have the final answer as

<h3>112.912 kPa</h3>

Hope this helps you

4 0
3 years ago
1. How many moles of oxygen gas are needed to form 21.8 liters of water vapor?
BaLLatris [955]

0.781 moles

Explanation:

We begin by balancing the chemical equation;

O₂ (g) + 2H₂ (g) → 2H₂O (g)

21.8 Liters = 21.8 Kgs

To find how many moles are in 28.1 Kg H₂O;

Molar mass of H₂O = 18 g/mol

28.1/18

= 1.56 moles

The mole ratio between water vapor and oxygen is;

1 : 2

x : 1.56

2x = 1.56

x = 1.56 / 2

x =  0.781

0.781 moles

7 0
2 years ago
1.How many mL of 0.401 M HI are needed to dissolve 5.97 g of BaCO3?
garri49 [273]

Answer:

The answer to your question is:

1.- volume = 0.151 l or 151 ml

2.- 0.241 l  or 241 ml of NaOH

Explanation:

1.-

Data

V = ? HI = 0.401 M

BaCO3 = 5.97 g

                     2HI(aq)    +    BaCO3(s)   ⇒   BaI2(aq) + H2O(l) + CO2(g)

MW BaCO3 = 137 + 12 + 48 = 197 g

                     197 g of BaCO3 ----------------- 1 mol

                     5.97 g                -----------------   x

                     x = (5.97 x 1) /197

                    x = 0.03 mol of BaCO3

                    2 moles of HI ----------------  1 mol of BaCO3

                    x                     ----------------  0.03 mol of BaCO3

                    x = (0.03 x 2) / 1

                   x = 0.060 mol of HI

Molarity = moles / volume

volume = moles / molarity

volume = 0.060 / 0.401

volume = 0.151 l or 151 ml

2.-

V = ?    NaoH 0.757 M

Co⁺² Volume = 167 ml   0.548 M

             CoSO4(aq) + 2NaOH(aq)   ⇒   Co(OH)2(s) + Na2SO4(aq)

Moles of Co = Molarity x  volume

Moles of Co = 0.548 x 0.167

Moles of Co = 0.092

                                 1 mol of CoSO4 -------------- 2 moles of NaOH

                                0.092 moles      ---------------   x

                                x = (0.092 x 2) /1

                               x = 0.183 moles of NaOH

Volume of NaOH = moles / molarity

                             = 0.183 / 0.757

                            = 0.241 l  or 241 ml of NaOH

6 0
3 years ago
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