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3 years ago

Explain why the gamma rays do not bend.

Chemistry

1 answer:

Natalka [10]3 years ago

6 0

They are electrically neutral, they are not deflected by magnetic fields

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Which of these elements has the smallest atomic radius?

shutvik [7]

Answer:

Its Br on Ap ex

Explanation:

Just got it wrong ;( it's B

4 0

3 years ago

Popeye wants to make a dilute spinach solution for Sweetpea's bottle. How much 3.0 M spinach solution should he add to the 500.0

Lynna [10]

Answer : The volume of 3.0 M spinach solution added should be, 50 mL

Explanation :

Formula used :

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the initial molarity and volume of spinach solution.

$M_2\text{ and }V_2$ are the final molarity and volume of diluted spinach solution.

We are given:

$M_1=3.0M\\V_1=?\\M_2=0.30M\\V_2=500.0mL$

Now put all the given values in above equation, we get:

$3.0M\times V_1=0.30M\times 500.0mL\\\\V_1=50mL$

Hence, the volume of 3.0 M spinach solution added should be, 50 mL

8 0

3 years ago

You place a large pear on a scale and it shows a mass of 200 grams. What is the mass of that pear in centigrams? Remember: King

statuscvo [17]

Answer:

20,000 centigrams

Explanation:

7 0

2 years ago

How many milliliters of a 0.285 M HCl solution are needed to neutralize 249 mL of a 0.0443 M Ba(OH)2 solution?

meriva

Answer:

$\large \boxed{\text{77.4 mL}}$

Explanation:

Ba(OH)₂ + 2HCl ⟶ BaCl₂ + H₂O

V/mL: 249

c/mol·L⁻¹: 0.0443 0.285

1. Calculate the moles of Ba(OH)₂

$\text{Moles of Ba(OH)$_{2}$} = \text{0.249 L Ba(OH)}_{2} \times \dfrac{\text{0.0443 mol Ba(OH)}_{2}}{\text{1 L Ba(OH)$_{2}$}} = \text{0.011 03 mol Ba(OH)}_{2}$

2. Calculate the moles of HCl

The molar ratio is 2 mol HCl:1 mol Ba(OH)₂

$\text{Moles of HCl} = \text{0.011 03 mol Ba(OH)}_{2} \times \dfrac{\text{2 mol HCl}}{\text{1 mol Ba(OH)}_{2}} = \text{0.022 06 mol HCl}$

3. Calculate the volume of HCl

$V_{\text{HCl}} = \text{0.022 06 mol HCl} \times \dfrac{\text{1 L HCl}}{\text{0.285 mol HCl}} = \text{0.0774 L HCl} = \textbf{77.4 mL HCl}\\\\\text{You must add $\large \boxed{\textbf{77.4 mL}}$ of HCl.}$