What trend in atomic radius occurs down a group on the periodic table

OlgaM077 [116]

oops pls forgive me I accidentally did the wrong question.

4 0

3 years ago

Some people must eat a low-sodium diet with no more than 2,000 mg of sodium per day. By eating 1 cracker, 1 pretzel, and 1 cooki

Leni [432]

Answer:

The amount of sodium is 32 mg per cracker, 49 mg per pretzel and 68 mg per cookie.

Explanation:

Let's assume amount of sodium is x mg per cracker, y mg per pretzel and z mg per cookie.

So, the following three equations can be written as per given information:

x+y+z = 149 ........(1)

8y+8z = 936 ........(2)

6x+7y = 535 .........(3)

From equation- (2), y+z = $\frac{936}{8}$ = 117

By substituting the value of (y+z) in equation- (1) we get,

x = 149-(y+z) = 149-117 = 32

By substituting the value of x into equation- (3) we get,

y = $\frac{535-(6\times 32)}{7}$ = 49

By substituting the value of y into equation- (2) we get,

z = (117-49) = 68

So, the amount of sodium is 32 mg per cracker, 49 mg per pretzel and 68 mg per cookie.

6 0

3 years ago

Which of the following do water, rust, and salt have in common?

Cloud [144]

The answer is "elements" :)

6 0

3 years ago

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What is an alkaline earth metal that has 5 shells?

brilliants [131]

Answer:

The elements in the alkaline earth metals group; beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra), have two electrons in their outer electronic shell.

Explanation:

3 0

3 years ago

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The equilibrium constant, Kc, for the following reaction is 5.10×10-6 at 548 K. NH4Cl(s) NH3(g) + HCl(g) Calculate the equilibri

levacccp [35]

Answer: The equilibrium concentration of HCl is $2.26\times 10^{-3}M$

Explanation:

We are given:

Moles of $NH_4Cl(s)$ = 0.564 moles

Volume of vessel = 1.00 L

Molarity is calculated by using the equation:

$\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}$

Molarity of $NH_4Cl=\frac{0.564}{1}=0.564M$

The given chemical equation follows:

$NH_4Cl(s)\rightleftharpoons NH_3(g)+HCl(g)$

Initial: 0.564

At eqllm: 0.564-x x x

The expression of $K_c$ for above equation follows:

$K_c=[NH_3][HCl]$

The concentration of pure solid and pure liquid is taken as 1.

We are given:

$K_c=5.10\times 10^{-6}$

Putting values in above equation, we get:

$5.10\times 10^{-6}=x\times x\\\\x=2.26\times 10^{-3}M,-2.26\times 10^{-3}M$

Negative sign is neglected because concentration cannot be negative.

So, $[HCl]=2.26\times 10^{-3}M$

Hence, the equilibrium concentration of HCl is $2.26\times 10^{-3}M$

5 0

3 years ago