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3 years ago

1. If the frequency of a wave increases from 2 Hz to 4 Hz without the wavelength changing, how will the wave speed change?

Chemistry

1 answer:

maw [93]3 years ago

8 0

Answer:

Given expression: 6(2b-4). To find the value of 6(2b-4) at b= 5, we need to substitute the b=5 in the expression, we get….Therefore, the value of 6(2b-4) is 36, when b=5.

Explanation:

Given expression: 6(2b-4). To find the value of 6(2b-4) at b= 5, we need to substitute the b=5 in the expression, we get….Therefore, the value of 6(2b-4) is 36, when b=5.

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How much water must be added to make a 1.0 M solution from 250. mL of a 2.75 M solution?

Stels [109]

700 mL ....................

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3 years ago

In a rocket motor fueled with butane, how many kg of liquid oxygen should be provided with each kilogram of butane to provide fo

Travka [436]

Answer

So this is the reaction that happens.
C4H10 + O2 = CO2 + H2O 

Balanced, it is 
2C4H10 + 8O2 = 8CO2 + 10H2O 

Given 1 kg or 1000 g of butane, use stoichiometry aka factor labeling aka conversions and mole ratios to get to grams of oxygen. 

I'll do an example. Let's form water. Hydrogen is diatomic too. 
2H2 + O2 = 2H2O 

Given 1000 g of Hydrogen, I need to know how many grams of oxygen to use. To convert grams to moles,
I know that 1 mol of H2 equals 2.02 g. Then, for every mole of O2, there are 2 moles of H2. Then converting moles of O2 to grams, I know that one mole of it equals 32 grams. 
[1000 g H2] x [1 mol H2/2.02 g H2] x [1 mol O2/2 mol H2] x [32 g O2/1 mol O2] 

My answer would be 7.9 kg

3 0

3 years ago

What is the volume, in liters, of 0.500 mol of c3h3 gas at stp?

jarptica [38.1K]

At STP conditions the volume of 1 mol of any ideal gas will be 22.4L

0.500 mol C3H3 x 22.4L / 1 mol = 11.2 L

8 0

3 years ago

3. Burns from boiling water can be severe, caused by the transfer of energy from the boiling water to the

Free_Kalibri [48]

Answer:

Q = 3937.56 J

Explanation:

Heat transferred due to change in temperature is given by :

$Q=mc\Delta T$

c is the specific heat of water, c=4.18 J/g-°C

We have, m = 15 g, $T_i=100^{\circ} C\ \text{and}\ T_f=37.2^{\circ} C$

So,

$Q=15\times 4.18\times (37.2-100)\\Q=-3937.56\ J$

Hence, 3937.56 J of heat is transferred.

8 0

3 years ago

7 f Find the volume in dm3 and in mole of 0.505m of NaoH required to react with 40ml of 0.505m

Anna [14]

The volume of NaOH required is 0.08 dm³

To solve this question, we'll begin by writing the balanced equation for the reaction between H₂SO₄ and NaOH. This is illustrated below:

H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O

From the balanced equation above,

Mole ratio of the acid, H₂SO₄ $(n_{A}) = 1$

Mole ratio of the base, NaOH $(n_{B}) = 2$

Next, we shall determine the volume of NaOH required to react with H₂SO₄. This can be obtained as follow:

Molarity of the base, NaOH $(M_{B}) = 0.505 M$

Volume of the acid, H₂SO₄ $(V_{A}) = 40 mL$

Molarity of the acid, H₂SO₄ $(M_{A}) = 0.505 M$

<h3>Volume of the base, NaOH $(V_{B})$ =? </h3>

$\frac{M_{A} * V_{A}}{M_{B} * V_{B}} = \frac{n_{A}}{n_{B}}\\\\\frac{0.505 * 40}{0.505 *V_{B}} = \frac{1}{2}\\\\\frac{20.2}{0.505 *V_{B}} = \frac{1}{2}$

Cross multiply

$0.505 * V_{B} = 20.2 * 2\\0.505 * V_{B} = 40.4$

Divide both side by 0.505

$V_{B} = \frac{40.4}{0.505}\\\\V_{B} = 80 mL$

Finally, we shall convert 80 mL to dm³. This can be obtained as follow:

$1000 mL = 1 dm^{3}\\\\Therefore,\\\\80 mL = \frac{80 mL * 1dm^{3}}{1000 mL}\\\\80 mL = 0.08dm^{3}$

Therefore, the volume of NaOH required is 0.08 dm³

Learn more: brainly.com/question/19053582

3 0

2 years ago