Answer:
0.0014 moles is present in 40cm³ of 0.035M of HCl solution
Explanation:
Molarity = 0.035M
V = 40.0mL
1mL = 1cm³
V = 40cm³
0.035 moles = 1000cm³
X moles is present in 40cm³
X = (40 * 0.035) / 1000
X = 0.0014moles
0.0014 moles is present in 40cm³ of solution
Answer:
C
Explanation:
<em>The </em><em>specific</em><em> </em><em>heat </em><em>capacity</em><em>=</em><em>quantity</em><em> of</em><em> </em><em>heat</em><em> </em><em>in </em><em>joule/</em><em>mass×</em><em>c</em><em>h</em><em>a</em><em>n</em><em>g</em><em>e</em><em> </em><em>in </em><em>temperature</em>
<em>from </em><em>this </em><em>question</em><em> </em><em>the </em><em>quantity</em><em> of</em><em> </em><em>heat</em><em> </em><em>is </em><em>7</em><em>5</em><em>2</em><em>5</em><em>0</em><em>,</em><em>the </em><em>mass </em><em>is </em><em>2</em><em>0</em><em>0</em><em>0</em><em> </em><em>and </em><em>the </em><em>change </em><em>in </em><em>temperature</em><em> </em><em>is </em><em>5</em><em>0</em><em>-</em><em>3</em><em>0</em>
<em>which </em><em>is </em><em>2</em><em>0</em>
<em>therefore</em>
<em>c=</em><em>7</em><em>5</em><em>2</em><em>5</em><em>0</em><em>/</em><em>2</em><em>0</em><em>0</em><em>0</em><em>×</em><em>2</em><em>0</em>
<em>c=</em><em>7</em><em>5</em><em>2</em><em>5</em><em>0</em><em>/</em><em>4</em><em>0</em><em>0</em><em>0</em><em>0</em>
<em>c=</em><em>1</em><em>.</em><em>8</em><em>8</em>
<em>I </em><em>hope </em><em>this </em><em>helps</em>
Answer: The new volume be if you put it in your freezer is 1.8 L
Explanation:
To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,

where,
are the initial volume and temperature of the gas.
are the final volume and temperature of the gas.
We are given:

Putting values in above equation, we get:

The new volume be if you put it in your freezer is 1.8 L
Answer: The correct answer is Option 1.
Explanation:
Control group is the group used in the experiment by the researchers in which no change in the variable is done. It is then set as a benchmark for the groups which are being tested.
We are given 4 groups which are being experimented for the flotation of an egg in water. As in Cup 1, there is no addition of salt and hence there is no change in the variables. So, this is set as a benchmark fro the cups which are further used in the experiment conducted.
Hence, the correct answer is Option 1.
Answer: The mowed lawn is the one from where the grasses are removed by using the machines or tools.
Explanation:
The mowed lawn is expected to have low number of species as the grasses may be few or scanty thus can support the population of few species like insects, mice, birds, and small number of grazing animals. On the other hand the weedy field can be hub of insects, reptiles like snakes, small mammals, and large mammals. Large weed field can provide food, and habitat to the large number of species. This will support the increase in biodiversity as compared to the mowed lawn.